Welcome to SRM Gateway Dear Student,
Welcome to SRM Gateway. We thank you very much for joining SRM Gateway. We are all committed to your satisfaction and fulfill your expectations in training, quality in our services, which is our primary goal. We are sure that you will become very skilled and will be positioning yourself well in the IT industry. We shall always ensure that you are guided well during the course. We wish you all the best for the complete learning of the technology Stay ahead in the technology. Way ahead of others.
Warm Regards, Management SRM Gateway
Document Number : SRMGW - IIT-JEE - MATH-01 Course Code : IIT-JEE
SRM Gateway This document is the property of SRM Gateway. This document has been prepared exclusively for the use of the students. No part of this document shall be copied or transferred in any form or by any means which would initiate legal proceedings, if found.
SRM Gateway No. 9, 3rd Avenue, Ashok Nagar, Chennai – 600 083
Preface About the book
This book will be an effective training supplement for students to master the subject. This material has in-depth coverage and review questions to give you a better understanding. A panel of experts have complied the material, which has been rigorously reviewed by industry experts and we hope that this material will be a value addition. We also would value your feedback, which will be useful for us to fine-tune it better.
Wishing you all success.
SRM Gateway
IIT - MATHS SET - 1
INDEX
1.
BASIC MATHEMATICS ...................................................................................
2
2.
ALGEBRA PROGRESSION .............................................................................
54
3.
BASIC TRIGONOMETRY ..............................................................................
132
4.
TRIGONOMETRY EQUATIONS ...................................................................
164
5.
INVERSE TRIGONOMETRY FUNCTION ...................................................
186
6.
PROPERTIES OF TRIANGLE .......................................................................
210
7.
CO - ORDINATE GEOMENTRY ...................................................................
232
IIT- MATHS
1
BASIC MATHEMATICS
2
BASIC TRIGONOMETRY
Number System (i) Natural numbers
: N = {1, 2, 3, 4, . . . . . }
(ii) Whole numbers
: W = {0, 1, 2, 3, 4, . . . . . }
(iii) Integers
: Z or I = {. . . . . –3, –2, –1, 0, 1, 2, 3, . . . . .}
Natural numbers are also called positive integers (denoted by Z+ or I+) Whole numbers are also called non–negative integers. The set of negative integers, Z– or I– = {. . .. . –3, –2, –1}. The set of non positive integers is {…, –3, –2, –1, 0}. Zero is neither positive nor negative but it is non–positive as well as non–negative. (iv)Rational numbers: Numbers of the form p/q where p, q Î Z and q 0 (because division by zero is not defined). ‘Q’ represents their set. All integers are rational numbers with q = 1 When q 1 and p, q have no common factor except 1, the rational numbers are called fractions. Rational numbers when represented in decimal form are either ‘terminating’ or ‘non– terminating’ but repeating. e.g., 5/4 = 1.25 (terminating) 5/3 = 1.6666 . . . . . (non terminating but repeating) p
(v) Irrational numbers: Numbers, which cannot be represented in q form. In decimal representation, they are neither terminating nor repeating all surds fall into this category e.g.,
2 , 151 / 3 , p, etc.
Note : p 22/7, 22/7 is only an approximate value of p in terms of rational numbers, taken for convenience Actually p = 3.14159 . . . . . (vi)Real numbers: All rational and irrational numbers taken together form the set of real numbers, represented by R. This is the largest set in the real world of numbers. Also note that Integers which give an integer on division by 2 are called even integers otherwise they are called odd integers. Zero is considered as even number. The set of natural numbers can be divided in two ways. (i) Odd and even natural numbers. (ii) Prime numbers (which are not divisible by any number except 1 and themselves) and composite numbers (which have some other factor apart from 1 and themselves).
3
IIT- MATHS 1 is neither prime nor composite 2 is the only even number which is prime
Set Theory Basic Concept Set: A set is a well–defined collection of objects or elements. Each element in a set is unique. Usually but not necessarily a set is denoted by a capital letter e.g. A, B, . . . . . U, V etc. and the elements are enclosed between brackets {}, denoted by small letters a, b, . . . . . x, y etc. For example: A = Set of all small English alphabets = {a, b, c, . . . . . x, y, z} B = Set of all positive integers less than or equal to 10 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} R = Set of real numbers = {x : – ¥ < x < ¥} The elements of a set can be discrete (e.g. set of all English alphabets) or continuous (e.g., set of real numbers). The set may contain finite or infinite number of elements. A set may contain no elements and such a set is called Void set or Null set or empty set and is denoted by (phi). The number of elements of a set A is denoted as n(A) and hence n () = 0 as it contains no element.
Union of sets Union of two or more sets is the set of all elements that belong to any of these sets. The symbol used for union of sets is ‘’ i.e., AB = Union of set A and set B = {x : x Î A or x Î B(or both)} e.g. If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C {1, 2, 6, 8}, then ABC = {1, 2, 3, 4, 5, 6, 8}
Intersection of sets It is the set of all the elements, which are common to all the sets. The symbol used for intersection of sets is ‘’ i.e., AB = {x : xA and xB} e.g. If A = {1, 2, 3, 4} & B = {2, 4, 5, 6}and C = {1, 2, 6, 8}, then ABC = {2}. Remember that n(AB) = n(A) + n(B) – n (AB)
Difference of two sets The difference of set A to B denoted as A – B is the set of those elements that are in the set A but not in the set B i.e., A – B = {x : x A and xB} Similarly B – A = {x : x B and x A}. In general A – B B – A e.g.
If A = {a, b, c, d} and B = {b, c, e, f}, then A – B = {a, d} and B – A = {e. f}
4
BASIC TRIGONOMETRY
LOGARITHM If a is a positive real number other then 1 and ab = c, then we write logac = b obviously c is positive. For example log3 81 = 4 34 = 81 Note The expression logb a is meaningful for a > 0 and for either 0 < b < 1. or b > 1 a = b log b a log c b
logab = log a c a1 a 2 0 if b 1 logba1 ³ logba2 0 a a if 0 b 1 1 2 Brain Teaser 1 : log x2 = 2logx, is it true or false? Formulae (i) loga|mn| = loga|m| + loga |n| (ii) loga m = loga|m| – loga|n| n
(iii) loga|mn| = n log a|m| (iv) logab = logcb × logac (v) log a k N =
1 logaN k
Modulus Function Let x Î R, then the magnitude of x is called it’s absolute value and in general, denoted by |x| and x , x 0 x 0
defined as |x| = x,
Note that x = 0 can be included either with positive values of x or with negative values of x. As we know all real numbers can be plotted on the real number line, |x| in fact represents the distance of number ‘x’ from the origin, measured along the number–line. Thus |x| ³ 0 secondly, any point ‘x’ lying on the real number line will have it’s coordinate as (x, 0). Thus it’s distance from the origin is x 2 Hence |x| =
x 2 . Thus we can define |x| as |x| =
x2
e.g if x = –2.5 then |x| = 2.5, if x = 3.8 then |x| = 3.8 Brain Teaser 2 : If x R– then
x 4x
2
is
1 1 1 , , or not defined. 2 2 2
Basic Properties :
5
||x|| = |x| |x| > a x > a or x < –a if a R+ and x R if a R– |x| < a – a < x < a if a R+ and no solution if a R––{0} |x + y| |x| + |y| |xy| = |x||y|
IIT- MATHS x |x| ,y0 y | y|
Intervals
Intervals are basically subsets of R and are of very much importance in calculus as you will get to know shortly. If there are two numbers a, b R such that a < b, we can define four types of intervals as follows: Open interval : (a, b) = {x : a < x < b} i.e., end points are not included. Closed interval: [a, b] = {x : a x b} i.e., end points are also included. This is possible only when both a and b are finite. Open–closed Interval: (a, b] = {x : a < x b} Closed–open interval:[a, b) = {x : a x < b} The infinite intervals are defined as follows (a, ) = {x : x > a} [a, ) = {x : x ³ a} (–, b] = {x : x b} intervals are particularly important in solving inequalities or in finding domains etc.
Inequalities The following are some very useful points to remember a b Þ either a < b or a = b. a < b and b < c a < c. a < b a + c < b + c cR a < b –a > –b i.e., inequality sign reverses if both sides are multiplied by a negative number. a < b and c < d a + c < b + d and a – d < b – c. a < b ma < mb if m > 0 and ma > mb if m < 0. 0 < a < b Þ ar < br if r > 0 and ar > br if r < 0.
1
1
a ³ 2 a > 0 and equality holds for a = 1. a a –2 a < 0 and equality holds for a = –1. a If a1 > b1, a2 > b2, a3 > b3 . . . . . , where ai > 0, bi > 0, i = 1, 2,… Then a1 + a2 + a3 + . . . > b1 + b2 + b3 + . . . and a1a2a3 . . . > b1b2b3 . . . If a > b, p and q are some positive integers, then following results are evident.
6
BASIC TRIGONOMETRY a > b an > bn Þ a–n < b–n where n Î N a > b al/q > bl/q Þ ap/q > bp/q
Wavy Curve Method In order to solve inequalities of the form Px Px 0, 0 , where P(x) and Q(x) are polynomials, we use the following method: Qx Qx
If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation, then within this interval the polynomial itself takes on values having the same sign. Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write
x 1 x 2 x 3 . . . . . x n Px f x x 1 x 2 x 3 . . . . . x m , Qx Where a1, a2, . . . . . an, b1, b2, . . . . . , bm are distinct real numbers. Then f(x) = 0 for x = a1, a2, . . . . . , an and f(x) is not defined for x = b1, b2, . . . . . , bm apart from these (m + n) real numbers f(x) is either positive or negative. Now arrange a1, a2, . . . . . , an, b1, b2, . . . . . , bm in an increasing order say c1, c2, c3, c4, c5, . . . . . , cm+n. Plot them on the real line. And draw a curve starting from right of cm+n along the real line which alternately changes its position at these points. This curve is known as the wavy curve.
The intervals in which the curve is above the real line will be the intervals for which f(x) is positive and intervals in which the curve is below the real line will be the intervals in which f(x) is negative.
Factor Theorem Let p(x) be a polynomial of degree greater than or equal to 1 and a be a real number such that p(a) = 0, then (x – a) is a factor of p(x). Conversely, if (x–a) is a factor of p(x), then p(a) = 0.
Remainder Theorem Let p(x) be any polynomial of degree greater than or equal to one and a be any real number. If p(x) is divided by (x–a), then the remainder is equal to p(a).
DETERMINANTS Consider the equations a1x + b1y = 0 and a2x + b2y = 0. These give a1 y a 2 a a 1 2 Þab –ab =0 1 2 2 1 b1 x b2 b1 b 2 a1
b1
We express this eliminant as a b = 0. 2 2
7
IIT- MATHS a1
b1
The expression a b is called a determinant of order two, and equals a1b2 – a2b1. 2 2 A determinant of order three consisting of 3 rows and 3 columns is written as a1
b1
c1
a2
b2
c2
a3
b3
b 2c 2
a 2c 2
a 2b2
and is equal to a1 b c – b1 a c + c1 a b 3 3 3 3 3 3 c3
= a1 (b2c3 – c2b3) – b1 (a2c3 – c2a3) + c1(a2b3 – b2a3) The numbers ai, bi, ci (i = 1, 2, 3) are called the elements of the determinant.
Function In the study of natural phenomena and the solution of technical and mathematical problems, it is necessary to consider the variation of one quantity as dependent on the variation of another. For example, in studies of motion, the path traversed is regarded as a variable, which varies with time. Here we say that the distance traversed is a function of time. The area of a circle, in terms of its radius R, is pR2. If R takes on various numerical values, the area assumes different numerical values. So the variation of one variable brings about a variation in the other. Hence area of the circle is a function of the radius R. If to each value of variable x (within a certain range) there corresponds a unique value of another variable y, then we say that y is a function of x, or, in functional notation y = f(x). The variable x is called the independent variable or argument. And the variable y is called the dependent variable. The relation between the variable x and y is called a functional relation. The letter f in the functional notation y = f(x) indicates that some kind of operation must be performed on the value of x in order to obtain the values of y. f(x)
f(x) y = f(x)
L y = f(x)
C y3
y2
B
y2 y1
y1
A
x2 x3
x1 Fig (a)
x
x0
x
Fig (b)
These figures show the graph of two arbitrary curves. In the figure any line drawn parallel to yaxis would meet the curve at only one point. That means each element of X would have one and only one image. Thus the figure (a) would represent the graph of a function. In the figure (b) certain line (e.g. line L) would meet the curve in more than one points (A, B and C). Thus element x0 of X would have three distinct images. Thus this curve will not represent a function. The set of all possible values which the independent variable (here ‘x’) is permitted to take for a given functional dependence to be defined is called the domain of definition or simply the domain of the 8
BASIC TRIGONOMETRY function. e.g. The function y = sin x is defined for all values of x. Therefore its domain of definition is the infinite interval – < x < . The function y =
1 x 1
is defined for all x > 1 its domain is (1, ).
Elementary Functions: (i) Constant function: y = c where c is a constant, defined for all real x. (ii) Power function: y x (a) a is positive integer. The function is defined in the infinite interval – < x < . (b) a is negative integer. The function is defined for all values of x except for x = 0. (iii) General exponential function: y = ax, where a is positive not equal to unity. This function is defined for all values of x. (iv)Logarithmic function: y = logax, a > 0 but a 1. This function is defined for all x > 0. (v) Trigonometric function: y = sinx, y = cosx defined for all real x y = tanx, y = secx, defined for R – (2n + 1)
. 2
y = cotx, y = cosecx, defined for R – n , where nl It must be noted that in all these function the variable x is expressed in radians. All these function have a very important property that is Periodicity. `
Is sec2 θ – tan2 θ = 1 valid for all θ R (real) ? (vi) Algebraic function: (a) Polynomial function: y = a0xn + a1xn–1 + … + an, where a0, a1 … an are real constants (a0 0) and n is a positive integer, called the polynomial of degree n. e.g. y = ax + b, a 0 (a linear function) y = ax2 + bx + c, a 0 (a quadratic function) A polynomial function is defined for all real values of x. a 0 x n a 1 x n 1 ... a n
(b) Rational Function y = b x m b x m 1 ... b 0 1 m e.g:y = a/x (inverse variation) The rational function is defined for all values of x except for those where the denominator becomes zero. (c) Irrational function e.g. y =
9
2x 2 x 1 5x 2
IIT- MATHS
Differential Calculus Let y = f(x) be a function. Putting the values of ‘x’ in this relation, we obtain the corresponding values of ‘y’. Suppose we start putting some values of ‘x’ in increasing order. The respective values of ‘y’ that we obtain may turn out to be in increasing order, or in decreasing order, or they may remain constant, or they may even have a mixed trend, depending upon the type of function. Let us take two values of x: x1 and x2(x1 < x2). So, y1 = f(x1) and y2 = f(x2) y y
2 1 Then, the quantity x x will tell us the average rate of change of y w.r.t. x in the interval [x1, x2] . 2 1
y 2 y1
Let y2 > y1 x x is positive Function is increasing on an average. 2 1 y 2 y1
if y2 < y1 x x is negative Function is decreasing on an average. 2 1 y 2 y1
If y2 = y1 x x is zero Function is constant on an average. 2 1 y y
2 1 As you can see, if x1 and x2 are sufficiently far apart, the quantity x x can not give the exact 2 1
idea of the variation of y w.r.t. x in the interval [x1, x2]. it just provides an overall information. For example if y2 = y1 it does not necessarily mean that y is same for all x in the interval [x1, x2]. Thus, to obtain a sufficiently accurate information, we have to choose x1 and x2 sufficiently close to each other. This sufficiently close is the key word here. To know the rate of change of y w.r.t. x at x = x1, we take x2 y 2 y1
very near to x1 (as much as possible), i.e., x 2 tends to x1 and then calculate x x . In the limiting case, 2 1 dy
we say that x2 nearly coincides with x1 and represent it as x2 ® x1. We use the notation dx
y 2 y1
x x1
for x x 2 1
as x2 ® x1. dx means small change in x (near x = x1) and dy means the corresponding change in y. We call dy the derivative or the differential coefficient of y w.r.t. x. dx
y y2 – y1 x2 – x 1
x1
x2
x
(You can understand it physically by taking x as time and y as displacement of a body, 10
BASIC TRIGONOMETRY Then
dy denotes the magnitude of velocity). dx
dy df ( x ) is also represented as f ¢ (x) or dx dx dy
Graphically, dx
x x1
(i.e.,
dy computed at x = x1) denotes the slope of the tangent to the curve y = dx
f(x) at x = x1 We will not here derive the formulae for ¢(x) of various functions, but we give the results of the derivations here,
Basic Differentiation Formulae y = constant
dy 0 dx
y = tan–1x
dy 1 dx 1 x 2
y = xn
dy nx n 1 dx
y = cot–1x
dy 1 dx 1 x 2
y = sinx
dy cos x dx
y = cosec–1x
dx | x | x 2 1
y = cosx
dy sin x dx
y = sec–1 x
dx | x | x 2 1
y = tanx
dy sec 2 x dx
y = ax
dy a x ln a dx
y = cotx
dy cos ec 2 x dx
y = ex
dy x e dx
y = sin–1x
dx 1 x2
y = logax
dx x ln a
y = cos–1x
dx 1 x2
y = ln x
dy
dy
1
dy
1
dy
1
dy
1
1
dy 1 dx x
Some Important Theorems The following are very important theorems, which can be applied directly. Theorem 1: If a function is of the form y = k f(x), where k is a constant, then
dy df ( x ) k dx dx
Theorem 2: The derivative of the sum or difference of a finite number of differentiable functions is equal to the sum or difference of the derivatives of these functions.
11
IIT- MATHS i.e., if y = u (x) + v (x) + w(x) then y = u(x) + v(x) + w(x). Theorem 3 The derivative of the product of two differentiable functions is equal to the product of the derivative of the first function with the second function plus the product of the first function with the derivative of the second function: i.e., if y = uv, then y = uv + uv. This formula can be extended for the derivatives of the product of any (finite) number of functions. Theorem 4 u(x)
dy
If y = v( x ) , then y¢ = dx Theorem 5
u v uv v2
If y = uv, where u and v are functions of x, then y = vuv-1 u + uv ln u.v Derivative of a Composite Function Given a composite function y = f(x), i.e., a function represented by y = F(u), u = f (x) or y = F[f(x)], then y =
dy dF du dx du dx
This is called the chain rule. The rule can be extended to any number of composite function; e.g. if y = f(u(v)), then y =
dy df du dv . dx du dv dx
Parametric Representation of a Function and it’s Derivatives We find the trajectory of a load dropped from an aeroplane moving horizontally with uniform velocity v0 at an altitude y0. We take the co-ordinate system as shown and assume that the load is dropped at the instant the aeroplane cuts the y-axis. Since the horizontal translation is uniform, the position of the load at any time t, is given x = v0t, y = y0 –
gt 2 2
Y v0 (x, y) X
Those two equations are called the parametric equations of the trajectory because the two variables x and y have been expressed in terms of the third variable t (parameter) i.e. two equations x = (t), y = (t)
12
BASIC TRIGONOMETRY where t assumes values that lie in a given interval (t1, t2) dy dy / dt
(d / dt ) ( t )
Then dx dx / dt (d / dt ) ( t )
Second Derivative of a Function The second derivative of y w.r.t. x is the function obtained by differentiating
It is represented as
So,
dy w.r.t. x. dx
d2 y dy 5 = 5x4 2 or y or f (x). e.g. If y = x then dx dx
d 2 y d dy d = (5x 4 ) = 5.4 x3 = 20x3 2 dx dx dx dx
The acceleration ‘a’ of a particle is the second derivative of the distance ‘s’ (given as a function of time). i.e. if s = f(t) then v =
ds d 2s dv = f (t) and a = 2 = = f (t) dt dt dt
THE BEHAVIOUR OF FUNCTIONS The following behaviours of a function are important to study
Increasing and Decreasing Functions (i) Increasing Functions If y = f(x) and x2 > x1 implies y2 > y1 for any x belonging to the interval [a, b], then y is said to be an increasing function of x. (x) increases in [a, b] f (x) > 0 x in (a, b). (ii) Decreasing Function If x2 > x1 y2 < y1 for any x belonging to [a, b], then y is said to be a decreasing function of x. f(x) decreases in [a, b] f (x) < 0 x in (a, b) For a constant function, f (x) = 0 For a non-decreasing function f (x) 0 For a non increasing function, f (x) 0
Maxima and Minima of Functions A function ‘f ’ is said to have a maxima at x = x0 if f(x) < f(x0), x in the immediate neighbourhood of x0. Similarly, a function ‘f’ is said to have a minima at x = x0 if f(x) > f(x0), x in the immediate neighbourhood of x0 We have used the word immediate here because a given function may have any number of high 13
IIT- MATHS and low points. It is just like moving on an uneven surface (which has many bumps and depressions). Mathematically, these bumps are called the points of local maxima and the depressions are called the points of local minima. The highest of all the bumps is the global maxima and the lowest of all the depressions is the global minima. We state here the preliminary methods only to find the maxima and minima of functions. dy or f ¢ (x) dx
(a)
Find
(b)
Find the points at which it becomes zero. These points are called critical points.
To find the points of maxima and minima we resort to either of the following tests. (a) First Derivative Test Suppose x = x0 is a critical point i.e., f (x0) = 0. If f (x) changes sign from positive to negative in the neighbourhood of x = x0
Maxima at x0
If f ¢ (x) changes sign from negative to positive in the neighbourhood of x = x0
Minima at ‘x0’
(b) Second Derivative Test (i)
d2y Find 2 or f (x) dx
(ii)
Compute the value of f (x) at the critical points
If it is positive Minima at those values of x. If it is negative maxima at those values of x. If the function is defined in an interval [a, b], then to find the maxima and minimum values i.e., global maxima and global minimum of the function in that interval we com pare the values of the function (i.e., y) at all the critical points and also the end points (i.e., y = f(a) and f(b)). Then the largest among them gives the global maximum values and smallest gives the global minimum values.
Integral Calculus: The Antiderivative of Function A function F(x) is called the antiderivative of the function f(x) on the interval [a, b] if, at all points of the interval f(x) = F(x). x3 ' x3 For example, the antiderivative of the function f(x) = x is , as 3 = x2. The function 3 3 x3 x3 x 2 and 1 are also antiderivatives of f(x) = x2. Infact, C , where C is an arbitrary constant, is 3 3 3 2
the antiderivative of x2. So if a function f(x) possesses an anti-derivative F(x), then it possesses infinitely many antiderivatives, all of them being contained in the expression F(x) + C, where C is a constant.
14
BASIC TRIGONOMETRY If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is called the indefinite integral of the function f(x) and is denoted by the symbol òf(x) dx. Thus, by definition òf(x) dx = F(x) + C, if F(x) = f(x). If a function f(x) is continuous on an interval [a, b], then this function has an antiderivative. The process of finding the antiderivative of a function f(x) is called integration. Two different integrals of a function differ by a constant.
Standard Elementary Integrals In the following integrals, C stands for an arbitrary constant. n x dx
x n 1 c, ( n 1) n 1
(f ( x )) n f ( x ) dx
(f ( x )) n 1 C (n 1 ) n 1
sec x dx = ln |sec x + tan x| + c dx
a2 x2 dx
1
1 x tan–1 c a a
=
x dx ln | x | c
1 x 2 tan
x x e dx e c
sin x dx = – cosx + c
cos x dx = sinx + c
x
dx a2 x2
1
x c
sin 1
x x c or –cos–1 + C a a
dx –1 1 x 2 = sin x + c
dx x 2 1
= sec–1x + c or – cosec–1 x + c
sec2 x dx = tanx + c cosec2 x dx = –cotx + c tan x dx = – ln|cosx| + c = ln |secx| + c cot x dx = ln |sin x| + c = – ln |cosecx| + c 1
e.g.
(i)
1 x2
1
x2 2 dx + c = . x3/2 + c 1 3 1 2
(ii)
1 x 21 1 2 dx x dx c c x2 2 1 x
The following points are to be noted:
1
x dx lnx + c if x is positive = ln (–x) + c if x is negative because 1) =
1 x
1 1 x2
d 1 (ln (–x)) = (– dx x
1
x dx ln | x | C
dx = sin–1x or –cos–1x. It does not mean that sin–1 x = –cos–1x.
The only legitimate conclusion is that they differ by some constant. In fact sin–1x – (–cos– 1 x) = sin–1x + cos–1x = /2.
15
If a is a constant, then a f(x) dx = a f ( x ) dx
IIT- MATHS
[f(x) g(x)] dx, = f(x) dx g(x) dx
Methods of Integration (i) Integration by Substitution This method consists of expressing the integral ò f(x) dx, where x is the independent variable, in terms of another integral where some other, say ‘t’, is the independent variable; x and t being connected by the relation x =f (t). i.e., f(x)dx = òf[j (t)] j¢ (t) dt. This method is useful only when a relation x = (t) can be so selected that the new integrand f(x)
dx is of a form whose integral is known dt
(ii) Integration by parts
f (x) (x) dx (i )
( ii )
df ( x ) dx dx dx
= f (x) ( x ) dx
Integral of the product of two function = first function × integral of second–integral of (derivative of first × integral of second).
Definite Integral The difference in the values of an integral of a function f(x) for two assigned values of the independent variable x, say a, b, is called the definite integral of f(x) over the interval (a, b) and is denoted by b
b
f ( x) dx. Thus
f ( x) dx F(b) F (a ),
a
a
where F(x) is the antiderivative of f(x). Or, we write
b b
f (x ) dx | F( x) dx |a F(b) F(a ). a is called the lower limit and b the upper limit of integration. a
Note: b
a
f (x ) dx f (x) dx a b
b c
b
f (x ) dx f ( x) dx f ( x) dx a
a
c
where c is any point inside or outside the interval (a, b).
Geometrically definite integral represents area under curve.
16
BASIC TRIGONOMETRY
SOLVED SUBJECTIVE EXAMPLES Example 1 : In a DABC, the medians AD, BE and CF pass through G. (a) If BG = 6, what is BE ? (b) If FG = 4, what is GC ? Solution: (a) We have, BG = 6=
2 BE 3
2 BE BE = 9 3
(b) We have, GC = 2 FG GC = 2 × 4 = 8 Example 2 : Triangles ABC and DBC are on the same base BC with A, D on opposite sides of lne BC, such that ar. (DABC) = ar. (DDBC). Show that BC bisects AD. Solution: Since D’s ABC and DBC are equal in area and have a common side BC. Therefore the altitudes corresponding to BC are equal i.e.,AE = DF Now, in D’s AEO and DFO, We have 1 = Ð2 (vertically opposite angles) AEO = ÐDFO (90° each) and AE = DF A
1 B
DAEO @ DDFO (by A.A.S) AO = OD 17
E
O
F
C
2
D
IIT- MATHS BC bisects AD Example 3 : Solve |x2 – 3x – 4| = x2–3x – 4 Solution: We know |x| = x when x ³ 0 So, x2 – 3x – 4 ³ 0 (x – 4) (x + 1) ³ 0 x ³ 4 or x –1. Example 4 : Solve 184x–3 = (54 2 )3x–4 Solution: Given equation is 184x–3 = (54 2 )3x–4 Taking log on both the sides, we get (4x – 3)log 18 = (3x – 4) log(18.3
2 ) (since 3 2 =
18 )
or, (4x – 3) log 18 = (3x – 4) log (18)3/2 or, 4x – 3 = (3x – 4)
3 2
or, 8x – 6 = 9x – 12, or x = 6 Example 5 : Solve for x if log3x + log9(x2) + log27 (x3) = 3 Solution: log3x + log9(x2) + log27(x3) = 3 log x
2 log x
3 log x
3 log x
log 3 2 log 3 3 log 3 3 log 3 3 logx = log3 x=3
Example 6 : If r be the ratio of the roots of the equation ax2 + bx + c = 0, show that
(r 1) 2 b 2 r ac
Solution: Let a and ra be the roots of the equation ax2 + bx + c = 0 So, a + ra =
b b b a(1 + r) = a = a (r 1) a a
Also, a × ra =
r.
… (1)
c c ra2 = a a
b2 c [Using (1)] 2 2 a ( r 1) a
18
BASIC TRIGONOMETRY
(r 1) 2 b 2 r ac
Example 7 : If x is real, prove that 3x2 – 5x + 4 is always positive Solution:
5
4
2 3x2 – 5x + 4 = 3 x 3 x 3
2 2 5 48 25 5 23 = 3 x 3 x 0 since square of real number is always non-negative. 6 36 6 36
Example 8 : Find the area of the largest circle that can be inscribed in a square of side 14 c.m. Solution: BC = 14 c.m. D 7
14 7 c.m. radius of circle = 2
7 A
Now the area of circle = r 2 = (7) 2 = 49 c.m2 Example 9 : x2 If f(x) = , x R, find the range of f(x) 1 x2
Solution: f(x) =
1 x2 x2 11 = = 1 2 2 1 x2 1 x 1 x
Clearly f(x) [0, 1) Example 10 : If x = 2 ln cot t and y = tan t + cot t, find Solution: Since
19
dy dy / dt dx dx / dt
C
dy dx
B
IIT- MATHS dx cos ec 2 t 2 4 2 Now dt cot t cos t sin t sin 2t
dy sin 2 t cos 2 t cos 2 t 2 2 sec t cos ec t – 4 Also 2 2 dt sin t cos t sin 2 2 t dy 4 cos 2t sin 2 t cos 2 t
Hence dx = cot 2t 2 sin 2t 4 sin 2 t
Example 11 : sin 3 x cos 3 x sin 2 x cos 2 x dx
Solution: sin 3 x cos 3 x sin 2 x cos 2 x dx =
sin 3 x cos 3 x dx sin 2 x cos 2 x sin 2 x cos 2 x dx
= tanx secx dx + cotx cosecx dx = secx – cosec x + c Example 12 : 1/ 2
Evaluate:
0
sin 1 x dx (1 x 2 ) 3 / 2
Solution: Let x = sinq dx = cos q dq and q = sin–1x; when x = 0, sinq = 0 When x = 1 / 2 , sinq = p/4 1 2
4
1
sin x
1 x
2 3/ 2
dx
0
0
4
cos d sec 2 d cos 3 0
/ 4 /4 = tan 0 1. tan
d (Integrating by parts) =
0
1 ln 2 4 2
Example 13 : Solve : log| x | | x | = 0 Solution: We have log| x | | x | = 0 |x| = 1
but | x | 1 (being in base of logarithm) x
Example 14 : 20
BASIC TRIGONOMETRY The maximum and minimum value of f (x) = 2x3 – 24x + 107 in the interval [1, 3]. Solution: We have f(x) = 2x3 – 24x + 107 So, f (x) = 6x2– 24 Now, f (x) = 0 Þ 6x2– 24 = 0 x = ± 2 But x = –2 Ï [1, 3] So x = 2 is the only stationary point. Now, (1) = 2 – 14 + 107 = 85, f(2) = 2(2)3 – 24 (2) + 107 = 75 And, (3) = 2 (3)3 – 24 × 3 + 107 = 89. Hence, the maximum value of f (x) is 89 which attains at x = 3 and the minimum value is 75 which is attained at x = 2. Example 15 : Find area bounded by y = cosx, xcos , y-axis and x = Solution: /2
The represented area =
cos x dx 0
/2
sin x 0 1 0 = 1 sp. unit.
21
. 2
IIT- MATHS
SECTION - I SUBJECTIVE LEVEL - I REVIEW YOUR CONCEPTS 1.
( x 3) 2 ( x 3) 5 ( x 1) Find the solution to the inequality >0 ( x 5) ( x 4)
2.
Solve for x (a) |x –4| > 7
3.
(b) |x| > x
(a)
Solve for x: log2x > 3
(b)
Which is greater: log23 or log1/25
4.
If the roots of (1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0 are equal, then find the value of m.
5.
For every x R, prove that 2x2 – 6x + 9 is always positive.
6.
Differentiate the following with respect to ‘x’. (i)
sinx + cosx
(ii)
xlogx
(iii)
Differentiate sin2x + cos2x with respect to x
7.
d2 y 2 If y = acos t + sin t , then prove that 2 y 0 dt
8.
(a)
Find domain of the definition of the following: (i)
(b)
(ii)
x
log e x
Draw the graph of the following: (i) f(x) = x3
(ii) f(x) = logex
9.
Find the intervals of increase and decrease of (x – 3) (x + 1).
10.
Evaluate : (i) (x 2 2x 1) dx .
2
(ii) (2xe x ) dx
/2
(ii)
(sin x cos x) dx 0
22
BASIC TRIGONOMETRY
LEVEL - II BRUSH YOUR CONCEPTS 1.
3 2 5 Find the solution common to both the inequalities ( x 1) ( x2 3x 2)7 | x 4 | 0 & 1 < |x – 3| < 5
2.
Solve for x
( x 4 x 4)
(i)
|x – 3| + |x + 2| = 3
(ii)
(a) Solve for x,
x2 5 2x 7
1 (log a log b 4 log 2) . 2
3.
If a2 + 4b2 = 12 ab, then prove that log (a + 2b) =
4.
Find Domain of definition
5.
Differentiate cos (4x3–3x) with respect to x.
6.
If for the function h, given by h(x) = kx2 + 7x – 4, h (5) = 97, find k.
7.
Find the intervals of increase and decrease of the function y = cosx, – x
(i) sin
x3 x 1
(ii)
Integrate the following sin x . cos 2 x dx . 0
9.
In the figure given below, find the value of angle P A 24° R
P
36° B
10.
54° C
D
In the figure given below, find the value of x T 5 P
23
4
B
x
A
( x 2) ( x 4)
2
2
8.
loge ( x 1) ( x 3)
IIT- MATHS
LEVEL - III CHECK YOUR SKILLS 1.
Solve: log0.3 (x – 1) < log0.09 (x – 1).
2.
Find the set of all solution of the equation 2|y| = 2 y1 1
3.
Evaluate: 7 log3 5 3log 5 7 5log 3 7 7 log 5 3
3
5
1 log 7 5
1 . log10 0.1
4.
Evaluate:
5.
Find domain of definition (i) f(x) =
log 2 ( x 3) x 2 3x 2
(ii) f(x) =
5x x 2 y log 10 (iii) 4
q
5 cos x 3sin x cos
(iv) y
| x | 1 2 | x | x2 1 x x2 1 x
dx . x
6.
Evaluate:
7.
Find the intervals of decrease and increase of (x + 2) e–x .
8.
In the figure given below, ABCD is square and triangles BCX and DYC are equilateral triangle. Find the value of y.
2
B A
y
x
C
D
9.
(i) In the figure given below. Find QSR Q
S
O
T
50°
P
R
24
BASIC TRIGONOMETRY (ii) In the given figure, O is the centre of the circle. If OCA = 26°, then find ODB C
B
O A
10.
25
D
ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108° and EAB = 124°. Then find ABC.
IIT- MATHS
SECTION - II OBJECTIVE LEVEL - I 1.
If log16x + log4x + log2x = 14, then x = (a) 16 (c) 64
2.
If | 4 – 3x | £
(b) 32 (d) none 1 then x is equal to 2
7 3
(a) , 6 2 7 3
(c) , 6 2 3.
(d) none of these
The product of all the roots of the equation x 2 - |x| – 6 = 0 is
(a) – 9 (c) 9 4.
7 3 6 2
(b) ,
(b ) 6 (d) 36
The value of p for which (x – 1) is a factor of x3 + (p + 1)2 x2 – 10 is given by (a) 4, 2 (c) –2, 4
(b) 2, –4 (d) none of these 1
5.
The domain of definition of the function f(x) = (a) R (c) (–, 0)
6.
log x x
(a) log x
(b)
(c) (xlogx)–1
(d) xlogx
The function f(x) = tanx – x, (a) always increases (c) never decreases
8.
(b) (0,) (d) none of these
The differential coefficient of f(x) = log (logx) with respect to x is x
7.
x x is
(b) always decreases (d) some times increases and some times decreases
The function f(x) = x3 – 3x is (a) increasing on (–, –1] [1, ) and decreasing on (–1, 1) 26
BASIC TRIGONOMETRY (b) decreasing on (–, –1] [1, ) and increasing on (–1, 1) (c) increasing on (0, ) and decreasing on (–, 0) (d) decreasing on (0, ) and increasing on (– , 0) 9.
The minimum value of 2( x 2 3)2 27 is (a) 227 (c) 1
10.
(b) 2 (d) none of these
The maximum value of x3 – 3x in the interval [0, 2] is (a) – 2 (c) 2
(b) 0 (d) 1 x/4
11.
Evaluate
0
12.
sin x dx cos 3 x 3 cos x
(a)
1 2
(b)
1 6
(c)
1 8
(d)
1 12
If log10 3 = 0.477, the number of digits in 340 is (a) 18 (c) 20
13.
(b) 19 (d) 21
The interior and its adjacent exterior angle of a triangle are in the ratio 1 : 2. What is the sum of the other two angles of the triangle? (a) 112° (c) 120°
14.
(b) 110° (d) 90°
In the figure PAQ is a tangent of the circle with centre O at a point A if ÐOBA = 32°. The value of x and y is P
C y
A
O x B
Q
27
IIT- MATHS (a) 30°, 50° (c) both 58° 15.
(b) both 40° (d) 30°, 60°
The difference between the interior and exterior angles of a regular polygon is 132°. The number of sides of the polygon is (a) 12 (c) 15
(b) 8 (d) 20
28
BASIC TRIGONOMETRY
LEVEL - II 1.
If A = log2log2log4256 + 2 log 2 2, then A equals to (a) 2 (c) 5
2.
(b) 3 (d) 7
If logkA . log5k = 3, then A = (a) 53 (c) 12
3.
(b) k3 (d) 243
If the product of the roots of the equation x 2 5 x 4 log 2 0 is 8, then l is (a) 2 2 (c) 3
(b) 2 2 (d) none of these
4.
The number of real roots of the equation (x –1)2 + (x – 2)2 + (x – 3)2 = 0 (a) 3 (b) 2 (c) 1 (d) 0
5.
If a and b are roots of the equation x2 + x + 1 = 0 then a2 + b2 = (a) 1 (b) 2 (c) –1 (d) 3
6.
x2 9 The domain of definition of the function f (x) = log x
(b) (1, ) (d) [1, )
(a) R (c) (0, 1) (1,
2
7.
dy If x = a cos , y = a sin , 1+ = dx (a) tan2 (b) sec2 (c) sec (d) |sec| 3
3
8.
The function y = x3 – 3x2 + 6x – 17 (a) increases everywhere (b) decreases everywhere (c) increases for positive x and decreases for negative x (d) increases for negative x and decreases for positive x
9.
The largest value of 2x3 – 3x2 – 12x + 5 for – 2 x 4 occurs at x = (a) – 2 (b) – 1 (c) 2 (d) 4
10.
The greatest value of f(x) = cos (xex + 7x2 – 3x), x [– 1, ) is (a) –1 (b) 1 (c) 0 (d) none of these
29
IIT- MATHS
11.
ex 1 e 2 x dx
(a) cot–1ex+c (c) tanex + c
(b) tan–1ex + c (d) sin–1ex + c
x/4
12.
Eva luate
tan 2 x dx
0
13.
(a) 3
2
(b) 2
(c) 1–
4
(d)
6
5 2
If the area of parallelogram ABCD is 32 sq. cm. Area of DAMN is equal to A
B
N
D
C
M
(a) 8 cm2 (c) 4 cm2
14.
(b) 2 cm2 (d) none of these
In the given figure if AX = 5 cm, XD = 7 cm, CX = 10 cm find BX B
A x C
D
O
P
T
(a) 3 cm (c) 4 cm
15.
(b) 3.5 cm (d) 4.5 cm
If A, B, C are three consecutive points on the arc of a semicircle such that the angles subtended by the chords AB and AC at the centre O is 90° and 100° respectively. Then the value of angle BAC is equal to (a) 5° (c) 20°
(b) 10° (d) 30°
30
BASIC TRIGONOMETRY
SUBJECTIVE SOLUTIONS LEVEL - I
(CBSE Level)
CHECK YOUR SKILLS 1.
( x 3)2 ( x 3)5 ( x 1) 0 Find the solution to the inequality ( x 5)( x 4)
Solution:
(x 3)2 (x 3)5 (x 1) 0 (x 5) (x 4)
By wavy curve method,
+
–
+
–3
–5
–
–
1
3
x ( , 5) ( 3 1) (4 )
2.
Solve for x (a) |x – 4| > 7 Solution: (a)
(b) |x| > x
|x – 4| > 7 x – 4 > 7 or x – 4 < – 7 x > 11 or x < – 3 –3
x (– , – 3) (11, ) (b) |x| > x Case I : x > 0, x > x which is not possible. Case II: x < 0 x < – x 2x < 0 x < 0 x ( – , 0) or R–
3.
(a) (b) Solution: (a)
Solve for x: log2x > 3 Which is greater: log23 or log1/25 log2x > 3
x > 23 [ base is greater than 1] 31
11
+ 4
IIT- MATHS (b)
x > 8 x (8, ) log23 or log1/2 5 log1/2 5 = – log25 < 0. [ log1/a x = – logax] log23 is greater than log1/2 5.
4. If the roots of (1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0 are equal then find the value of m. Solution: Given equation is (1 + m) x2 – 2(1 + 3m) x + (1 + 8m) = 0 roots are equal , then D = 0 4 (1 + 3m)2 – 4 (1 + m) (1 + 8m) = 0 1 + 9m2 + 6m – 1 – 9m – 8m2 = 0 m2 – 3m = 0 m (m – 3) = 0 m = 0, 3. 5. For every x R, prove that 2x2 – 6x + 9 is always positive. Solution: Let f(x) = 2x2 – 6x + 9 = 2(x2 – 3x + 9/2) = 2( x – 3/2)2 + 9/4 > 0 [ square of real number is always non–negative] Hence f(x) is always positive. 6.
Differentiate with respect to ‘x’ (i) sinx + cosx (ii) xlogx Solution: (i)
(ii)
Let y = sinx + cosx
dy cos x sin x dx
d (sinx + cosx) = cosx – sinx dx
Let y = x logx dy d d 1 x (log x) log x (x) = x log x1 dx dx dx x d (x log x) = 1 + logx dx
8.
(a)
Find domain (i)
(b)
x
Draw graph (i) f(x) = x3
(ii)
log e x
(ii) f(x) = logex 32
BASIC TRIGONOMETRY Solution: (a)
(i) Let f(x) =
x f(x) is real for all x 0 Df = [0, ] (ii) Let f(x) =
log e x
f(x) is real for all x 1 Df = [1, )
(b) Let f(x) = x3 It is odd function so graph will be symmetric about the origin (ii) Df R (iii) Rf R (iv)
y
x
f (x) = 3x2 f (x) = 0 x = 0
(v)
f (x) = 6x
(vi)
ve, x 0 f (x) 6x ve, x 0
(ii)
Let f(x) = logx
y
y = logx
(1, 0) x
1 1 (i) f (x) = , f (x) = – 2 x x
(ii) Df R {0} (iii) Rf R 9.
Differentiate with respect to x (i) sin2x + cos2x (ii)
Solution: (i)
x = a (cost + log tan
t ), y = a sint. 2
Let y = sin2x + cos2x
dy d d d d (sin 2 x) (cos 2 x) (cos x) (sinx) + dx dx dx dx dx
= 2sinx cosx + 2cosx (–sinx) = 2sinx cosx – 2sinx cosx = 0
33
d (sin 2 x cos 2 x) 0 dx
d d (1) 0 [ (constant) = 0] dx dx
(ii)
x = a (cost + log tan
t ), 2
y = asint
IIT- MATHS Diff. w.r.to t 1 t 1 sin t sec 2 dy dx t 2 2 , a a cos t tan dt dt 2 = a[– sint + cosect] dy cos t dx cos ect sin t
cos t.sin t cos t.sin t = = tant 2 1 sin t cos 2 t
10. Find the intervals of increase and decrease of (x – 3) (x + 1). Solution: Let f(x) = (x – 3) (x + 1) = x2 – 2x – 3 f (x) 2x 2 = 2(x – 1)
for increasing, f (x) 0
2(x – 1) 0 x 1 x [1, )
for decreasing, f (x) 0
x ( ,1]
11.
Find the point of local maxima and minima of the following (i) x2 + 3x (ii) logex + x Solution: Let f(x) = x2 + 3x = x(x + 3) f (x) = 2x + 3, for local maxima or local minima we must have, f (x) = 0
x
3 2
f (x) = 2 > 0
–3/2
f(x) is local minima at x
(ii)
f(x) = logx + x f (x)
3 2
1 1 x
for local maxima or local minima, we must have f (x) 0
1 1 0 x = –1 x 1 0 x2 f(x) is local maxima at x = –1. f (x)
34
BASIC TRIGONOMETRY 12. Integrate the following (i)
(ii) 2 xe x2
x2 + 2x + 1 x/2
(iii)
2
(sin x cos x ) dx
(iv)
x dx
1
0
Solution: (i)
Let I = (x 2 2x 1) dx x 2 dx 2 x dx 1.dx
(ii)
x3 x2 2 x C 3 2
=
x3 x 2 x C, where C is integral constant. 3
Let I =
2x e
x2
dx
Put x2 = t 2xdx = dt 2
x t I e dt = et + C = e C, where C is integral constant. /2
(iii)
I
(sin x cos x) dx 0
/2
=
/2
sin x dx
0
cos x dx 0
= cos x 0 / 2 [sin x]0 / 2
= – cos cos 0 sin sin 0 = –[0 –1] + [1 –0] = 2 2 2 2
(iv)
Let I | x |dx 1
x, x 0 Let f(x) = |x| = x, x 0 0
2
I f (x)dx f (x) dx 1
0
0
2
0
2
2 2 xdx xdx x x 2 1 0 1 2 0
1 1 [0 1] (4 0) 2 2
13. 35
5 2
In the figure given below. Find QSR
IIT- MATHS Q
S
O
50°
T
P
R
Solution: QSR + PRQ + QPR + QOR = 360° 90° + 90° + 50° + QOR = 360° QOR = 130° 1 QSR = QOR = 65° 2 14.
In the given figure, O is the centre of the circle. If OCA = 26°, then find ODB C
B
O A
D
Solution: OCA = 26° = DBA OB = OD OBD = ODB = 26°
C
B 26°
26°
O A
D
ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108° and EAB = 124°. Then find ABC. Solution: DCA = 108° E ACB = 72° 124° A 56° EAB = 124° 108° BAC = 56° D C B ABC = 180° – (72° + 56°) = 62° 15.
36
BASIC TRIGONOMETRY
LEVEL - I BRUSH UP YOUR CONCEPTS 1.
Find the solution common to both the inequalities ( x 1)3 ( x 2 3x 2) 5 | x 4 | 0 & 1 < |x – 3| < 5 ( x 2 4 x 4) 7
Solution: Case I : x < – 4
(x 1)3 (x 2)5 (x 1)5 (x 4) 0 (x 2)14
(x 1)3 (x 2)5 (x 1)5 (x 4) 0 (x 2)14
–
+
–2
–4
+
–
+
1
–1
x ( , 4) Case II : x > –4
–
+ –4
–2
+
–
+
1
–1
Similarly, x (4 2) ( 1,1) ... (i) x ( , 4) ( 4, 2) ( 1,1) And 1 < | x – 3| < 5 Case I : x < 3 1 < – x + 3 < 5 – 2 < –x < 2 – 2 < x < 2 x (–2, 2) Case II: x > 3 1
... (ii)
x ( 2, 2) (4,8) from (i) & (ii) –4
2.
–1
x ( 1,1)
Solve for x (i)
|x – 3| + |x + 2| = 3
Solution: (i) |x – 3| + |x + 2| = 3 Case I: 37
–2
(ii)
x2 5 2x 7
2
4
8
IIT- MATHS x<–2 –x + 3 – x – 2 = 3 –2x = 2 x = – 1, which is impossible Case II: –2 x < 3 –x + 3 + x + 2 = 3, which is not possible Case III: x3 x–3+x+2=3 2x = 4 x = 2, which is impossible x (ii)
x2 5 2x 7
x 2 5(2x 7) x 2 10x 35 0 0 2x 7 2x 7
9x 37 0 (9x + 37) (2x + 7) > 0 2x 7
+ –37/9
3.
–
x ( ,
+ –7/2 37 7 ) ( , ) 9 2
If a2 + 4b2 = 12 ab, then prove that log (a + 2b) =
1 (log a log b 4log 2) 2
Solution: a2 + 4b2 = 12ab a2 + 2a2b + 4b2 = 16 ab (a + 2b)2 = 16ab taking log on both the sides, 2 log (a + 2b) = log16 + log a + logb
4.
log (a + 2b) =
Find Domain
1 (loga + logb + 4log2) 2
(i) sin
x3 x 1
(ii)
( x 1)( x 3) loge ( x 2)( x 4)
Solution:
x 3 x 1 f(x) is defined for
(i) Let f(x) = sin
x 3 0 (x – 1) (x + 3) 0 x 1
38
BASIC TRIGONOMETRY x
+
– –3
1
x ( ,3] (1, )
(ii)
Let f(x) = log e
(x 1) (x 3) (x 2) (x 4)
f(x) is defined for (x 1) (x 3) 0 (x 2) (x 4) x ( , 3) ( 2,1) (4, )
5.
+
–
–3
+ –2
–
+ 4
1
Differentiate with respect to x (i)
(ii) (iii) Solution: (i)
x sin x + log (1 x 2 )
(1 x 2 )
(sinx)x cos (4x3–3x)
Let y
x sin x log 1 x 2 2 1 x
dy (1 x 2 ) (sin x x cos x) (2x)x sin x 1 (x 2x) 2 2 dx (1 x ) 1 x2 2 1 x2
(ii)
(1 x 2 ) x cos x (1 x 2 ) sin x 2x 2 sin x 1 2 (1 x )2 1 x2
(1 x 2 ) x cos x (1 x 2 ) sin x 1 2 2 (1 x ) 1 x2 Let y = (sinx)x. Taking log on both the sides, we get logy = x log sinx 1 dy x cos x . log sin x = x cotx + log sinx y dx sin x dy = (sinx)x [x cotx + log sinx] dx d {(sin x) x } (sin x) x[x cot x log sin x] dx Let y = cos(4x3 – 3x)
(iii)
dy = –sin (4x3 – 3x) [12x2 – 3] dx
d (cos 4x3 – 3x) = 3 (1 – 4x2) sin (4x3 – 3x) dx
6. If for the1 function h, given by h(x) = kx2 + 7x – 4, h (5) = 97, find k. Solution: 39
IIT- MATHS
7.
h(x) = kx2 + 7x – 4 h(x) 2kx + 7
h(5) = 10k + 7 97 = 10k + 7 k = 9 [ h (5) = 97]
(i)
Find the intervals of increase and decrease of the function y = cosx,
x 2
(ii)
Find the point of local maxima & minima of the function, f (x) = x
1 . x
Solution: (i)
Let f(x) = cosx, x 2 f (x) = –sinx f(x) increases, if f (x) 0 – sinx 0 sinx 0
x , 0 2
/2
f(x) decreases, if f (x) 0 – sinx 0 sinx 0 x [0, ] (ii)
1 f (x) x x
f (x) 1
f (x)
O
O
/2
1 x2
2 x3
For local maximum or local minimum we must have f (x) 0
1 0 x = 1 x2 at x = 1, f (x) 1 0
1–
Hence local minima at x = 1 at x = – 1, f (x) 1 0 Hence local maxima at x = – 1. 2
8.
Integrate the following (i)
sin x . cos 0
8
2
x dx
3
e x (ii) x 2 / 3 dx . 1
Solution: / 2
(i)
Let I =
sin x.cos
2
x dx
0
0
put cosx = t = t 2 dt –sinx dx = dt when x = 0, t = 1, when x = 1
, t 0 2
40
BASIC TRIGONOMETRY 1
1
t3 1 t dt = 3 3 0 0 2
8
(ii)
3
e x 1 1 Let I = 2 / 3 dx put x1/3 = t . 2 / 3 dx dt when x = 8, t = 2 , x 3 x 1 2 t t 2 x = 1, t = 1 = 3 e dt 3(e ) , = 3(e2 – e) = 3e (e – 1) 1
9.
In the figure given below, find the value of angle P A 24° R
P
36° B
54° C
D
Solution: ACD = ABC + BAC = 36 + 24° = 60° P = RCD + RDC = 60 + 52 = 104°.
10.
In the figure given below, find the value of x T 5 P
Solution: PT2 = PA.PB 5 5 25 4 4 AB = PA – PB
PA =
x=
25 9 4 units 4 4
Hence x =
41
9 units. 4
4
B
x
A
IIT- MATHS
LEVEL - III CHECK YOUR SKILLS 1. Solve, log0.3 (x – 1) < log0.09 (x – 1) Solution: log0.3 (x – 1) < log0.09 (x – 1) 1 log0.3 (x – 1) < log0.3 (x – 1) 2
1 log0.3 (x – 1) < 0 2 x–1>1 x>2 x (2, )
Find the set of all solution of the equation 2 y 2 y 1 1 Solution: 2|y| = 2y–1 + 1 Case I: y < 0 2–y = 2y–1 + 1 2.
2 2y 2 y (2y)2 + 2.2y – 2 = 0 2 a2 + 2a – 2 = 0 , where a = 2y 1=
a
2 4 8 2 2 3 2 2
a = 1 3 , – 1 +
a=–1–
a=
2y =
log22y = log2 ( 3 1 )
3
3 , which is not possible
3 1 3 1
y = log2 ( 3 1 ) Case II: y < 0 2y = 2y–1 + 1 2.2y = 2y + 2 2y = 2 y=1
set of solution 1, log2 ( 3 1)
3. Evaluate: 7 log 3 5 3log 5 7 5log 3 7 7 log 5 3 Solution: 5
7
7
Let y = 7log3 3log5 5log3 7 log5
3
= 7log3 5 3log5 7 7 log3 5 3log 5 7 [ alogb c clogb a ] 42
BASIC TRIGONOMETRY =0 Hence the result. 1
4.
Evaluate: 3 2 log 75
1 log 100.1
Solution: 3
Let y 5
1 1 log 7 5 log10 0.1
1/ 3
1 log 7 5 5 log10 10 Hence the result.
5.
= (7 + 1)1/3 = (8)1/3 = 2
Find domain (i) f(x) =
(ii) f(x)
x 1 2 x
Solution: log 2 (x 3) x 2 3x 2 Let g(x) = log2 (x + 3) and h(x) = x2 + 3x + 2 g(x) is defined for x+3>0 x>–3 Dg = (–3, ) And h(x) is defined for x2 + 3x + 2 0 (x + 2) (x + 1) 0 Dh = R – {–2, – 1} Df = Dg Dh = (–3, ) R – {–2, – 1} = (–3, ) {–2, –1}
(i) f(x) =
(ii)
| x | 1 f(x) is defined for 2 | x |
f(x) =
| x | 1 (| x | 1) (2 | x |) 0 0 2 | x | (2 | x |) 2
6.
(2 – |x|) (|x| – 1) 0 2 – |x| > 0 or |x| – 1 0 |x| < 2 or |x| 1 – 2 < x < 2 or x 1 or x – 1 Df = (–2, –1] [1, 2)
Find the domain 5x x 2 4
(i) y log10 (iii) y
x2 1 x x2 1 x
Solution: (i)
43
2 y log10 5x x f (x) 4
(ii) y cos ecx (sin x )1 / 3
IIT- MATHS f(x) is defined for 5x x 2 log10 0 4
(ii)
5x x 2 1 5x x 2 4 0 4 x2 – 5x + 4 0 (x – 1) (x – 4) 0 either x 1 or x 4 Df = [1, 4] Let f(x) =
cos ecx + (sinx)1/3
Let g(x) = cos ecx and h(x) = (sinx)1/3 g(x) is defined for cosecx > 0 Dg = (2n , (2n + 1) ). And h(x) is defined for R Df = Dg Dh = (2n , (2n + 1) ) (iii)
Let f (x)
x2 1 x x2 1 x
x2 and h(x) = x2 g(x) is defined for x 2 0 either x – 2 0 x2 or x + 2 < 0 x 2 or x + 2 < 0 Dg = (– , – 2) [2, ) h(x) is defined for Let g(x) =
1 x 1 x
1 x x 1 0 0 1 x x 1 either 1 – x 0 or 1 + x > 0
x 1 or x > – 1 Dn = (–1, 1] Df = Dg Dh = 7.
Integrate the following (a)
q 5 cos x – 3 sinx + cos 2 x
4
(b)
2 3 sin x dx cos 2 x 0
Solution: (a) Let I = (5 cos x 3sin x
a )dx cos 2 x
5 cos x dx 3 sin x dx 9 sec 2 x dx 44
BASIC TRIGONOMETRY = 5 sinx + 3cosx + 9 tanx + C, where C is integral constant. /4
(b)
Let I
0
2 3sin x dx cos 2 x
/4 2 = 2 sec xdx 3 tan x sec x dx 0
= 2[tan x]0 / 4 3[sec x]0 / 4 = 2[1 – 0] + 3 [ 2 1] = 3 2 1 8. Differentiate e Solution: Let y = e dy e dx
9.
(a) (b)
x 2 1
w.r.t. ‘x’.
x 2 1
x 2 1
1
×
2 x2 1
× (2x – 0)
x.e
x 2 1
x2 1
Find the intervals of decrease and increase of (x + 2) e–x . Find the greatest and least values of the following functions on the intervals y = – 3x4 + 6x2 – 1 (–2 x 2) y=
x3 2 x 2 3x 1 3
(–1 x 5)
Solution: (a) Let f(x) = (x + 2) e–x f (x) = –e–x (x + 2) + e–x = – e–x (x + 1) for decreasing f (x) 0 – e–x (x + 1) 0 e–x (x + 1) 0 ( e–x 0] x+1>0 x > – 1 x [ 1, ) for increasing , f (x) 0 –e–x (x + 1) 0 e–x (x + 1) 0 x+1 0 [ e–x 0] x 1 x (, 1] (b) (i) y = –3x4 + 6x2 – 1, – 2 x 2 dy 12 x 3 12x dx
d2 y 36x 2 12 12(x 2 1) 0 2 dx
45
dy 0 – 12x (x2 – 1) = 0 x = 0, 1 2 dx least value y = 2, at x = 0
IIT- MATHS greatest value y = –25 at x = 2 (ii)
y
x3 –– 2x2 + 3x + 1 , – 1 x 5 3
dy d2 y x 2 4x 3 2x 4 2(x 2) dx dx 2
dy 0 x2 – 4x + 3 = 0 dx (x – 1) (x – 3) = 0 x = 1, 3, 5
d2 y 20 dx 2 maximum at x = 1
maximum value of y =
at x = 1,
7 3
d2 y 20 at x = 3, dx 2 minimum at x = 5
10.
minimum value of y
23 3
In the figure given below, ABCD is square and triangles BCX and DYC are equilateral triangle. Find the value of y. B A
y
x
D
Solution:
C
BCX and DYC are equilateral AB = BC = CD = DA = CX = DY = CY = BX CBX = 60° ABX = 150° BAX = 15° A [ AD DY] DAY = 75° = DYA
46
BASIC TRIGONOMETRY Illustration 1: If A = {a, b, c} and B = {b, c, d} then evaluate AB, AB, A – B and B – A Solution: AB = {x : x A or x B} = {a, b, c, d} AB = {x : x A and x B} = {b, c} A – B = {x : x A and x B} = {a} B – A = {x : x Î B and x Ï A} = {d} Illustration 2: Find the logarithms of 0.0625 to the base ‘2’. Solution: Suppose 2x = 0.0625 = or
1 1 4 16 2
2x = 2–4 or x = –4
Illustration 3: Solve the equation a2x b3x = c5, where a, b, c R+ Solution: Equation is a2x.b3x = c5 Taking log on both sides, we have 2x log a + 3x logb = 5 logc or x (2loga + 3 logb) = 5logc
5 log c
x = 2 log a 3 log b
Illustration 4: Solve for x, log1/2 (x – 2) > 2. Solution: 1 4
1 4
9 4
log1/2(x – 2) > log1/2 0 < x – 2 < 2 < x < So
9 4
x 2,
Illustration 5: Solve |2x – 1| < 3. Solution: |2x – 1| < 3 – 3 < 2x – 1 < 3 – 2 < 2x < 4 – 1 < x < 2. Illustration 6:
47
IIT- MATHS
x 3x 2 x 5 x 1x 7 then find x such that
If f(x) =
(i) f(x) > 0. Solution: Given f(x) =
(ii) f(x) < 0.
x 3x 2 x 5 x 1x 7
Illustration 7: x
5
If x x = 24,, then find the value of x. Solution: x
5
Here, x x = 24 x2 – 5x = 24 x2 – 8x + 3x – 24 = 0 (x – 8) (x + 3) = 0 x = – 3, 8 Illustration 8: 2
3
x
y
7
Find the value of x and y when y x = 4 and 2 1 2 Solution: 2 3
Since, y x 4 2x 3y 4 x
y
7
… (1)
7
and 2 1 2 x 2 y 2
… (2)
Solving (1) and (2), we get x =
5 , y = –3. 2
Illustration 9: Find the derivative of y =
3x ( x )1 / 3
1 x
Solution: Here we have y = u + v + w, where u = 3x , v x 1 / 3 and w =
1 x
Hence we can use theorem 2 1
1
dy 1 1 1 1 1 3 x 2 x 3 1.x 11 dx 2 3 2
3 1 1 1 2 2/3 x 3 (x) x
Illustration 10: Find the derivative of y = (a + x) ex w.r.t. x . Solution: 48
BASIC TRIGONOMETRY Using theorem 3 dy d d (a x ) e x e x (a x ) dx dx dx
= (a x ) e x e x .1 e x (a x 1) Illustration 11: Find the derivative of y =
ax w.r.t. x ax
Solution: Illustration 12: Differentiate the following w.r.t x. (i) y = x x
(ii) y = (sin x ) x Solution: (i) y = xx
2
dy x.x x 1 .1 x x ln x.1 x x (1 ln x ) dx
(ii)
y (sin x ) x
2
2 2 dy d dx 2 x 2 (sin x ) x 1 (sin x ) (sin x ) x ln sin x dx dx dx
= x2 (sin x )
x 2 1
2
cos x (sin x ) x ln sin x. 2x.
Illustration 13: Find the derivative of the following functions w.r.t. x (i) y = sinx2 (ii) y = (lnx)3 (iii) y = sin (lnx)3 (iv) y = cos–1 (lnx) Solution: (i) y = sin (x)2. Let u = x2 y = sin u
dy d( u 3 ) du 1 3 3u 2 (ln x ) 2 = dx du dx x x
(ii)
y = (ln x)3. Let u = ln x y = u3
(iii)
y = sin (lnx)3 Let u = ln x, v = u3 y = sin v
(iv)
dy dy dv du d(sin v) d( u ) 3 d 1 3(ln x ) 2 (ln x ) = cos v. 3u 2 cos [(ln x ) 3 ] dx dv du dx dv du dx x x
Let u = ln x = cos–1 u
49
dy d d 2 (sin u ) ( x ) cos u.2 x 2x cosx2 dx du dx
dy 1 du 1 1 1 dx 1 u 2 dx 1 u 2 x x 1 (ln x ) 2
IIT- MATHS Illustration 14: The function y of x is given by, x = a cos t, y = a sin t. Find the derivative of y w.r.t. x. Solution: dy (a sin t ) cos t – cot t dx (a cos t ) sin t
If we want to compute
dy dy cot 1 at a particular t, say t = , then dx 4 x / 4 dx 4
Illustration 15: d2y Find , where y = sin2x dx 2
Solution: dy = 2sinxcosx = sin2x dx
Illustration 16: Find the interval of increase and decrease of the function y = x4. Solution: y = x4 y = 4x3 For x > 0, y > 0 the function increases in (0, ). For x < 0, y < 0 the function decreases in (–, 0) Illustration 17: Separate the intervals in which f(x) = 2x3 – 15x2 + 36x + 1 is increasing or decreasing Solution: We have f (x) = 6x2 – 30x + 36 = 6 (x – 2) (x – 3) Thus for x < 2, f (x) > 0 over 2 < x < 3, f (x) < 0 and for x > 3, f (x) > 0 Hence the given function is increasing in (–, 2) and (3, ), and decreasing in (2, 3) Is tanx always increasing x R ? Illustration 18: Test y = 1–x4 for maximum and minimum Solution: Here y = –4x3 = 0 for x = 0 x = 0 is the critical point. Now y = –12 x2 = 0 at x = 0 It is thus impossible to determine the character of the critical point by means of the sign of the second derivative. Thus we investigate the character of the given function in an interval containing point x = 0. For x < 0, y > 0 the function is increasing for x < 0 For x 0, y < 0 the function is decreasing for x > 0. Consequently, at x = 0, the function has a maximum i.e., yx = 0 = 1
50
BASIC TRIGONOMETRY Illustration 19: Determine the maximum and minimum of the function y = x3 – 3x + 3 on the interval [–3,
3 ]. 2
Solution: For the given function, y = 3x2 – 3 For the critical points, 3x2 – 3 = 0 x = 1 Then y = 6x > 0 at x = 1, y’’ < 0 at x = – 1 Hence there is maximum at x = –1 at which y = –1 + 3 + 3 = 5 Also there is minimum at x = 1 at which y = 1–3 + 3 = 1 Now at x = –3, y = –27 + 9 + 3 = –15 and at x = 3/2, y = 15/8 Hence the minimum value of the given function is –15 at x = –3 and the maximum value is 5 at x = –1. It should be noted that the values are actually the largest and smallest values of the function in the given interval. Illustration 20: Evaluate:
2
(i)
(a0 + a1x + a2x2 ) dx
(ii)
cos x x e
(iii)
x2 1 x 2 dx
(iv)
x4 x 2 1 dx
Solution: (i)
x
dx
(a0 + a1x + a2x2) dx = a0 dx + a1 x dx + a2 x2 dx x2 x3 a c = a0 x + a1 2 2 3
2
x
1 x dx cos x dx 2 dx e dx = sin x + 2 log |x| –ex + c x
(ii)
cos x x e
(iii)
x2 x2 11 1 dx dx 1 x 2 1 x 2 dx 1 1 x 2 dx dx 1 x 2 = x – tan–1 x + c
(iv)
x4 x4 11 1 2 dx x 2 1 1 x 2 dx x 1 1 x 2 dx 2 = x dx dx
dx x3 x tan 1 x c 1 x2 3
Illustration 21: Integrate the following w.r.t. x. (i) sin2x cosx (ii) Solution: (i) 51
x3 1 x8
Let sin x = t cos x dx = dt
IIT- MATHS 2 2 sin x cos x dx t dt
(ii)
t3 sin 3 x c c 3 3
Let x4 = t = 4 x3 dx = dt x3 dx = ¼ dt x3 1 1 dt 1 1 dx dt = tan 1 t c tan 1 x 4 c 8 2 2 1 x 4(1 t ) 4 1 t 4 4
Illustration 22: 1
Evaluate
2
a x2
dx
Solution: Let x = a sin dx = a cos d a2 – x2 = a2 cos2
Note:
dx 2
a x
2
a cos d x 1. d sin 1 c a cos a
1 bx c sin 1 a a 2 ( bx c) 2 b dx
Illustration 23: Evaluate (i)
xe
x
(ii)
dx
sin
1
x dx
Solution: (i)
let f(x) = x, (x) = ex xe x dx x.e x 1.e x dx xe x e x c
(ii)
Let f(x) = sin–1x. (x) = 1 1 1 1 sin x dx sin x. 1 dx sin x.x
1 1 x2
x dx
= x sin–1 x – x (1 x 2 ) 1/ 2 dx x sin 1 x 1 x 2 c Note: e x (f ( x ) f ( x )) dx e x f ( x ) dx e x f ( x ) dx = f(x) ex – f ( x ) e x dx e x f ( x ) dx f ( x ) e x c Illustration 24: Evaluate : / 2
cos x
1 sin 2 x dx 0
Solution: Let sin x = t cos x dx = dt When x = 0, t = 0, x = /2, t = 1 /2
0
1
cos xdx dt (tan 1 t )10 = tan–1 1– tan–1 0 = – 0 = /4 2 2 1 sin x 0 1 t 4
52
BASIC TRIGONOMETRY
53
IIT- MATHS
2
ARITHMETIC PROGESSION
54
ARTHMETIC PROGESSION
SQUENCE AND SERIES A succession of numbers a1, a2, a3..., an, ... formed, according to some definite rule, is called a sequence.
ARITHMETIC PROGESSION (A.P.) A sequence of numbers {an} is called an arithmetic progression, if there is a number d, such that d = an-an–1 for all n. d is called the common difference (C.D.) of the A.P.
(i) Useful Formulae If a = first term, d = common difference and n is the number of terms, then (a) nth term is denoted by tn and is given by tn = a + (n – 1) d. (b) Sum of first n terms is denoted by Sn and is given by n Sn [2a (n 1)d] 2 n or S n (a l ) , where l = last term in the series i.e., l = tn = a + (n – 1) d. 2 (c) Arithmetic mean A of any two numbers a and b a b A . 2 (d) Sum of first n natural numebrs ( n ) n(n 1) n 2 , where n N . (e) Sum of squares of first n natural numbers ( n 2 )
n
2
n( n 1) (2n 1) 6
(f) Sum of cubes of first n natural numbers (n 3 ) n(n 1) n 3 2
2
(g) Middle term: If the number of terms is n, and th
n 1 n is odd, then term is the middle terms 2 th
th
n n n is even, then and 1 terms are middle terms. 2 2
(h) If terms are given in A.P., and their sum is known, then the terms must be picked up in following way For three terms in A.P., we choose them as (a – d), a, (a + d) For four terms in A.P. , we choose them as (a – 3d), (a – d), (a + d), (a + 3d) 55
IIT- MATHS For five terms in A.P., we choose them as (a – 2d), (a – d), a, (a + d), (a + 2d) etc.
(ii) Useful Properties (a) If tn = an + b, then the series so formed is an A.P. (b) If Sn = an2 + bn + c, then series so formed is an A.P. `
(c) If every term of an A.P. is increased or decreased by some quantity, the resulting terms will also be in A.P. (d) If every term of an A.P. is multiplied or divided by some non-zero quantity, the result ing terms will also be in A.P. (e) In an A.P. the sum of terms equidistant from the beginning and end is constant and equal to sum of first and last terms. (f) Sum and difference of corresponding terms of two A.P.’s will form an A.P. (g) If terms a1, a2, ..., an, an+1, ..., a2n+1 are in A.P., then sum of these terms will be equal to (2n + 1)an+1. (h) If terms a1, a2, ..., a2n–1, a2n are in A.P. The sum of these terms will be equal to a n a n 1 (2n) . 2
GEOMETRIC PROGRESSION (G.P.) A sequence of the numbers {an}, in which a1 0 , is called a geometric progression, if there is a an number r 0 such that a r for all n.r is called the common ratio (C.R.) of the G.P.. n 1
(i) Useful Formulae If a = first term, r = common ratio and n is the number of terms, then (a) nth term, denoted by tn , is given by
tn = arn–1
(b) Sum of first n terms denoted by Sn is given by a rl a(1 r n ) a(r n 1) Sn or or , 1 r 1 r r 1 where l is the last term (the nth term) in the series, r 1 In case r = 1, Sn = na. (c) Sum of infinite terms (S ) a S (for | r | 1) 1 r Note: When |r| 1, the series is divergent and so its sum is not possible. (d) Geometric mean (G.M.) G ab is the geometric mean of two positive numbers a and b.
(e) If terms are given in G.P. and their product is known, then the terms must be picked up in the following way. 56
ARTHMETIC PROGESSION a For three terms in G.P., we choose them as , a, ar r a a 3 For four terms in G.P., we choose them as 3 , , ar, ar r r a a 2 For five terms in G.P., we choose them as 2 , , a, ar, ar etc. r r (ii) Useful Properties (a) The product of the terms equidistant from the beginning and end is constant, and it is equal to the product of the first and the last term. (b) If every term of a G.P. is multiplied or divided by the some non-zero quantity, the resulting progression is a G.P. (c) If a1, a2, a3 ... and b1, b2, b3, ... be two G.P.’s of common ratio r1 and r2 respectively, then a1b1 , a2b2 ... and
a1 a 2 a 3 r , , ... will also form a G.P. Common ratio will be r1r2 and 1 b1 b 2 b3 r2
respectively. (d) If a1, a2, a3, ... be a G.P. of positive terms, then loga1, loga2, loga3, ... will be an A.P. and conversely.
HARMONIC PROGRESSION (H.P.) (a) A sequence is said to be in harmonic progression, if and only if the reciprocal of its terms form an arithmetic progression. For example 1 1 1 , , ... form an H.P., because 2, 4, 6, ... are in A.P.. 2 4 6
(b) If a, b are first two terms of an H.P. then tn
1 1 1 1 (n 1) a b a
(c) Harmonic mean H of any two numbers a and b H
2 1 1 a b
2ab , where a, b are two non-zero numbers. ab
(d) If terms are given in H.P. then the terms could be picked up in the following way (i) For three terms in H.P, we choose them as 1 1 1 , , a d a a d
(ii) For four terms in H.P, we choose them as
57
IIT- MATHS 1 1 1 1 , , , a 3d a d a d a 3d
(iii) For five terms in H.P, we choose them as 1 1 1 1 1 , , , , a 2d a d a a d a 2d
Insertion of Means Between two numbers Let a and b be two given numbers. (i) Arithmetic Means If a, A1, A2 , ... An, b are in A.P., then A1, A2, ... An are called n A.M.’s between a and b. If d is the common difference, then b = a + (n + 2 – 1) d d = Ai = a + id = a + i
ba n 1
b a a(n 1 i) ib , i = 1, 2, 3, ..., n n 1 n 1
Note: The sum of n-A. M s, i.e., A1 + A2 + ... + An =
n (a b) 2
(ii) Geometric means If a, G1, G2 ... Gn, b are in G.P., then G1, G2 ... Gn are called n G.M.s between a and b. If r is the common ratio, then 1
b = a.r
n+1
(n 1) r = b a
i
b n 1 Gi = ar = a a a i
n 1 i n 1
.b
i n 1
, i = 1, 2, ..., n
Note: The product of n-G. M s i.e., G1 G2 ... Gn =
ab
n
(iii)Harmonic Means If a, H1, H2... Hn, b are in H.P., then H1, H2 ... Hn are called n H.M.s between a and b. If d is the common difference of the corresponding A.P., then
1 1 a b (n 2 1) d d b a ab(n 1) 1 1 1 ab ab(n 1) id i , i 1, 2, 3, ..., n Hi a a ab(n 1) b(n i 1) ia
58
ARTHMETIC PROGESSION
MEANS OF NUMBERS Let a1, a2, ... an be n given numbers (i) Arithmetic Means (A.M.) =
a1 a 2 ... a n n
(ii) Geometric Means (G.M.) = (a1 a2 ... an)1/n n (iii) Harmonic Mean (H.M.) = 1/ a 1/ a ... 1/ a 1 2 n
If weights of a1, a2, ... , an are w1, w2, ..., wn respectively, then their weighted arithmetic mean, weighted geometric mean and weighted harmonic mean are respectively defined by 1
a1w1 a 2 w 2 ... a n w n (a1w1 .a 2 w 2 ...a n w n ) w1 w 2 ...w n w1 w 2 ... w n and
w1 w 2 ... w n . w1 w 2 wn ... a1 a 2 an
RELATION BETWEEN A, G AND H If A, G and H are A.M., G.M. and H.M. of positive numbers a1, a2 ... an ( a1 a 2 ... a n ) then a1 H G A a n ...(1) Note: (i) The equality at any place in (1) holds if and only if the numbers a1, a2, ..., an are all equal (ii) (1) is true for weighted means also (iii) G2 = AH, if n = 2.
ARITHMETIC MEAN OF mth POWER Let a 1 , a 2 a1m a m2 ... a mn n
... , a n be n positive real numbers
and let m be a real number, then
m
a a ... a n 1 2 , if m R [0,1]. n
a1m a m2 ... a mn a1 a 2 ... a n However if m (0, 1), then n n
m
a m a m2 ... a mn a1 a 2 ... a n Obviously if m {0,1}, then 1 n n
m
ARITHMETIC-GEOMETRIC SERIES A series whose each term is formed, by multiplying corresponding terms of an A.P. and a G.P., is called an Arithmetic-geometric series. 59
IIT- MATHS e.g.
1 + 2x + 4x2 + 6x3 + ..... ; a + (a + d) r + (a + 2d)r2 + ..... (i) Summation of n terms of an Arithmetic-Geometric Series Let Sn = a + (a + d) r + (a + 2d)r2 + ... + [a + (n – 1)d] rn–1, d 0 , r 1 Multiply by ‘r’ and rewrite the series in the following way rSn = ar + (a + d)r2 + (a + 2d)r3 + ... + [a + (n – 2)d]rn–1 + [a + (n – 1)d ]rn on subtraction, Sn (1 – r) = a + d(r + r2 + ... + rn–1) – [a + (n– 1)d]rn or, Sn (1 r) a
or, Sn
dr(1 r n 1 ) [a (n 1)d].r n 1 r
a dr(1 r n 1 ) [a (n 1)d] n .r 1 r (1 r)2 1 r
(ii) Summation of Infinite Series If |r| < 1, then (n –1)rn, rn–1 0, as n . Thus S = S =
a dr 1 r (1 r)2
SUM OF MISCLENIOUS SERIES (i) Defference Method Let T1, T2, T3 ... Tn be the trms of a sequence and let (T2 – T1) = T1 , (T3 – T2) = T2 ... , (Tn – Tn–1) = T n 1 . Case I: If T1, T2 ,....Tn1 are in A.P. then Tn is a quadratic in ‘n’. If T1 – T2 , T2 T3 , ... are in A.P., then Tn is a cubic in n. Case II: If T1, T2 ,....Tn1 are not in A.P., but if T1 , T2 ,..., Tn1 are in G.P., then Tn = arn + b, where r is the C.R. of the G.P. T1 , T2 , T3 .....and a, b R. Again if T1,T2,...Tn1 are not in G.P. but T2 T1, T3 T2 ,...Tn1 Tn2 are in G.P., then Tn is of the form arn + bn + c; r is the C.R. of the G.P. T2 T1, T3 T2 T4 T3 ... and a, b, c R. (ii) Vn – Vn–1 Method Let T1, T2, T3 , ... be the terms of a sequence. If there exists a sequence V1, V2, V3 ... satisfying Tk = Vk – Vk–1, k 1, n
n
k 1
k 1
then S T (V V ) V V . k k k 1 n 0 n
60
ARTHMETIC PROGESSION
OBJECTIVE SECTION - I LEVEL - I Multiple Choice Questions with Single Answer: 1.
2.
If a1, a2, a3 ... are in AP, then ap, aq, ar are in AP, if p, q, r are in (a) AP
(b) GP
(c) HP
(d) none of these
Let tr denote the rth term of an AP. If tm =
(a)
1 mn
(c) 1
3.
1 1 and t n = , then tmn equal to m n
(b)
1 1 –1 m n
(d) 0
If p, q, r, s N and they are four consecutive terms of an AP, then the pth, qth, rth, sth terms of a GP are in
4.
5.
6.
61
(a) AP
(b) GP
(c) HP
(d) none of these
Let a1, a2, a3, ... be in AP and ap, aq, ar be in GP. Then aq : ap is equal to (a)
rp qp
(b)
qp rq
(c)
rq qp
(d) none of these
If a, b, c are in G.P., then a + b, 2b, b + c are in (a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
If a, b, c, d are nonzero real numbers such that (a2 + b2 + c2) (b2 + c2 + d2) (ab + bc + cd)2, then a, b, c, d are in (a) AP
(b) GP
(c) HP
(d) none of these
IIT- MATHS 7.
8
9.
10.
11.
12.
13.
If 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca), where a, b, c are nonzero numbers, then a, b, c are in (a) AP
(b) GP
(c) HP
(d) none of these
If a, b, c are in H.P., then c, c – a, c – b are in : (a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
Let S be the sum, P be the product and R be the sum of the reciprocals of n terms of a GP. Then P2Rn: Sn is equal to (a) 1 : 1
(b) (common ratio)n : 1
(c) (first term)2 : (common ratio)n
(d) none of these
If a1, a2, a3 are in AP, a2, a3, a4 are in GP and a3, a4, a5 are in HP, then a1, a3, a5 are in (a) AP
(b) GP
(c) HP
(d) none of these
1 1 1 If x > 1, y > 1, z > 1 are three numbers in GP then 1+ ln x , 1+ ln y , 1+ ln z are in
(a) AP
(b) HP
(c) GP
(d) none of these
If a, a1, a2, a3, ... a2n–1, b are in AP, a, b1 , b2 , b3 ..., b2n–1, b are in GP and a, c1, c2, c3, ... , c2n–1, b are in HP, where a, b are positive, then the equation anx2 – bnx + cn = 0 has its roots (a) real and unequal
(b) real and equal
(c) imaginary
(d) none of these
The product of n positive numbers is unity. Then their sum is (a) a positive integer (c) equal to n +
14.
1 n
(b) divisible by n (d) never less than n
If x > 0 and a is known positive number, then the least values of ax + (a) a2
(b) a
(c) 2a
(d) none of these
a is x
62
ARTHMETIC PROGESSION 15.
16.
17.
If p, q, r be three positive real numbers, then the value of (p + q) (q + r) (r + p) is (a) > 8 pqr
(b) < 8 pqr
(c) 8 pqr
(d) none of these
a1 a2 an If a1, a2, a3 ….. an are in H.P., then a 2 a 3 ....an , a 2 a 3 ....an ,.... a 2 a 3 ....an 1 are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these
1 1 1 1 Sn = 1 + 2 3 4 ..... 2 n 1 , then
(a) S100 < 100
(b) S200 < 200
(c) S200 > 100
(d) S50 > 25
n
18.
Given Sn r0
19.
1 1 1 ,s . , then least value of ‘n’ is r r If S – Sn 2 1000 r 0 2
(a) 8
(b) 9
(c) 10
(d) 11
If x15 – x13 + x11 – x9 + x7 – x5 + x3 – x = 7,(x > 0), then (a) x16 is equal to 15
(b) x16 is less than 15
(c) x16 greater than 15
(d) none of these
20.
1
(n 1) (n 2) (n 3) ......(n k)
is equal to
n 1
1 (a) (k 1) k 1
1 (b) k k
1 (c) (k 1) k
1 (d) k
n
21.
63
n 1 1 Maximum value of n for which n is 2 14 1
(a) 4
(b) 5
(c) 6
(d) 7
IIT- MATHS 22.
1 7 1 20 ..... is The nth term of the series 2 1 1 2 13 9 23 20 (5n 3)
(b)
20 (5n 3)
(b) 20 (5n + 3)
(d)
20 (5n 2 3)
(a)
23.
24.
If
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of these
If 2p + 3q + 4r = 15, the maximum value of p3q5 r7 will be
(a) 2180
54.35 (b) 15 2
55.7 7 217.9
(d) 2285
(c)
25.
1 1 1 1 + + + = 0 and a c b then a, b, c are in a a-b c c-b
The minimum value of x4
1 is x2
1/ 3
1 (a) 3 4
(b)
1 2
1/ 3
1 (c) 2 3
(d) 2
64
ARTHMETIC PROGESSION
LEVEL - II Multiple Choice Questions with one or more than one correct Answers: 1.
Total number of positive real values of x such that x, [x], {x} are in H.P, where [.] denotes the greatest integer function and {.} denotes fraction part, is equal to (a) zero (c) 2
2.
(b) 1 (d) none of these
In the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ..... , where n consecutive terms have the value n, the 1025th term is (a) 210 (c) 29
3.
(b) 211 (d) 28
Sr denotes the sum of the first r terms of an AP. Then S3n : (S2n – Sn) is (a) n (c) 3
4.
(b) 3n (d) independent of n
a, b, c are distinct real numbers such that a, b, c are in A.P. and a2, b2, c2 are in H.P. then (a) 2b2 = – ac (c) 2b2 = ac
(b) 4b2 = – ac (d) 4b2 = ac n-1
5.
If x1, x2, ... xn are in H.P. then
x
r
x r+1 is equal to
r=1
(a) (n – 1)x1xn (c) (n + 1) x1 xn 6.
(b) nx1 xn (d) none of these
If ax = by = cz and x, y, z are in GP then logc b is equal to (a) logba (c) z/y
(b) logab (d) none of these n
7.
The value of
r=1
8.
65
(a)
n a a nx
(c)
n( a nx a) x
1 a + rx + a + (r -1)x
is
(b)
a nx a x
(d) none of these
5c Sides a, b, c of a triangle are in G.P. If ln , ln a must be
3b a and ln are in A.P., then triangle 5c 3b
IIT- MATHS (a) Isosceles (c) Obtuse angled 9.
(b) Equilateral (d) None of these
G13 + G 32 If A1 be the A.M. and G1, G2 be two G.Ms between two positive numbers a and b, then G1G 2 A1 is equal to (a) 1 (c) 3
10.
(b) 2 (d) none of these
{bi}, i = 1, 2, ..., n is an arithmetic sequence. If b1 + b5 + b10 + b15 + b20 + b24 = 255, then 24
b
i
is equal to
i=1
(a) 600 (c) 300 11.
(b) 900 (d) none of these
If 2. nP1, nP2, nP3 are three consecutive terms of an AP then they are (a) in GP (c) equal
12.
(b) in HP (d) none of these
In a GP the product of the first four terms is 4 and the second term is the reciprocal of the fourth term. The sum of the GP up to infinite terms is (a) 8 (c) 8/3
13.
(b) -8 (d) –8/3
If a1, a2, ... an are positive real numbers whose product is a fixed number c, then the minimum value of a1 + a2 + ... an–1 + 2an is (a) n(2c)1/n (c) 2nc1/n
14.
If a, b, c, d are four positive numbers then a b c d (a) 4. b c d e (c)
15.
(b) (n + 1)c1/n (d) (n + 1) (2c)1/n
a b c d e 5 b c d e a
a e
a c b d (b) 4. b dc e (d)
a e
b c d e a 1 a b c d e 5
If a2 + b2 + c2 = 1 then ab + bc + ca lies in the interval. 1 (a) , 2 2
(b) [–1, 2]
1 (c) ,1 2
1 (d) 1, 2
66
ARTHMETIC PROGESSION 16.
Let f(x) =
2 3 n +1 1- x n +1 and g(x) = 1 – + 2 – ... + (–1)n . Then the constant term in x x xn 1- x
f (x) × g(x) is equal to
17.
18.
n(n 2 1) (a) , when n is even 6
(b)
n (c) – (n 1) , when n is even 2
(d)
(a)
1 an , 1 a 1 a
(b)
1 an , 1 a 1 a
(c)
1 an , 1 a 1 a
(d)
1 an , 1 a 1 a
Let S1, S2, S3, ... be squares such that for n 1, the length of a side of Sn equals the length of a diagonal of Sn+1. If the length of a side of S1 is 10 cm then for which of the following values of n is the area of Sn less than 1 cm2? (b) 8 (d) 10
If a, b, c, d are distinct integer in A.P. such that d = a4 + b4 + c4, then a + b + c + d is (a) 0 (c) 2
20.
(b) 1 (d) none of these
Three positive numbers form a GP. If the middle number is increased by 8, the three numbers form an AP. If the last number is also increased by 64 along with the previous increase in the middle number, the resulting numbers form a GP again. Then (a) common ratio = 3 (c) common ratio = –5
21.
67
n(n 1) , when n is odd 2
a n 1 A B , then S = , where A & B are respectively If tn = n 1 n n (1 a x) (1 a x) 1 x 1 an
(a) 7 (c) 9
19.
n(n 1) , when n is odd 2
(b) first number = 4/9 (d) first number = 4
If a, b, c are in GP and a, p, q are in AP such that 2a, b + p, c + q are in GP then the common difference of the AP is (a)
2a
(b) ( 2 + 1) (a – b)
(c)
2 (a + b)
(d) ( 2 - 1) (b - a)
IIT- MATHS
22.
23.
1 1 1 Value of 1 + 1 2 1 2 3 .... 1 2 3 ....n is equal to
(a)
2n n 1
(b)
(c)
4n 3n 1
(d) none of these
If x, y, z are positive numbers in AP then (a) y2 xz (c)
1 r 1 n r 1
(a) 2n2 (c) n2
25.
(b) y 2 xz
xy yz xy yz 4 has the minimum value 2(d) 2y x 2y z 2y x 2y z
n
24.
3n 2n 1
r 1
is equal to (b) 3n2 (d) none of these
Between two unequal numbers, if a1, a2 are two AMs; g1, g2 are two GMs and h1, h2 are two HMs then g1.g2 is equal to (a) a1h1 (c) a2h2
(b) a1h2 (d) a2h1
68
ARTHMETIC PROGESSION
SECTION - II COMPREHENSIVE PASSAGE I
An A.P. is a sequence whose terms increase or decrease by a fixed number, called the common difference of the A.P.. If ‘a is the first term and ‘d’ the common difference, the A.P. can be written as a, a+d, a + 2d,…… The nth term an is given by an = a + (n – 1)d. The sum sn of the first n terms of such an A.P. is given by : sn =
n n (2a + (n – 1)d) = (a + l) where l is the last term (i.e., the nth 2 2
a c is the A.M. of a and c. The n numbers A1, 2 A2,….., An are said to be A.M.’s between the numbers a & b if term of the A.P. ). If a, b, c are in A.P., then b
a, A1, A2, …..,An, b are in A.P. If ‘d’ is the common difference of this A.P., then d Ar a r
1.
2.
3.
(b a) th n 1 , where Ar is the r mean
9 If 6 A.M.’s are inserted between 1 and , then the 4th arithmetic mean is equal to 2 (a)
3 2
(b)3
(c)
2 3
(d)
The pth term of an A.P. is a and qth term is b, then sum of its (p + q) terns is (a)
pq 2
pq (b) a b pq
(c)
pq a b ab 2 p q
(d)none of these
If log 2, log (2x – 1) and log (2x + 3) are in A.P., then the value of x is (a)5/2 (c)log35
4.
(b)log25 (d)log53
2 2 If am be the mth term of an A.P., then a12 a 22 a 32 a 42 ...... a 2n 1 a 2n is equal to
(a)
69
13 5
2n 1 2 a1 a 2n2 n
(b)
n a12 a 2n2 2n 1
ba n 1
IIT- MATHS (c) II
n a12 a 22n 2n 1
(d)
2n 1 2 2 a1 a 2n n
If x1, x2,.. . . . . xn are ‘n’ positive real numbers; then A.M. ³ G.M. ³ H.M.
x1 x 2 .............xn n (x1 x 2....xn ) 1/n 1 1 1 n ......... equality occurs when numbers are same using x1 x 2 xn this concept.
5.
If a > 0, b > 0, c > 0 and the minimum value of a(b2 + c2) + b(c2 + a2)+ c(a2 + b2) is labc, then l is (a) 1 (c) 3
6.
(b) 2 (d) 6
If a, b, c, d, e, f are positive real numbers such that a + b + c + d + e + f = 3, then x = (a + f)(b + e)(c + d) satisfies the relation (a) 0 < x 1
(b) 1 x 2 (d) 3 x 4
(c) 2 x 3 7.
8.
If a and b are two positive real numbers, and a + b = 1, then the greatest value of a3b4 is (a)
32 43 75
(b)
(c)
77 33 44
(d) none of these
If x2 + y2 = 9, then the value of x + y lies in the interval (a) [0,3] (c) [-3 2 , 3 2 ]
III.
33 44 77
(b) [-3,3] (d) [0,3 2 ]
If nth term Tn of a given sequence is of the form Vn – Vn – 1, where V1, V2, V3, . . . is some other n
sequence, then Sn T k Vn V 0 . Similarly if the nth term Tn of a given sequence is of the form K 1
Vn , then Pn = Vn 1
9.
n
Vn
T V n
k 1
.
0
n 4 Lt tan -1 2 = n®¥ 4n + 3 n =1
(a) 1 (c) 3
(b) 2 (d) 4
70
ARTHMETIC PROGESSION 10.
If nth term of a sequence is (a) more than (c) equal to
11.
12.
IV.
1 n 1 n , then sum of first n terms of the sequence is (b) less than
n 1
n 1
(d) can not be determined
n 1
If cos 1 + cos 2 + cos 3 + . . . + cos n = Sn, then lim n
Sn n
(a) equals 1
(b) equals sin 1
(c) does not exist
(d) none of these
1 1 1 Sum of the infinitely many terms of the series 2 3 4 ..... is equal to
(a)
2 3
(b) 1
(c)
3 2
(d) does not exist as the sum tends to infinity
The sum of n terms of a series each term of which is composed of the reciprocal of the product of r factors in arithmetical progression, can be done by following rule: Write down the nth term, strike off a factor from the beginning, divide by the number of factors so diminished and by the common difference, change the sign and add a constant. The sum of n term of a series each term of which is composed of r factors in arithmetical progression. Write down the nth term, affix the next factor at the end, divide by the number of factors thus increased and by the common difference and add a constant.
13.
14.
1 1 1 The sum of n terms of the series 1.2.3.4 2.3.4.5 3.4.5.6 .... is
1 1 (a) 18 3(n 1) (n 2) (n 3)
1 1 (b) 18 3(n 1) (n 2) (n 3)
1 1 (c) 18 3(n 1) (n 2) (n 3)
1 1 (d) 18 3(n 1) (n 2) (n 3)
The sum of n terms of the series
1 1 ..... is 1.3.5.7.9 3.5.7.9.11
1 1 (a) 840 8(2n 1) (2n 3) (2n 5)
71
1 1 (b) 840 8(2n 1) (2n 3) (2n 5) (2n 7)
IIT- MATHS
1 1 1 1 (c) 840 8(2n 1) (2n 3) (2n 5) (2n 7) (d) 840 8(2n 1) (2n 3) (2n 5) (2n 7)
15.
If
1 b c 3 4 5 ... = a n 3 (n 2) (n 3) (n 1) (n 2) (n 3) , then 1.2.4 2.3.5 3.4.6
(a) a
29 3 4 ,b ,c 36 2 3
(b) a
29 3 4 ,b ,c 36 2 3
(c) a
29 3 4 ,b ,c 36 2 3
(d) a
29 3 4 ,b ,c 36 2 3
72
ARTHMETIC PROGESSION
SUBJECTIVE SECTION - III LEVEL - I 1.
The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5. Find the number of sides of the polygon.
2.
The ratio between the sum of n terms of two A.P.’s is 7n + 1 : 4n + 27. Find the ratio between their nth terms.
3.
The r th , s th and t th terms of a certain G..P. are R, S and T respectively. Prove that Rs–t . St–r. Tr–s = 1.
4.
(i) (ii)
If one G.M. G and two arithmetic means p and q be inserted between any given numbers, then show that G2 = (2p – q) (2q – p). If H be the H.M. between a & b, then show that (H – 2a) (H – 2b) = H2
5.
If pth, qth, rth terms of an A.P. be a, b, c respectively, then prove that p(b – c) + q (c – a) + r (a – b) = 0.
6.
(i) (ii)
The sum of three numbers in G.P. is 42. If the first two numbers are increased by 2 and third is decreased by 4, the resulting numbers form an A.P. Find the numbers of G.P. The sum of an infinite geometric series is 162 and the sum of its first n terms is 160. If the reciprocal of its common ratio is an integer, find all possible values of the common ratio, n and the first term of the series.
7.
Evaluate: 1 + 2.2 + 3.22 + 4.23 + ... + 100.2 99 .
8.
Find the sum of n terms of the series, the rth term of which is (2r + 1) 2r.
9.
Let x = 1 + 3a + 6a2 + 10a3 + ..., |a| < 1; y = 1 + 4b + 10b2 + 20b3 + ..., |b| < 1. Find S = 1 + 3 (ab) + 5 (ab)2 ... in terms of x and y.
10.
If a, b, c are the sides of a triangle , then prove that a2 + b2 + c2 > ab + bc + ca.
73
IIT- MATHS
LEVEL - II BRUSH UP YOUR CONCEPT 1.
Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn , then show that the ratio S3n/Sn is equal to 6.
2.
Show that the number 1111.....1 is a composite number.. 91 digits
3.
(i)
If a1, a2, a3, ... are in A.P. with common difference d, then prove that tan 1
4.
d d d nd tan 1 ... tan 1 tan 1 1 a 1a 2 1 a 2a 3 1 a n a n 1 1 a 1a n 1
(ii)
Sum to n terms the series 12 – 22 + 32 – 42 + 52 – 62 + ...
(i)
If the A.M. of a and b is t wice as great as their G.M., then show t hat
(ii)
a : b = (2 3):(2 3) . If the (m + 1)th, (n + 1)th and (r + 1)th terms of an A.P. are in G.P., m, n, r are in H.P. show 2 that the ratio of the common difference to the first term in the A.P. is . n
13 13 23 13 23 33 ...16 terms. 1 1 3 1 3 5
5.
Evaluate:
6.
Find the sum Sn of the cubes of the first n terms of an A.P. and show that the sum of first n terms of the A.P. is a factor of Sn.
7.
If the roots of 10x3 – cx2 – 54x – 27 = 0 are in harmonic progression, then find c and all the roots.
8.
An A.P. and a G.P. with positive terms have the same number of terms and their first terms as well as last terms are equal. Show that the sum of the A.P. is greater than or equal to the sum of the G.P.
9.
1 1 1 If x, y, z are postive and x + y + z = 1, prove that 1 1 1 8 x y z
10.
In a triangle ABC prove that
3 a b c 2 2 bc ca ab
74
ARTHMETIC PROGESSION
LEVEL - III 1.
The sum of the squares of three distinct real numbers, which are in G.P. is S2. If their sum is S,
1 2 show that ,1 (1, 3) . 3
2.
[1986]
If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means 2
n 1 between the same two numbers, prove that the value of q cannot be between p and p. n 1
[1991] 3.
If S1, S2, S3, ... Sn are the sums of infinite geometric series whose first terms are 1, 2,3, ..., n and whose common ratios are
1 1 1 1 , , ,..., respectively, then find the value of 2 3 4 n 1
2 S12 S 22 S32 ... S 2n 1 .
[1991]
4.
The real numbers x1, x2, x3 satisfying the equation x3 - x2 + x + = 0 are in A.P. Find the intervals in which and lie. [1996]
5.
Let a1, a2, ... be positive real numbers in geometric progression. For each n, let An, Gn, Hn be respectively, the arithmetic mean, geometric mean and harmonic mean of a1, a2, ... an. Find an expression for the geometric mean of G1, G2, ..., Gn in terms of A1, A2, ... , An, H1, H2, ... Hn. [2001]
6.
The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer. [2002]
7.
Let a, b, be positive real number’s. If a, A1, A2, b are in arithmetic progression, a, G1 , G2, b are in geometric progression and a, H1, H2 , b are in harmonic progression, show that G1G 2 A1 A 2 (2a b) (a 2b) . H1H 2 H1 H 2 9ab
[2002]
8.
If a, b, c are in A.P. and a2, b2, c2 are in H.P., then prove that either –a/2, b, c are in G.P. or a = b = c. [2003]
9.
Prove that (a + 1)7 (b + 1)7 (c + 1)7 > 77 a4 b4 c4, where a, b, c, R
[2004]
10.
An infinite G.P has first term x and sum 5, then find the exhaustive range of x ?
[2004]
11.
For n = 1, 2, 3, … , let A n
2
3
n
3 3 3 3 ..... (1) n 1 , and Bn = 1 – An . 4 4 4 4
Find the smallest natural number n0 such that Bn > An for all n n0. 75
[2006]
IIT- MATHS
ANSWER OBJECTIVE SECTION - I LEVEL - I 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
a
b
b
c
c
b
c
c
a
b
b
16.
c
17. 18. 19.
b
c
c
20.
c
12. 13.
c
d
14. 15.
c
a
21. 22. 23. 24. 25.
b
a
c
c
a
LEVEL - II 1.
2.
3.
4.
5.
6.
b
a
c,d
a
a
a,c a,b
16.
17. 18. 19.
b,c
c b,c,d c,
20.
7.
8.
9.
10.
11.
12. 13.
14. 15.
c
b
b
a,b c
a,b c,d
a,b c,d
12. 13.
a
c
21. 22. 23. 24. 25.
a,d b,d
a
a,d
c
b,d
SECTION - II 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
a
b
b
c
c
b
c
c
a
b
b
c
d
14. 15.
c
a
76
ARTHMETIC PROGESSION
SUBJECTIVE SECTION - III LEVEL - I 1. 2. 6.
9 14n 6 8n 23 (i) 6, 12, 24 OR 24, 12, 6 (ii) n = 4, 2 or 1 and a = 108, 144 or 160
7. 8.
99.2100 + 1 n2n+2 – 2n+1 + 2
9.
S
1 ab where a = 1 – x–1/3 & b = 1 – y –1/4 (1 ab)2
LEVEL - II
n(n 1) n(n 1) , when n be even & , when n be odd 2 2
3.
(ii)
5.
446
6.
1 n Sn . 2a (n 1)d [2a 2 2ad (n 1) nd 2 (n 1)] 2 2
7.
3 3 c = 9, roots are 3, , 2 5
LEVEL - III LEVEL-III 3.
1 n (2n 1) (4n 1) 1 3
5.
1 1 and 3 27 (A1 A2 .... An . H1 H2 ..... Hn)1/2n
10.
0 < x < 10.
11.
least value of n0 = 2
4.
77
IIT- MATHS
SUBJECTIVE SOLUTIONS LEVEL - I (CBSE LEVEL) REVIEW YOUR CONCEPTS 1.
Find the sum of first 24 terms of the A.P. a 1 , a 2 , a 3 , ... if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225. Solution : Let s = a1 a2 ... a24 are in A.P. Given a1 + a5 + a10 + a15 + a20 + a24 = 225 First term = a1 commoun ratio = d a1 + a1 + 4d + a1 + ad + a1 + 14d + a1 + 19d + a1 + 23d = 225 6a1 + 69d = 225 2a1 + 23d = 75 = t24 Sn = n/2 (a + l) 524 =
2.
24 (75) = 900 2
The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5. Find the number of sides of the polygon. Solution : First term a = 120° common ratio = 5 Given : sum of enterior angles of a polygon ‘n’ is (2n - 4) = (n – 2) (n – 2) 180° 2 n Sn = [2a (4 1) d] = (n – 2) 180° 2 n [240 5n 5] = (n - 2)180° 2 n2 - 25n + 144 = 0 n = 16 n = a
3.
The ratio between the sum of n terms of two A.P.’s is 7n + 1 : 4n + 27. Find the ratio between their nth terms. Solution : Let a1 a2 ... a4 are om AP b1b2 ... bn are in A.P. n n Sn = {2a1 + (n – 1) d1} Sn = {2b1 + (n – 1)d2} 2 2
78
ARTHMETIC PROGESSION 7n 1 2a1 (n 1)d ratio is 4n 27 2b (n 1)d 1
2a1 (n 1)d 2 4 7(n 1) 31 2 4(n 1) 2b1 (n 1)d 2 2a1 (n 1)d 8 – 1 (7 – 7n) = 2b (n 1)d 1
2 4 7(n 1) 2a1 (n 1)d 31 2 4(n 1) 2b1 (n 1)d 2
comparing the values a1, b1 and d. a1 = 4 b1 = ratio of nth term is
4.
31 d = 7 d2 = 4 2 1
14n 6 a1 (n 1)d1 4 (n 1)7 = 31 23 8n b1 (n 1)d 2 (n 1) 4 2
The rth, sth and tth terms of a certain G..P. are R, S and T respectively. Prove that Rs–t . St–r. Tr–s = 1. Solution : Given Tr = arr – 1 = R TS = ars–1 = S Tt = ar t – 1 = T S–t
then R
t t St t RS a s .r rs s St r S a .r = t tr t ; Sr a r .r rs t Rt a .r
Tr–s =
5.
T r a r .r tr S Ts a s .r ts s
RS–t . St – r . Tr – s = r° = 1
The sum of three numbers in G.P. is 42. If the first two numbers are increased by 2 and third is decreased by 4, the resulting numbers form an A.P. Find the numbers of G.P. Solution : Let a, ar, ar2 are in G.P.
79
IIT- MATHS 2
Given a + ar + av = 42 a(1 + r + v2) = 42
... (1)
and a + 2, ar + 2, ar2 – 4 are in A.P.
2(ar + 2) = a + 2 + ar2 – 4
a (1 + r2 – 2r) = 6 6 a= 2 1 r 2r Put ‘a’ in 0 6 (1 + r + r2) = 42 2 (1 r 2r) 6r2 - 15r – 6 = 0 1 r = 2, 2 a = 6, 24 If a = 6, r = 2 6, 12, 24 If
a = 24, r = 1/2 24, 12, 6
6.
If one G.M. G and two arithmetic means p and q be inserted between any given numbers, then show that G2 = (2p – q) (2q – p). Solution : x, p, q, y in these A.M. of x, p, q 2p = x + q x = 2p - q and
p, q, y 2q = p + y y = 2q – p G2 = 2y (G - Geometric mean) G2 = (2p – q) (2q – p)
7.
The sum of an infinite geometric series is 162 and the sum of its first n terms is 160. If the reciprocal of its common ratio is an integer, find all possible values of the common ratio, n and the first term of the series. Solution : Sum of an infinite geometric series 80
ARTHMETIC PROGESSION a Sn = = 162 a = 162 (1 – r) 1 r sum of first in terms is Sn =
a(1 r n ) 160 1 r
162 (1 r) (1 r n ) 160 (1 r) 1
160 n r r is an integer.. 162
1 rn 3
4
if n = 4, r = 1/3, then a = 108 1 if n = 1; r = then a = 160 81 1 n = 2, r then a = 144. a a = 108, 160, 144
8.
Evaluate: 1 + 2.2 + 3.22 + 4.23 + ... + 100.2 99 . Solution : Let s = 1 + 2.2 + 3 . 22 + ..... + 100.299 2.5 = 2 + 2.22 + 3.23 + ..... 99.299 + 100.2100 s - 25 = (1 + 2 + . . . . . + 299) + 100.2100 –8 = 1 + 2 (299 - 1) - 100.2100 s = –99.2100 – 1 = 99.2100 + 1
9.
If p th , q th , r th terms of an A.P. be a, b, c respectively, then prove that p(b – c) + q (c – a) + r (a – b) = 0. Solution : Given Tp = a1 + (p - 1) d1 = a Tq = a1 + (q - 1) d1 = b Tr = a1 + (r - 1) d1 = c
ab a - b = (p - q) d1 = = d1 pq
81
IIT- MATHS Similarly
bc ca d1 ; = d1 qr rp
p(b - c) + q (c - a) + r (a - b) pd, (q - r) + qd1 (r.p) + rd1 (p – q) 0. 10.
Find S of the G.P. whose first term is 28 and the fourth term is
4 . 49
Solution : First term a = 28 tH =
4 ar 3 r = 1/7 49
S =
11.
a 1 r
If H be the H.M. between a & b, then show that (H – 2a) (H – 2b) = H2 Solution : a, H, b are in H.P. H
2ab ab
Ha + Hb = 2ab Ha – 2ab = –Hb a(H 2b) b(H 2a) H; and H b a
(–H).(–H) =
a b (H 2b) (H.2a) b a
H2 = (H – 2a) (H – 2b)
12.
Find the sum of n terms of the series, the rth term of which is (2r + 1) 2r. Solution :
13.
Let x = 1 + 3a + 6a2 + 10a3 + ..., |a| < 1; y = 1 + 4b + 10b2 + 20b3 + ..., |b| < 1. Find S = 1 + 3 (ab) + 5 (ab)2 ... in terms of x and y. Solution : Series can be written as s = 1 + 3z + 5z2 + .... 82
ARTHMETIC PROGESSION sz = z + 3z2 + ...
s(1 – z) = 1 + 2 z + 2z2 + ... 1 + 2z [1 + z + z2 + ... ]
1 z (1 ab) 1 1 z s (1 – z) = 1 + 2 s= = 2 (1 z) (1 ab)2 1 z 1 z where a = 1 – x–1/3 and b = 1 – y–1/4 x = 1 + 3a + 6a2 + ..... |a| < 1 b(1 - b) y =
b 2b 2 .... 1 b b2 ...
ax = a + 3a2 + ..... x(1 - a) = 1 + 2a + 3a2 + ... ax(1 – a) = a + 2a2 + .... (1 - b)3 y =
(1 – a)2a = 1 + a + a2 + .... =
1 y = (1 - b)-1/4 1 – y–1/4 = b 1 b
1 1 x = (1 a)3 = (1- a)–3 1 a
a = 1 – x–1/3 . y = 1 + 4b + 10b2 + 20b3 + ... |b| < 1 by = b + 4b2 + 10b3 + ... y (1 - b) = 1 + 3b + 6b2 + 10b3 + ... b(1- b)y = b + 3b2 + .... y(1 - b)2 = 1 + 2b + 3b2 + ...
14.
If a, b, c are the sides of a triangle , then prove that a2 + b2 + c2 > ab + bc + ca. Solution : a, b, c are sides of a triangle (a 2 b 2 c 2 ) a b c 3 3
2
a 2 b 2 c 2 a 2 b 2 c 2 2(ab bc ca) 3 9 9 3a 2 3b 2 3c 2 a 2 b 2 c2 2(ab bc ca) 9 9 a2 + b2 + c2 > ab + bc + ca
83
IIT- MATHS 15.
1 1 1 2 If xi > 0, (i = 1, 2, ... n), then prove that (x1 + x2 + ... + xn) x + x + ... + x ³n . 1 2 n Solution : xi > 0 (Given) x1 + x2 + ..... + xn > 0 x1 x 2 .... x n (x1. x2. x3 ... xn)n n
(x1 x 2 .... x n ) n n 1 1 1 ... xn x1 x 2 1 1 1 (x1 + x2 + ... + xn) x x ... x 2 n 1
2 n
84
ARTHMETIC PROGESSION
LEVEL - I BRUSH UP YOUR CONCEPTS 1.
Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn , then show that the ratio S3n/Sn is equal to 6. Solution : Given S2n = 3Sn 2n n {2a (2n 1)d} 3 {2a (n 1)d} 2 2 2a = (n + 1)d n Sn {2a (n 1)d} {(n 1)d (n 1)d} n 2 d 2 3n 3n S3n = {2a + (3n - 1)d} {(n + 1)d + 3nd – d} = 6n2d 2 2 S3n = 6Sn S3n = 6 Sn
2.
If the roots of the equation x3 – 12x2 + 39x – 28 = 0 are in A.P., then find the common difference. Solution : Let f(x) = x3 – 12x2 + 39x – 28 = 0 if x = 1 f(x) = 0
(x – 1) is a factor
x2 – 11x + 28 = 0 (x – 7) (x – 4) = 0 x = 7, 4
series 1, 4, 7 are in A.P. (or) 7, 4, 1, ...
common ratio is d = 3
3.
If a1, a2, a3, ... are in A.P. with common difference d, then prove that tan 1
d d d nd tan 1 ... tan 1 tan 1 1 a 1a 2 1 a 2a 3 1 a n a n 1 1 a 1a n 1
Solution :
85
IIT- MATHS a1 a2 ... are in AP also tan–1a1 tan–1a2 .... are in AP tan-1 a2 – tan–1 a1 tan-1 a3 – tan–1 a2 tan-1 an + 1 – tan–1 an
1 (a n 1 a 1 ) tan–1 an+1 – tan–1 a1 tan (1 a a ) = tan–1 1 n 1 d = n (an + 1 – an)
nd 1 a1 a n 1
n=1 d = a2 – a1 nd = an+1 – an nd = an + 1 – a1
4.
If x = 1 + a + a2 + a3 + ... to (|a| < 1) and y = 1 + b + b2 + b3 + ... to (|b| < 1), then prove that 1 + ab + a2b2 + a3b3 + ... to =
xy . x + y -1
Solution : x = 1 + a + a2 + .... x=
1 x 1 a 1 a x
y = 1 + b + b2 + ...
y
1 y 1 b 1 b y
S = 1 + ab + (ab)2 + ..... S =
5.
xy 1 = x y 1 1 ab
Show that the number 1111.....1 is a composite number.. 91 digits
Solution : We have 111 ..... 1 (91 digits) = 1090 + 1089 + ..... + 102 + 10 + 1
1091 1 1091 1 107 1 10 1 107 1 1 0 1
= (1084 + 1077 + 1070 + ... + 1) (106 + 105 + ..... + 10 + 1)
86
ARTHMETIC PROGESSION [ 91 = 13 × 7, (91 divide and multiply by (10–1 – 1) or (1013 – 1] Thus, 111 ... 1 (91 digits) is a compossite number.
13 13 + 23 13 + 23 + 33 + +...16 terms. 6. Evaluate: + 1 1+ 3 1+ 3 + 5 Solution : Tn
13 23 33 ... n 3 n 2 (n 1)2 1 2 (n 2n 1) 1 3 5 ...n terms 4.n 2 4
16 16 1 16 1 1 1 S16 Tn n 2 1 (16.17.33) 2. .16.17 16 446 4 1 4 6 2 1 1
7.
Sum to n terms the series 12 – 22 + 32 – 42 + 52 – 62 + ... Solution : 12 – 22 + 32 – 42 + ... we consider two cases (i) Let n be even (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + ... [(n – 1) – n] [(n – 1) + n] (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + ..... + [(n – 1) – n] [(n – 1) + n]
= – 1 (1 + 2 + ... + n) =
n(n 1) 2
Case II n is odd (12 – 22) + (32 – 42) + .... + [(n – 2)2 – (n –1)2 ] + n2 = – 1 [1 + 2 + .... + (n – 1)] + n2 =
8.
(n 1)n n(n 1) n2 2 2
If the A.M. of a and b is twice as great as their G.M., then show that a : b = (2 + 3 ):(2 - 3 ) . Solution : Am =
ab GM = 2
ab and A = 2G
ab 2 ab a b 4 ab 2
a a 1 4 0 put b b
a =x b
x2 + 1 – 4x = 0 x = 2 3
87
IIT- MATHS taking positive sign
a 2 3 (2 3)2 b 2 3
taking negative sign a : b = 2 – 9.
3/2 3
a, b, c are the first three terms of a geometric series, If the harmonic mean of a and b is 12 and that of b and c is 36, find the first five terms of the series. Solution : a, b, c are in G. .P., Let choose them to br, b, b/r
12 =
Diving , r =
10.
2.br.b 2 br b(1 r) 1 r
36 =
2b.b / r 2 b 1 1 r b(1 ) r
1 and hence from any b = 24 3
five numbers are 8, 24, 72, 216, 648
In a triangle ABC prove that
3 a b c £ + + <2. 2 b +c c+a a+b
Solution : Ist part adding 3 to both sides a bc a bc a b c 3 3 bc ca ab 2 9 (b + c)–1 + (c + a) –1 + (a + b)–1 2(a b c)
Applying A.N of nth powers (A.M)m (b c) (c a) (a b) then LHS 3 3 2 3 (a b c)3 3
1
3.
1
2 (a b c)3 3
3
2 9 3(a b c) 2(a b c)
2nd part b + c > a in a triangle add (b + c) to both sides 2 (b + c) > (a + b + c)
1 1 2(b c) a b c 88
ARTHMETIC PROGESSION a 2a write similar inequalities and add ac abc
a
a bc
a c 2 a b c 2
89
IIT- MATHS
LEVEL - II CHECK UP YOUR SKILLS 1.
The sum of the squares of three distinct real numbers, which are in G.P. is S2. If their sum is 1 α 2 Î ,1 È (1, 3) . α S, show that 3
Solution : Let the numbers are ar, a1 a/r 1 a(r 1 ) S r 1 2 2 2 and a (r 1 2 ) S r 1 1 2 2 put r t, r 2 t 2 r r a (t + 1) = S and a2 (t2 – 1) = S2 Eliminating S, a2 (t2 – 1) =
(t – 1) 2 = (t + 1)
t
2 1 d2 1
Now t = r
1 r
a 2 (t 1) 2 2
... (1)
r2 – r t + 1 = 0
for t to be real t2 – 4 > 0
t < – 2 or t > 2
from (1)
2 1 2 1 2 or >2 2 1 2 1
In an enequality we can multiply only by a +ve quantity.
2 1 is we whether it is +ve or –ve
3 2 1 3 2 2 3 0 0 0 2 1 2 1 2 1
1 3( 2 ) ( 2 1) 3 0 ( 2 1)2
... (2)
and
( 2 1) ( 2 3) <0 ( 2 1) 2
... (3)
(3) implies that 2 lies between 1, and 3 90
ARTHMETIC PROGESSION
2.
(2) implies that 2 lies between 1/3 and 1. 1 2 ( ,1) (1,3) 3 Find the sum Sn of the cubes of the first n terms of an A.P. and show that the sum of first n terms of the A.P. is a factor of Sn. Solution : Let A.P. be (a + d) + (a + d) + ... + (a + nd) n n [2a (n 1)d] Sn = [2a + (n – nd] 2 2 Sn = (a + d)3 + (a + 2d)3 + ..... + (a + nd)3 = na3 + 3a2 d n + 3ad2 n2 + d2 n3 = na3 + 3a2 d
n(4 1) n(n 1) (2n 1) 3 n 2 (n 1) 2 3ad 2 d 2 6 4
1 n . [2a + (n + 1)d] [2a2 + 2ad + (n + 1 + d2 n (n + 1)] 2 2
3.
If the roots of 10x3 – cx2 – 54x – 27 = 0 are in harmonic progression, then find c and all the roots. Solution : , , be the roots in H.P..
then p, q, r be the roots of the equation 10 c 54 27 0 x3 x2 x
27x3 + 54x2 + (x – 10 = 0 are in AP ... (1) 2q = p + r
or 3q = p + q = r =
54 2 27
q = –2/3 is roots of (1)
27x2 + 54x2 + 9x – 10 = 0 has a root
(3x + 2) (9x2 + 12x – 5) = 0
x = 1/3, –2/3, –5/3 are in A.P
2 or 3
3x + 2 is its factor
3, –3/2, –3/5 are in H.P.
4.
91
If the (m + 1)th, (n + 1)th and (r + 1)th terms of an A.P. are in G.P., m, n, r are in H.P. show that
IIT- MATHS 2 the ratio of the common difference to the first term in the A.P. is - . n
Solution : a + md, a + nd, a + rd are in G.P
(a + nd)2 = (a + md) (a + rd) a (2n – m – r) = d (mr – n2)
d 2n (m r) a mr r 2
... (1)
but m, n, r are in H.P. 2mr mr
n
d 2n (m r) a (m r) n n 2 2
... (2)
2 2n (m r) 2 by () & (2) (m r) 2n n n
1 1 1 If x, y, z are positive and x + y + z = 1, prove that - 1 - 1 - 1 ³ 8 x y z Solution :
5.
A.M G.NM
yz 2
yz
... (i)
zn 2
zx
... (ii)
xy 2
xy
... (iii)
(i), (ii) & (iii)
(x y) (z x) (y z) xyz 8
(1 – x) (1 – y) (1 – z) 8 xyz
6.
Prove that x2 + y 2 + z 2 x+y+z
x+ y + z
x+y +z > xx y y zz > 3
x+ y + z
Solution : Consider x numbers each equal to x, y numbers each equal to y and z numbers each equal to z. then A.M. > G.M. On these x + y + z numbers 92
ARTHMETIC PROGESSION x x x... x times) (y y ... y times) (z z ... z times) x yz [x.x ... x factors) (x. y ... y factors) (z.z ... z factors]1/x + y + z But x + x ... x times = x.x = x2 and x.x ... . . x factors = xx
x 2 y2 z 2 x 2 y2 z 2 x y z 1/ x y z (x y z ) or xyz xyz
x y z
x x y y z z ... (A)
2nd part : consider x number each equal to 1/x y is 1/y 1 z is z AM > GM 1 1 1 1 1 1 ...x times ... y times ... z times y y x x z z x yz 1
1 1 1 1 ... y factors 1 1 x yz ...z factors . ...x factors x x y y z z
1 1 1 1 xyz x . z 3 1 x y z 1 1 1 xyz x y z xyz x . y . z x yz xyz x y z
x yz 3
x y z
x x y y z z or
xyz x y z > 3 x
y
z
x y z
... (3)
Both (A) & (B) prove the required results.
7.
a 2 + b2 Prove that a+b Solution :
a+b
> aa b b
Apply A.M > G.M
(a a .... a times a 2 . a 2 aa.a ...a factors a 2 a 2 b2 ab
8. 93
ab
> aa. bb
The sum of first ten terms of an A.P. is equal to 155, and the sum of the first two terms of a
IIT- MATHS G.P. is 9, find these progressions, if the first term of A.P. is equal to common ratio of G.P. and the first term of G.P. is equal to common difference of A.P. Solution : Let the AP be a + (a + d) + (a + 2d) + ... By given condition G.P is d + da + da2 + ... S1o of A.P = 155
S2 of G.P = 9
2a + ad = 31 d + da = 9 Solving a – 2, d = 3; or a =
A.P is 2 + 5 + 8 + ...
G.P. is 3 + 6 + 12 + 24 + ... 2 25 625 ... G.P. is 3 3 6 9.
25 2 d 2 3
or A.P
25 79 83 ....and 2 6 6
The series of natural numbers is divided into groups (1); (2, 3, 4); (5, 6, 7, 8, 9); ... and so on. Show that the sum of the numbers in the nth group is (n – 1)3 + n3. Solution :
The number of terms in successive groups are 1, 3, 5, ... and hence nth group will be nth terms of this A.P = 2n –n = N. The last terms of successive groups are 12, 22 ... hence nth groups is n2 and of (n – 1)th group is (n – 1)2 . Hence 1st term of nth group is one more than the last term of (n – 1) th group
A = (n – 1)2 + 1 = n2 – 2n – 2
Also terms in each group are in A.P. whose D = 1.
10.
n sum of terms is nth group is sum of an A.P = [2A (n 1)D] 2 2n 1 2 = [2n – 4n + 4 + (2n – 2)] 2 = 2n3 – 3n2 + 3n – 1 n3 + (n – 1)3 .
An A.P. and a G.P. with positive terms have the same number of terms and their first terms as well as last terms are equal. Show that the sum of the A.P. is greater than or equal to the sum of the G.P. Solution : Let ‘a’ be the first terms, ‘b’ the last term and n the numbers of terms of A.P and G.P. The C.d of
94
ARTHMETIC PROGESSION 1/ n 1
ba b A.P = and C.r of G.P = n 1 a
Let S be the sum of n terms of A.P and S the sum of n terms G.P.. n Now S (a b) 2 S = a (1 + r + r2 ... + rn–1)
S = a(rn–1 +rn–2 + ... + 1) a S [1 + rn–1) + (r + rn–1) + ... (rx + rn-k ) + ..... + (rn–1 + 1)] ... (2) 2 k (r + rn–k–1) – (rn–1 + 1) = (rk – 1) + (rn–1 (r–x – 1) r n 1 k n k 1 )0 = (r – 1) 1 k (r 1) (1 r r k
[for 0 r 1 and also for 1 < r < ]
rk + rn–k–1
from (2) S
95
n–1 + 1 for k = 0, 1, 2, .... , n – 1 r
a b ab an (1 r n 1 ) = n 1 = n S [ from (1)] 2 a 2 2
S S .
IIT- MATHS
PROBLEMS ASKED IN IIT - JEE 1.
If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means between the same two numbers, prove that the value of q cannot be between 2
n +1 p and p. n -1
[1991]
Solution : Let two numbers be a and b Since n A.M.’s have been inserted between a and b common difference d
ba h 1
Now P = first A.M. = 2nd terms of A.P = a + d = a +
ba n 1
Since n + 1 m’s have been inserted a and b C.d of A.P d1 =
ab ab(n 1)
q = first H.M = 2nd term of H.P
1 1 1 ab bn a q a d1 a ab(n 1) ab(n 1) q =
p
ab(n 1) ... (ii) bx a
an b ... (iii) n 1
x not will between and is ( x ) ( x ) > 0 n(a b)2 Now p – q = ... (A) (bn a) (n 1) 2 2 n 1 p ab(n 1) n 1 an b (n 1) (a b) n Again q – – = = (n 1)2 (bn a) ... (B) bn a n 1 m 1 n 1
n 1 2 n 2 (a b) 2 (a b) 2 Now (q – p) q p = (n 1)2 (bn a) 2 0 n 1 2
n 1 p Hence q can not lie between p and n 1
96
ARTHMETIC PROGESSION 2. If S1, S2, S3, ... Sn are the sums of infinite geometric series whose first terms are 1, 2,3, ..., n and whose common ratios are
1 1 1 1 , , ,..., respectively, then find the value of 2 3 4 n +1
S12 +S 22 +S 32 + ...+ S 2n-12 .
[1991]
Solution : 1
S1
1
1 2
2
2, S2
1
1 3
3,
...(i)
Now S12 S22 + ... + S22 + ... S22 (n – 1) = 22 + 32 + .... + (24)2 = S1 12 + s2 .... + (24)2}– 12 =
3.
2n (24 1) (4n 1) n(2n 1) (4n 1) 1 6 3
Let a1, a2, ... be positive real numbers in geometric progression. For each n, let An, Gn, Hn be respectively, the arithmetic mean, geometric mean and harmonic mean of a1, a2, ... an. Find an expression for the geometric mean of G1, G2, ..., Gn in terms of A1, A2, ... , An, H1, H2, ... Hn. [2001] Solution : a1 a2 ... a4 are in G.P. a2 = a1r : a3 = a1r2 ... an = a1rn–1
An =
... (1)
a 1 a1 (1 r r 2 ...r n 1 ) by (1) n n
or An
a1 r 1 ... (2) n r 1
Gn = (a1 a2 ... an )1/n = a1(1.r..r2 ...n) ... rn–1)1/n orGn = a1 [r(1/2)
(n – 1)
]1/4 = a1r1/2(n– 1) ... (3)
1 1 1 1 1 1 1 . [1 r 2 ... n 1 ] = H n n a1 n a1 r r 1 1 1 1 1 1 1 1 r n 1 1 1 . 1 ... = b.a r 1 r n 1 H n x a1 = n a1 r r 2 r n 1 1
na1 r n 1 (r 1) ... (4) Hn = r n 1 from (2) (3) & 4)
97
IIT- MATHS 2 n
2 1
G = Hn An = a
n–1
i(r ) above is true for each n, hence
G12 . G22 ... Gn2 = (–A1 A2 ... A4) (H1 ... H4) (G1 ... G4)1/4 = [(A1A2 ... An) (H1 H2 ... Hn)]1/2x
LHS is G.M/. of G1, G2, G3, ... Gx whose value we have determined in terms of A1A2 ... An and H1 H2 ... Hn .
4.
The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer. [2002] Solution :
Any four consecutive integers in A.P be taken as (a – 3d), (a – d), (a + d), (a + 3d) so that common difference is 2d and their product is (a2 – 9d2) (a2 – d2) we have to evaluate (2d)4 + (a2 – 9d2) (a2 – d2) a4 – 10a2 d2 + 25d2 = (a2 – 5d2)2 (a2 – 5d2)2 is also an enteger.
5.
Let a, b, be positive real number’s If a, A1, A2, b are in arithmetic progression, a, G1 , G2, b are in geometric progression and a, H1, H2 , b are in harmonic progression, show that G1G 2 A1 + A2 (2a + b)(a + 2b) = = . H1 H 2 H1 + H 2 9ab
[2002]
Solution : a, H1, H2, b are in H.P 1 1 1 1 a , H , H , b are A.P.. 1 2
1 1 3D b a
11 1 D 33 a
1 1 1 1 1 2b a D 3 H a a b a 3ab 1 1 1 1 1 1 1 2b a D 3 3 H1 a a a b a 3ab 98
ARTHMETIC PROGESSION G1G 2 ab (ab a) (2a b) (2b a) (2a b) = 2 2 HH 9a b 9ab 1 2
6.
If a, b, c are in A.P. and a2, b2, c2 are in H.P., then prove that either –a/2, b, c are in G.P. or a = b = c. [2003] Solution : 2b = a + c ab, b, c are in H.P ... (1)
b2
2a 2 c 2 a2, b2, c2 are in A.P. a 2 c2
... (2)
b2 [(a + c)2 – 2ac] = 2a2c2
b2 [(4b2 – 2ac)] = 2a2 c2 2b4 – b2ac – a2 c2 = 0 factorize b2 – ac = 0 or 2b2 + ac = 0 2
a c ac if b = ac then by 2 2
(a + c)2 – 4ac = 0 or (a – c)2 = 0 a = c and 2b = a + c = a + a = 2a b=a
b=a a=b=c if 2b2 + ac = 0 then b2 =
1 c 2
a , b, c are in G.P.. 2
7.
Prove that (a + 1)7 (b + 1)7 (c + 1)7 > 77 a4 b4 c4, where a, b, c, R Solution : (a + 1) (b + 1) (c + 1) = abc + ab + bc + ca + a + b + c + 1 > abc + ab + bc + (a + a + b + c = 7 4 4 4 1/7 7 (a b c )
99
abc ab bc ca a c 7
[2004]
IIT- MATHS Thus (a + 1) (b + 1) (c + 1) > 7 (a4 b4 c4)1/7 (a + 1)7 (b + 1)7 (c + 1)7 > 77 (a4 b4 c4)
8.
An infinite G.P has first term x and sum 5, then find the exhaustive range of x ? [2004]
100
ARTHMETIC PROGESSION Illustration 1: The mth term of an A.P. is n and its nth term is m. Prove that its pth term is m + n – p. Also show that its (m + n) th term is zero. Solution: Given Tm = a + (m – 1) d = n and Tn = a + (n – 1) d = m Solving we get, d = –1 and a = m + n – 1
Tp = a + (p –1)d = m + n – 1 + (p – 1) (–1) = m + n – p
Now, Tm + n = a + (m + n – 1)d = (m + n –1) + (m + n–1) (–1) = 0. Illustration 2: 1 2 Find the number of terms in the series 20, 19 , 18 , ..... of which the sum is 300. Explain the 3 3 double answer. Solution:
Clearly here a = 20, d
2 and Sn = 300. 3
n 2 2 20 (n 1) 300 2 3
Simplifying, n2 – 61n + 900 = 0 n = 25 or 36. Since common ratio is negative and S25 = S36 = 300, it shows that the sum of the eleven terms i.e., T26, T27 , ....., T36 is zero.
Illustration 3: The first term of an infinite G..P is 1 and any term is equal to the sum of all the succeeding terms. find the series. Solution: Given that Tp = (Tp+1 + Tp+2 + ..... ) or,
arp–1 = arp + arp+1 + arp+2 + ...
rp–1 =
1–r=r r=
rp [sum of an infinite G.P.] 1 r 1 2
1 1 1 Hence the series is 1, , , , ... . 2 4 8
Illustration 4: 101
IIT- MATHS If a1, a2, a3, ... an are in harmonic pregression, prove that a1a2 + a2a3 + ... + an–1 an = (n – 1) a1an . Solution: Since a1, a2 , ... , an are in H.P., 1 1 1 1 , , ,..., a1 a 2 a 3 a n are in A.P. having common difference d (say) .
1 1 1 1 1 1 d, d, ... d a 2 a1 a3 a 2 a n a n 1
or
a1 – a2 = d(a1a2), a2 – a3 = d (a2a3), ... , (an–1 – an) = d(an–1 an )
Adding the above relations, we get a1 – an = d (a1a2 + a2a3 +... + an–1 an)
... (1)
1 1 Now a a (n 1)d n 1
1 1 (n 1)d a n a1
or
(a1 – an) = (n – 1) d an a1
... (2)
Putting the value of a1 – an from (2) in (1), we get (n – 1) an a1d = d (a1a2 + a2a3 + ... + an–1 an)
(n – 1) ana1 = a1a2 + a2a3 + ... + an–1 an.
Illustration 5: If A1, A2; G1 , G2 and H1 , H2 be two A.M.s, G.M.s and H.M.s between two quantities ‘a’ and ‘b’ then show that A1H2 = A2 H1 = G1G2 = ab Solution: a, A1 , A2, b be are in A.P. ... (1) a, H1 , H2 , b are in H.P.
1 1 1 1 , , , a H1 H 2 b are in A.P..
Multiply by ab.
b,
ab ab , , a are in A.P. H1 H 2
take in reverse order. or
a,
ab ab , , b are in A.P.. H 2 H1
... (2)
Compare (1) and (2) ab ab and A 2 H2 H1
A1
A1H2 = A2H1 = ab = G1G2 102
ARTHMETIC PROGESSION Illustration 6: ab
a b Prove that 2
ab.ba , a,b N;a b.
Solution: Let us consider b quantities each equal to a and a quantities each equal to b. Then since A.M. > G.M. (a a a ...b times) (b b b ...a times) [(a.a.a...b times) (b.b.b. ... a times)]1/(a+b) ab
ab ab (a b ba )1/(a b) ab
2ab (a b ba )1/(a b) ab
Now
a b 2ab (A.M. > H.M.) 2 ab ab 2
a b
a b .b a .
Illustration 7: Prove that a4+ b4 + c4 abc (a + b + c), [a, b, c > 0] Solution: Using mth power inequality, we get a 4 b4 c4 a b c 3 3 a b c a b c 3 3
or
4
3
abc 1/ 3 3 [(abc) ] 3
( A.M G.M) .M)
a 4 b 4 c4 a b c abc 3 3 4 4 4 a + b + c abc (a + b + c).
Illustration 8: Prove that
s s s 9 , if s = a + b + c, [a, b, c > 0] sa sb sc 2
Illustration 9: Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ... . Solution: Clearly here the differences between the successive terms are 7 – 3, 14 – 7, 24 – 14, ... i.e., 4, 7, 10, ... which are in A.P.
Tn = an2 + bn + c Thus we have 3 = a + b + c, 7 = 4a + 2b + c and 14 = 9a + 3b + c
103
IIT- MATHS
3 2
Solving we get, a , b
1 ,c 2 . 2
1 Hence Tn (3n 2 n 4) 2
Sn =
1 [3n 2 n 4n] 2
1 n(n 1) (2n 1) n(n 1) n 3 4n (n 2 n 4) 2 6 2 2
Illustration 10: Find the sum of n terms of the series 3 + 8 + 22 + 72 + 266 + 1036 + ..... Solution: 1st difference 5, 14, 50, 194, 770, ... 2nd difference 9, 36, 144, 576, ..... They are in G.P. whose nth term is arn–1 = a4n–1 Tn of the given series will be of the form Tn = a4n–1 + bn + c T1 = a + b + c = 3 T2 = 4a + 2b + c = 8 T3 = 16a + 3b + c = 22 Solving we have a = 1, b = 2, c = 0. Tn = 4n–1 + 2n
1 Sn = 4n 1 2n (4 n 1) n(n 1) . 3
104
ARTHMETIC PROGESSION
PROGRESSION AND SERIES ARITHMETIC PROGRESSION (A.P.) If a is the first term and d the common difference, the A.P. can be written as a, a+d, a+2d,... The nth term a n is given by a n=a+(n-1)d. The sum S n of the first n terms of such an A.P. is given by Sn
n n ( 2a ( n 1)d) ( a l) where l is the last term (i.e., the nth term of the A.P.). 2 2
nth Term and Sum of n Terms : •
If a fixed number is added (subtracted) to each term of given A.P. then the resulting sequence is also an A.P. with the same common difference as that of the given A.P.
•
If each term of an A.P. is multiplied by a fixed number (say k) (or divided by a nonzero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k.
REMARKS : •
If a 1, a 2, a 3,.... and b1, b2, b3,..... are two A.P.’s with common differences d and d' respectively then a 1+b1, a 2+b2, a 3+b3 ,.... is also an A.P. with common difference d+d'.
•
If we have to take three terms in an A.P., it is convenient to take them as a-d, a, a+d. In general, we take a-rd, a-(r-1)d,...... a-d, a, a+d,.... a+rd in case we have to take (2r+1) terms in an A.P.
•
If we have to take four terms, we take a-3d, a-d, a+d, a+3d. In general, we take a-(2r-1)d, a-(2r-3)d,..... a-d, a+d,....a+(2r-1)d, in case we have to take 2r terms in an A.P.
•
If a 1, a 2, a 3,.... a n are in A.P. then a 1+a n= a 2+a n-1= a 3+a n-2 = ..... and so on.
•
ar
•
If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.)
a r k a r k k , k n r, k 0 2
between the other two i.e., if a, b, c are in A.P., then b
ac is the A.M. of a and c. 2
Arithmetic Mean(s): •
If a 1, a 2,.... a n are n numbers, then the arithmetic mean (A) of these numbers is A
.
1 (a 1 a 2 a 3 .....a n ) n
The n numbers A1, A2,...... An are said to be A.M.’s between the numbers a and b if a, A1, A2,....., Anb are in A.P. A n a
105
n ( b a ) a nb . n 1 n 1
IIT- MATHS
GEOMETRIC PROGRESSION (G.P.) nth Term and Sum of n Terms : If a is the first term and r the common ratio, then G.P. can be written as a, ar, ar 2,.... The nth term, a n, is given by a n = ar n-1. The sum S n of the first n terms of the G.P. is
a (r n 1) Sn , r 1 r 1 na , r 1 If -1
a 1 r
REMARKS : •
If each term of a G.P. is multiplied (divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same ratio as that of the given G.P.
•
If each term of a G.P. (with common ratio r) is raised to the power k, then the resulting sequence is also a G.P. with common ratio r k .
•
If a 1, a 2, a 3, .... and b1, b2 , b3,.... are two G.P.’s with common ratios r and r respectively, then the sequence a 1b1, a 2b2, a 3b3.... is also a G.P. with common ratio rr'.
•
If we have to take three terms in G.P., it is convenient to take them as a/r, a, ar. In general, we take
•
a a , k 1 ,..., a , ar ,..., ar k in case we have to take (2k+1) terms in a G.P.. k r r
If we have to take four terms in a G.P., it is convenient to take them as a/r 3, a/r, ar, ar 3. In general, we take
a r
2 k 1
,
a r
2 k 3
a ,....., , ar ,.....ar 2 k 1 , in case we have to take 2k terms r
in a G.P. •
If a 1, a 2,...., a n are in G.P., then a 1a n = a 2a n-1 = a 3a n-2 = ........
•
If a 1, a 2, a 3,...... is a G.P. (each a 1 > 0), then loga 1, loga 2 , loga 3...... is an A.P. The converse is also true.
Geometric Means : •
If three terms are in G.P., then the middle term is called the geometric mean (G.M) between the two. So if a, b, c are in G.P., then b ac is the geometric mean of a and c.
•
If a 1, a 2,...... a n are non-zero positive numbers, then their GM. (G) is given G = (a 1a 2a 3 ......a n) 1/n
•
If G 1, G 2,..... G n are n geometric means, between a and b, then a, G 1, G 2,...., G n, b
106
ARTHMETIC PROGESSION b will be a G.P. G n a n 1 a
n
HARMONIC PROGRESSION (H.P.) 1 1 1 1 The sequence a 1, a 2, a 3,.....a n (a i 0) is said to be an H.P. if the sequence a , a , a ,.... a ,.... 1 2 3 n
is an A.P. 1 1 1 1 The nth term, a n of the H.P. is a n a (n 1)d , where a a , and d a a . 1 2 1
NOTE : There is no formula for the sum of n terms of an H.P. •
If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have
1 11 1 2ab H . H 2a b ab
•
If a 1, a 2,..... a n are ‘n’ non-zero numbers, then the harmonic mean H of these numbers is given by 1 1 1 1 ..... 1 H n a 1 a 2 a n
•
The n numbers H 1, H 2,....., H n are said to be n-harmonic means between a and b, 1 1 1 1 1 if a, H 1, H 2......, H n, b are in H.P. i.e. if a , H , H ..... H , b are in A.P.. 1 2 n 1 1 n (a b ) b na H n a (n 1)ab ab(n 1)
ARITHMETIC - GEOMETRIC PROGRESSION : The sum of S n of first n terms of an A.G.P. is obtained in the following way : Sn = ab + (a + d)br + (a + 2d)br 2 +.....+ (a + (n - 2)d)br n-2 + (a + (n - 1)d)br n-1 Multiply both sides by r, so that rS n = abr + (a+d)br 2 +....+ (a + (n - 3)d)br n-2 + (a + (n - 2)d)br n-1 + (a + (n - 1)d)br n Subtracting, we get (1-r)S n=ab+dbr+dbr 2+.....+dbr n-2+dbr n-1-(a+(n-1)d)br n
dbr (1 r n 1 ) ab (a (n 1)d )br n (1 r )
Sn 107
ab dbr (1 r n 1 ) (a (n 1)d )br n 1 r (1 r ) 2 1 r
IIT- MATHS If -1 < r < 1, the sum of the infinite number of terms of the progression
lim n
Sn S
ab dbr 1 r (1 r ) 2
MISCELLANEOUS PROGRESSION : Some Important Results : n ( n 1) (sum of the first n natural numbers) 2
•
1 2 3 ..... n
•
12 2 2 32 ..... n 2
•
13 23 33 .... n 3
n ( n 1)(2n 1) (sum of squares of the first n natural numbers) 6
n 2 (n 1) 2 (1 2 3 .... n ) 2 (sum of cubes of first n natural 4
numbers) •
1 + x + x2 + x3 + ..... = (1 - x) -1 ,
if -1 < x < 1
•
1 + 2x + 3x2 + ..... = (1 - x) -2 ,
if -1 < x < 1
INEQUALITIES : Let a 1, a 2,.....,a n be n positive real numbers, then we define their arithmetic mean (A), geometric mean
H
(G)
and
harmonic
n 1 1 1 1 ..... an a1 a 2 a 3
mean
(H)
as
A
a 1 a 2 ..... a n , n
G=(a 1 a 2 .....a n ) 1/n
and
It can be shown that A>G>H. Moreover equality holds at either place if and only if a 1=a 2=...=a n .
Weighted Means : Let a 1, a 2,....., a n be n positive real numbers and m1, m2,....., mn be n positive rational numbers. Then we defined weighted Arithmetic mean (A*), weighted Geometric mean (G*) and Weighted mean (H*) as A*
m1a 1 ..... m n a n m m m , G* a 1 1 .a 2 2 ......a n n m1 m 2 .... m n
1 ( m1 m 2 .... m n )
and H*
m1 m 2 .... m n m1 m 2 m ..... n an a1 a 2
It can be shown that A* > G* > H*. Moreover equality holds at either place if and only if a1 = a 2 = ..... = a n .
Arithmetic Mean of mth Power :
108
ARTHMETIC PROGESSION
Let a 1, a 2,......, a n be n positive real numbers (not all equal) and let m be a real number, then m
m
m
m
a 1 a 2 ...... a n a a 2 ..... a n 1 if m R [0, 1] n n m
m
m
a 1 a 2 ..... a n a a 2 ..... a n 1 However if m (0,1), then n n m
Obviously if m {0,1}, then
m
m
m
a 1 a 2 ..... a n a a 2 ..... a n 1 n n
m
Cauchy’s Inequality : If a 1's and bis are reals, then (a 12 + ..... + a n2) (b12 + .... + bn2) > (a 1b1+a 2b2+....+a nbn) 2 .
Proving Inequalities : (i)
Any inequality has to be solved using a clever manipulation of the previous results.
(ii)
Any inequality involving the sides of a triangle can be reduced to an inequality involving only positive reals, which is generally easier to prove. For the triangle we have the constraints a + b > c, b + c > a, a + c > b Do the following : put
x = s-a, y=s-b, z=s-c
then, x+y+z=3s-2s=s and a=y+z, b=x+z, c=x+y Substitute a=y+z, b=x+z, c=x+y in the inequality involving a, b, c to get an inequality involving x, y, z. Also note that the condition a+b>c is equivalent to a + b + c > 2c i.e., 2s > 2c OR s - c > 0, i.e., z > 0 Similar b + c > a = x > 0, a + c > b = y > 0
109
The inequality obtained after the substitution is easier to prove. (involving only positive reals without any other constraints).
IIT- MATHS
WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 The sum of n terms of two series in A.P. are in the ratio 5n +4 : 9n + 6. Find the ratio of their 13th terms. (A)
129 131
(B)
127 132
(C)
125 134
(D)
121 139
Solution : Ans : (A) Let be the first terms of two A.P.s and are their respective common differences.
n 1 d n a1 2a1 n 1 d1 5n 4 1 5n 4 2 2 ==> n ==> n 1 d 9n 6 2a2 n 1 d 2 9n 6 a2 2 2 2
.................. (1)
a1 12d1 Now the ratio of 13th terms = a 12d 2 2
==> put
n 1 12 2
a1 12d1 5 25 4 129 i.e. n = 25 in equation (1) ==> a 12d 9 25 6 231 2 2
ILLUSTRATION : 02 At what values of parameter 'a' are there values of n such that the numbers : 51 x 51 x ,
a x , x 2 25 25
form an A.P. ? (A) a £ 8
(B) a £ 8
(C) a £ 12
(D) a £ 12
Ans : (C) Solution : For the given numbers to be in A.P., Let 5x k ==> a 5k
5 1 k2 2 k k
1 2 1 ==> a 5 k k 2 k k
As the sum of a positive number and its reciprocal is always greater than or equal to 2, k
1 1 2 and k 2 2 2 ; k k
Hence a 5 2 2 ==> a 12
110
ARTHMETIC PROGESSION ILLUSTRATION : 03 1 1 1 1 ............ 1.2.3 2.3.4 3.4.5 4.5.6
Find the sum of first n terms of the series :
1 1 (A) 2 2 n 1 n 2
1 1 (B) 4 n 1 n 2
1 1 (C) 2 n 1 n 2
1 1 (D) 4 2 n 1 n 2
Ans : (D) Solution :
1 1 1 1 Let S = 1.2.3 2.3.4 3.4.5 ............ n n 1 n 2
n 2 n 3 1 4 2 5 3 2S = 1.2.3 2.3.4 3.4.5 ............ n n 1 n 2 1 1 1 1 1 1 2S = 1.2 2.3 2.3 3.4 .......... n n 1 n 1 n 2
1 1 2S = 1.2 n 1 n 2
==>
1 1 S = 4 2 n 1 n 2
Note : You should observe that here, Tn
1 1 1 1 n n 1 n 2 2 n n 1 n 1 n 2
It is in the form f(n) - f(n+1). ILLUSTRATION : 04 Find the three digit number whose consecutive number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order. Now if we increase the second digit of the required number by 2, the resulting number will form an A.P. (A) 901
(B)931
(C)981
Ans : (B) Solution : Let the three digit be a, ar, ar2 then according to hypothesis 100a + 10ar + ar2 +792 = 100ar2 + 10ar + a ==> a r 2 1 8 ..................................................(1) 111
(D)991
IIT- MATHS and a, ar + 2, ar2 are in A.P. then 2 ar 2 a ar 2 ==> a r 2 2r 1 4
...............................................(2)
Dividing (1) by (2), a r 2 1
8 Then a r 2 2r 1 4
r 1 r 1 2 2 r 1
==>
r=3
r 1 2 r 1
==>
from (1), a =1
Thus digits are 1, 3, 9 and so the required number is 931. ILLUSTRATION : 05 Sum to n terms of the series.
n
cos ec 1 10 cos ec 1 50 cos ec 1 170 ........... cos ec 1 1 (A) tan n 1
4
1 (B) tan n 1
4
2
1 n 2 2n 2 is.............
1 (C) tan n 1
2
1 (D) tan n 1
2
Ans : (A) Solution :
n
Let cos ec1 ==>
2
1 n 2 2n 2
n
cosec2 = =
n
2
2
1 n 2 2n 2
=
2
1 2n n 2 1 n 2 1
==>
cot2 =
==>
tan
n
2
n 1
=
n
2
n
2
1 n 2 1 2n 1 2
n 1 1
2
n 1 n 1 n n 1 1 n 1 n 2
n 1 n 1 1 tan 1 tan n 1 tan n 1 n 1 n
==>
Thus, sum to n terms of the given series. =
tan
1
2 tan 1 1 tan 1 3 tan 1 2 +
tan
1
4 tan 1 3 ........ tan 1 n 1 tan 1 n
112
ARTHMETIC PROGESSION = tan 1 n 1 tan 1 1 tan
1
n 1
4
ILLUSTRATION : 06 If a, b, c are in G.P. ,and loga - log2b, log2b - log3c. and sides of a triangle which is (A) acute - angled
(B) obtuse - angles
are in A.P., then are the lengths of the
(C) right - angled
(D) equilateral
Ans : (B) Solution : We have b 2 ac and 2 log 2b log 3c = log a log 2b log 3c log a 2a 4a and 2b 3c ==> b and c 3 9
==> b 2 ac since a b
5a 10a 13a b , therefore are the sides of a > c, b c >a and c a 3 9 9
triangle. Also, as a is the greatest side, let us find angle A of ABC.
cos A
b 2 c 2 a 2 29 0 2bc 48
Hence, ABC is an obtuse-angled triangle. ILLUSTRATION : 07 If log x a, a x / 2 and log b x are in G.P. then is equal to (A) log a log b a
(B) log a log e a log a log e b
(C) log a log a b
(D) log a log e b log a log e a
Ans : (a) and (b) Solution : As log x a, a x / 2 , log b x are in G.P.. x/2 2
a
log x a.log b x
x ==> a
log a log x log a . log b a log x log b log b
==> x log a log b a log a log e a log a log e b
ILLUSTRATION : 08
113
IIT- MATHS If there unequal positive real numbers a, b, c are in G.P. and b c, c a, a b are in H.P. then the value of a b c 1is independent of (A) a
(B) b
(C) c
(D) None of these
Ans : (D) Solution : As are in G.P., b 2 ac and b c, c a, a b are in H.P..
ac 2 1 1 = b c a b c a b c a b ==> 2 b c a b a c ==> 2 ab ac b 2 bc = ==> 2 ab 2b 2 bc ==> 2b
a c
2
2
2
a c
a c
a c
2
2
a c
a c
a c
2
2
2
a c 0
==> 2b a c 2 ac a c 2b ==> a b c 3 ac Which is not independent of a, b and c.
ILLUSTRATION : 09 An A.P. whose first term is unity and in which the sum of the first half of any even number of terms to that of the second half of the same number of terms is in constant ratio, the common difference d is given by (a) 1
(b) 2
(c) 3
(d) 4
Ans: (B) Solution : Let denote the sum to n terms of the A.P. According to the given condition Sn k n 1 S 2n S n S1 S2 ==> S S S S 2 1 4 2
==> S1 S4 - S1S2 = -S1S2 114
ARTHMETIC PROGESSION ==> S1S 4 S22
2 4 ==> a 2a 4 1 d a a d 2
==> a 4a 6d 2a d
2
==> 4a 2 6ac 4a 2 4ad d 2
==> 2ad d 2 ==> 2a d As a =1, we get d = 2.
ILLUSTRATION : 10 n
If
t r 1
r
n 1 1 n n 1 n 2 n 1 , Then lim is n 6 r 1 t r
(a) 2
(b) 3
Ans : (A) Solution : We have, for n 1, n 1
n
t n t r t r r 1
r 1
1 1 n n 1 n 2 n 1 n n 1 6 6
1 1 n n 1 n 2 n 1 n n 1 6 2
1 2 1 1 Now, for r 1 , t r r 1 2 r r 1 r n
n 1 1 1 1 2 ==> t r r 1 2 1 n 1 r 1 r r 1 n
1 1 lim 2 1 ==> lim 2 1 0 2 z n n n 1 r 1 t r
115
(c) 3/2
(d) 6
IIT- MATHS
SECTION A
SINGLE ANSWER TYPE QUESTIONS 1.
The mth term of an A.P is n and its nth term is m. Its pth term is A) m +n+p
2.
B) m+n-p
B) x = y -1
If
B) (2p+1)(p+1)2
D) p3 + (p+1)3
D) p3,q3,r3 are in A.P
A.G.P consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, the common ratio will be B) 3
If b-c, 2b-x, b-a are in H.P then a A) A.P
7.
C) (p+1)3
B) p2,q2,r2 are in A.P C) 1/p, 1/q,1/r in A.P
A) 2 6.
D) z2 = x
1 1 1 , , are in A.P then q+r r+p p+q
A) p,q,r are in A.P 5.
C) z-3 = y
The first term of an A.P of consecutive integers is p2 + 1. The sum of (2p +1) terms of this series can be expressed as A) (p+1)2
4.
D) m-n-p
If 1, log y x , log z y -15log x z are in A.P then A) z3 = x
3.
C) m-n+p
C) 4
D) 5
x x x ,b,care in 2 2 2
B) G.P
If S 1,S 2 ,S 3 denote the sums of
C) H.P
D) none of these
n1 , n2 , n3 t erms respectively of an A.P then
s1 s s n 2 - n 3 + 2 n 3 - n1 + 3 n1 - n 2 is equal to n1 n2 n3
A) 0 8.
B) 1
10.
D) n1n2n3
If the interior angles of a polygon are in A.P with common difference 50 and smallest angle is 1200 then the number of sides of the polygon is A) 9 or 16
9.
C) S1,S2,S3
B) 9
C) 16
D) 13
If the arithmetic mean of two positive numbers a and b (a > b) is twice their geometric mean then a : b is A) 2 +
3 :2-
C) 6+
7 :6-
3 7
B) 5 +
6
: 5-
D) 1 +
2 :1- 2
6
If m is a root of the equation (1-ab)x2 - (a2+b2)x - (1+ab) = 0 and m harmonic means are inserted between a and b then the difference between the last and the first of the means equals A) b - a
B) ab( b - a )
C) a (b-a)
D) ab(a-b) 116
ARTHMETIC PROGESSION 11. If a1 , a 2 a 3 , .... a n are in H.P then a1a 2 + a 2 a 3 + a 3a 4 + . . . . . + a n-1a n is equal to A) n 12.
C) 2n
1 mn
B)
1 1 m n
C) 1
D) 0
If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the square of their reciprocals then
a b c , , are in c a b
A) Arithmetic - Geometric progression C) Geometric progression 14.
D) (n-2)
Let Tr be te rth term of an A.P for r =1,2,3. If for some positive integers m, n we have Tm= 1/n and Tn = 1/m, then Tmn is A)
13.
B) (n-1)
B) Arithmetic progression D) Harmonic progression
Suppose a,b >0 and x1 , x2, x3 (x1 > x2 > x3) are the roots of
x-a x-b b a + = + b a x -a x -b
and
then a,b,c are in A) A.P 15.
18.
B) 1
D) does not exist
tan 2 α x2 + x
C) 0 < x < 10
D) x > 10
is always greater than equal to D) sec2
C) 2
Let the positive numbers a,b,c,d be in A.P. Then abc, abd, acd, bcd are B) A.P
C) G.P
D) H.P
If a,b,c,d are +ive real no such that a+b+c+d =2 then M = (a+b) (c+d) satisfies the relation B) 1 M 2
C) 2 M 3
D) 3 M 4
Let S (0, ) denote the set of values of x satisfying the equations 81 |cosx|+ cos2x + |cos3x| +.... to = 43 then S = z 2 B) , 3 3
2 C) , 3 3
2 D) , 3 3
Sum of the series 1+2.2 + 3.22 + 4.23 +.....+100.299 is A) 100.2100+1
117
B) - 10 < x < 0
A) 2 tan
A) 3
21.
C) 3/4
x2 + x -
A) 0 M 1 20.
B) 1
If (0, π /2) then
A) Not in A.P/G.P/H.P 19.
D) none of these
An infinite G.P has first term 'x' and sum then x belongs to A) x < - 10
17.
C) H.P
If is the nth term of a G.P with first term 1 and common ratio r, then the minimum value of is A) 1/4
16.
B) G.P
B)99.2100+1
C)99.299-1
D)100.2100-1
IIT- MATHS 22.
If a,b,c are in arithmetic progression then A)A.P.
23.
B) G.P.
If
B) 17
D) None of these
C) 27/14
D) 56/15
1 π cosec 2 q, 2cotq, sec , 0 < θ < are in G.P. iff ? is equal to 2 2
A) /6 25.
C) H.P.
If the sum of first n terms of a series be 5n 2 + 2n , then its second term is A) 16
24.
1 1 1 , , are in : a+ b a+ c b+ c
If 3 +
B) /4
C) /3
D) None of these
1 1 3 + d + 2 3 + 2d + ....... to =8, then the value of is 4 4
A) 9
B) 5
C) 1
D) None of these
a n 1 n log n1 C) 2 b
a n 1 n log n 1 D) 2 b
ar log r-1 = The value of r=1 b n
26.
an n log n A) 2 b 27.
The roots of the equation A) form an A.P.
28.
- 4 | x -1 | +3 = 0
B) form a G.P.
C) Form a H.P.
B) -1
x -1
D) do not form any progression 2
- 4 | x -1 | +3 = 0
C) 1/2
B) are in A.P
1 1 1 π2 + + + ....to If 2 = 1 22 32 6
2 A) 8 31.
2
D) None of these
c A 3b If in a Δ ABC acos2 +ccos2 = then the sides a,b,c 2 2 2
A) satisfy a+b=c 30.
x -1
1 a n + bn 1 If , n+1 n+1 , are in A.P., then n is equal to a a +b b A) 0
29.
a n 1 n log n B) 2 b
If
2 B) 12
C) are in G.P
D) are in H.P
then equals :
2 C) 3
2 D) 2
a + be y b + ce y c + de y = = then, a,b,c,d are in a - be y b - ce y c - de y
A) A.P.
B) G.P.
C) H.P.
D) None of the above
118
ARTHMETIC PROGESSION 32. If f x is a function satisfying f x + y = f x f y for all x, y N , such that f 1 = 3 n
f x = 120. then the value of n is
and
x =1
A) 4 33.
34.
B) 5
C) 6
The sum of all the products of first n positive integers taken two at a time is A)
1 n 1 n n 1 3n 2 24
B)
C)
1 n n 1 n 2 n 5 6
D) None of the above
B) G.P only when x > 0 D) G.P only when x < 0
If the A.M of the roots of a quadratic equation is 8/5 and the A.M of their reciprocals is 8/7, then the quadratic equation is A) 7x2 + 16x +5 = 0
36.
1 n 2 n n 1 n 2 48
If a, b, c are in A.P then 10ax+10, 10bx+10,10cx+10 are in A) A.P C) G.P for all x
35.
D) None of these
B) 7x2 - 16x + 5 = 0
C) 5x2-16x + 7 =0
D) 5x2 - 8x + 7 = 0
α , β be roots of the equation x2 - 3x + a =0 and λ , δ be roots of x2 - 12x + b = 0 and numbers
, , , form an increasing G.P then A) a=3 b=12 37.
B) a=12 b=3
B) 11 : 10
C)
5 1 , 2
5 1 2
A) 0
B)
3 , 1/ 3
D)
3 1 , 2
B) 2+
sinx + cosx is sinx - cosx
3
C) 2 - 3
B) ab (b -a)
D) 1 +
C) a (b - a) 1
119
3 1 2
3
If m is a root of the equation (1-ab)x2 - (a2 + b2) x - (1+ab) = 0 and m harmonic means are inserted between a and b then the difference between the last and the first of the means equals A) b - a
41.
D) 1 : 2
If for 0 < x < /2 y = exp [ sin 2 x sin 4 x ..... ] is a zero of the quadratic equation x2 -9x + 8 =0 then the value of
40.
C) 17 : 15
If the sides of a right angles triangle are in A.P then the sines of the actute angles are A) 3/5, 4/5
39.
D) a=4 b=16
Let a,b, be roots of x2-3x+p=0 and Let c,d be the root of x2-12x + q= 0 where a,b,c,d form an increasing G.P. Then the ratio of p+q : q - p is equal to A) 8 : 7
38.
C) a=2 b=32
1
D) ab (a - b) 1
1
{an} and {bn} are two sequences given by an = x 2n + y 2n and bn = x 2n - y 2n
for all n =
IIT- MATHS N. The value of a1, a2 . . .an is equal to A) x - y 42.
xy D) b n
B) 20/9
C) 9/20
x2 -
D) 11/23
Suppose a,b,c are in A.P and a2,b2,c2 are in G.P. If a < b < c and a + b + c = 3/2 then the value of a is A)
44.
x y C) b n
If exp {tan2x - tan4x+tan6x - tan8 x +..} loge16 0 < x < /4, satisfies the quadratic equation 3x + 2 = 0 the value of cos2x + cos4x is equal to A) 21/16
43.
x y B) b n
1 2 2
B)
1
C)
2 3
1 1 2 3
D)
1 1 2 2
If . . . . . . . are in G.P then the value of the determinant
is A) -2 45.
46.
B) 1
C) 2
D) 0
If (a,b) (c,d) (e,f) are the vertices of a triangle such that a,c,e are in G.P with common ratio 'r' and b,d,f ax is G.P with common ratio 's' then area of the triangle is A)
ab r 1 ( s 2)( s r ) 2
B)
ab r 1 s 1 s r 2
C)
ab r 1 s 1 s r 2
D)
r 1 s 1 s r
The roots of equation x2 + 2(a-3) + 9= 0 lie between -6 and 1 and 2,h1,h2...h20[a] are in H.P, where [a] denotes the integral part of a and 2, a1a 2 ... a 20 ,[a] are in A.P then a3h18 = A) 6
47.
D) none of these
B) G.P
C) H.P
D) none of these
If the A.M of two numbers is twice of their geometric mean, then ratio of sum of numbers to the difference of numbers equals A) 2
49.
C) 18
If a,b,c,d and p are distinct real numbers such that (a2+b2+c2)p2 - 2(ab+bc+cd)p + (b2+c2+d2) 0 then a, b, c, d are in A) A.P
48.
B) 12
B) 3 / 2
C) 4
D) 2/
3
Let α , β be the roots of x2 - x + p = 0 and λ, δ be the roots of x2 - 4x + q =0. If α , β , λ, δ are in G.P, then the integral values of p and q respectively are A) -2, -32
B) -2,3
C) -6,3
D) -6, -32 n
50.
α r , β r α r < β r are the roots of x 2 - r 2 r +1 x + r 5 =0 The value of
3α
r
+ 2β r is
r=1
120
ARTHMETIC PROGESSION 1 2 A) n n 1 n 3n 1 2 C) 51.
3 n n 1 n 2 n 1 2
2 1 bc b 2
If for 0 < x <
A) 0
C) 12
B)
1 3 2 1 4 c 2 ca a 2
C)
3 2 2 b ab
π 2 4 6 , y = exp is a zero of the quadratic sin x + sin x + sin x + ....... to log2 2 sinx + cosx is : sinx - cosx
B) 2 3
C) 2 3
D) none of these
B) a b c a
C) 3ac
D) 3bd
B) bx 2 c
C) cx 2 d
B) 5000
C) -5050 2
58.
D) x a
The coefficient of x99in the expansion of (x - 1) (x - 2) ..... (x - 100) is A) 5050
57.
D) None of the above
If a,b,c,d are in G.P., then a factor of ax 3 + bx 2 + cx + d is A) ax 2 c
56.
D) None of these
If a,b,c,d are in H.P., then ab + bc + cd is A) 3ad
55.
1 n n 1 n 2 n 3 2
B) 10
equation then the value of the value of
54.
D)
1 1 1 1 1 1 If a , b,c are in H.P., then the value of + - + - is b c a c a b
A)
53.
1 n n 1 3n 2 n 1 2
1+x 1-x a x -x The least value of 'a' for which 5 + 5 , , 25 + 25 are three consecutive terms of an A.P. is 2 :
A) 5 52.
B)
D) -5000 2
2
1 2 1 3 1 The sum of the first 10 terms of the series x + + x + 2 + x + 3 + ..... is x x x
x 20 1 x 22 1 A) x 2 1 x 20 20
x18 1 x11 1 B) x 2 1 x 9 20
x18 1 x11 1 C) x 2 1 x9 20
D) None of these
When are in A.P., and when
1 1 1 1 1 1 1 1 5 , , , , are in A.P., + + = then the numbers a and b a x y z b x y z 3
are : A) 8,2
121
B) 9,1
C) 7,3
D) none of these
IIT- MATHS 59.
2
Let a and b be the roots of x - 3x + p = 0 and let c and d be the roots of x -12x + q = 0, where a,b,c,d form in increasing G.P. Then the common ratio of q + p : q - p is A) 8:7
60.
B) 8
B) 2
D) 4
C) 3a2
D) 4a2
r r 1 B) 2 x
r r 1 C) 2 x xy
r r 1 D) 2 x rx
B) second term D) last term
If the numbers a,b,c,d,e form an A.P then the value of a - 4b + 6c - 4d + e is B) 2
C) 0
D) 3
If ax = by = cz = dt and a,b,c,d are in G.P then x,y,z,t are in A) A.P
67.
D) 10
In an A.P the sum of terms equidistant from the beginning and end is equal to
A) 1 66.
C) 3
B) 2a2
A) first term C) sum of 1st and last terms 65.
C) 9
n A series whose nth terms is +y then sum of r terms will be x
r r 1 A) 2 x ry
64.
D) none of these
The length of a side of a square is 'a' metre. A second square is formed by joining the middle points of this square. Then a third square is formed by joining the middle points of the second square and so on. Then the sum of the area of the squares which carried upto infinity is A) a2
63.
C) 17 :15
If the roots of the equation are in A.P., then the common difference will be A) 1
62.
B) 11:10
The sum of the cubes of first n natural numbers does not exceed 1360, the maximum value of n is A) 7
61.
2
B) G.P
C) H.P
D) none of these
If α1 ,α2 ,α3 .....Α.Π such that a1 + a 5 + a10 + a15 + a 20 + a 24 = 225, then
a1 + a 2 + . . . . . a 23 + a 24
equal to A) 909 68.
C) 750
D) 900
C) 3ac
D) 3cd
If a,b,c,d are in H.P then ab+bc+cd is equal to A) 3ad
69.
B) 75
B) (a+b)(c+d)
A1 +A2 If A1, A2 be two A.Ms and G1,G2 be two G.Ms between a and b then G G is equal to 1 2
A)
a b 2ab
B)
2ab a b
C)
a b ab
D)
a b ab 122
ARTHMETIC PROGESSION 70.
a1 a2 a3 If a1,a2,a3. . .an are in H.P then a + a +. . . + a + a + a +. . . + a + a + a +. . . + a +. . . are in 2 3 n 1 3 n 1 2 n
A) A.P 71.
B) G.P
n
If
r=1
r=1
B) 3/2
D) a = 3d
1 is equal to tr
C) 3/4
D) 3/8
The sixth term of an A.P is 2 and its common difference is greater than one. The value of the common difference of the progrenion so that the product of the first, fourth and fifth terms is greatest is B) 2/3
C) 5/8
D) 3/2
Consider an infinite geometric series with first term 'a' and common ratio 'r'. If its sum is 4 and the second term is 3/4 then A) a = 7/4, r = 3/7
75.
C) 2a = d n
A) 8/5 74.
B) a = d
t r = 2 3 n - 1 " n³1 then lim x®¥
A) 3 73.
D) none of these
consider an A.P with first term 'a' and common difference 'd' Let Sk denotes the sum of the first k terms. If Skx/Sx is independent of x then A) a = 2d
72.
C) H.P
B) a = 2, r = 3/8
C) a=3/2, r =1/2
D) a=3, r = 1/4
If a1a 2 . . .. . a10 be in A.P and h1 h2 .....h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3 then a4 h7 is equal to A) 2
76.
B) 3
B) are in A.P D) are in H.P
If 2(y-a) is the H.M between y - x , y - z then x - a , y - z then x - a, y - a, z - a are in A) A. P
B) G.P
C) H.P
21
78.
If
j
= 693 where a , a ..........a are in A.P. then 1 2 21
A) 361
B) 396
a
2i+1
is
i=0
C) 363
D) data insufficient
If the roots of the equation x3-12x2+39x-28=0 are in A.P, then their common difference will be A) 1
80.
D) none of these 10
a j=1
79.
D) 6
If the system of linear equations x + 2ay + az = 0 ; x + 3by + bz = 0 ; x + 4cy + cz = 0 has a non-zero solution then a,b,c A) satisfy a+2b+3c =0 C) are in G.P
77.
C) 5
If
B) 2
C) 3
D) 4
are the +ve real numbers where product is a fixed number c, then the minimum value of
a1 + a 2 + a 3 + ........ a n -1 +2an is 1/ n
A) n 2c 123
B) n 1 c1/ n
C) 2nc1/ n
D)
1/ n
n 1 2c
81.
IIT- MATHS If the sum of the first 2n terms of the A.P. 2,5,8...... is equal to the sum of the first n terms of the A.P. 57,59,61....... Then n equals A) 10
82.
B) 12
B) H.P
C) G.P
D) none of these
The harmonic mean of 2 numbers is 4. Their arithmetic mean is A and geometric mean is G. If G satisfies 2A + G2 = 27, the numbers are A) 1,13
84.
D) 13
If the lines a1y +b1x - a1b1 = 0, a2 y + b2 x = a2 b2 cut the co-ordinate axes in co cyclic points then a1, b1, b2 a2 may be in A) A.P
83.
C) 11
B) 9,12
C) 3,6
D) 4,8
If a1 , a 2 , a 3 ,.......a n are in Arithmetic series with common difference 'd' .The value of sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + .......+ cosec an-1 cosec an ) is ..... A) sec a1 sec an
85.
B) cos eca1 cos ecan C) tan a1 tan an
D) cot a1 cot an
The first term of an A.P. of consecutive integers is p2 +1. The sum of (2p+1) term of this series can be expressed as: A) (p+1)2
86.
B) (2p + 1) (p + 1)2
The sum of the first n terms of the series A) 2n - n + 1
87.
B) 1 - 2-n
D) p3 + (p + 1)3
1 3 7 15 + + + + ....... = 2 4 8 16
C) n + 2-n - 1
D) 2n - 1
Let a, b, c form a G.P. of common ratio r with 0 < r < 1. If a, 2b and 3c form an A.P., then r equals : A) 1/2
88.
C) (p + 1)3
B) 1/3
C) 2/3
D) None of these
The geometric and harmonic means of two numbers x1 and x2 are 18 and 16
8 respectively. The 13
value of | x1 - x 2 | is A) 5 89.
B) 10
l 2 a2 B) 2S a l
l 2 a2 C) 2 S a l
D) None of these
If ax3 + bx2 + cx + d is divisible by ax2 + c then a, b, c, d are in A) A.P.
91.
D) 20
If S denotes the sum of first n terms of the A.P. a + (a + d) + (a + 2d) +........ whose last term is l, then the common difference 'd' of the A.P. is la A) n
90.
C)15
B) G.P.
The sum of A)
x 2
n2
x + 2
n-1
x 1
n
+ x + 2
C) H.P. n-2
n-3
D) None of these
x +1 + x + 2 x +1 B) x 2
n 1
2
+ ....... + x +1
x 1
n-1
is equal to :
n 1
124
ARTHMETIC PROGESSION n n C) x 2 x 1 92.
Given two numbers a and b, Let A denote the single A.M. and S denote the sum of n A.M.'s between a and b, then A) n, a, b
C) n, a c
b
B) G.P
C) H.P
A)
n digits
4 10n 1 9
B)
4 102n 1 9
C)
2 4 10n 1 9
Given that 0 < x < ?/4, ?/4 < y < ?/2 and
tan
2k x cot 2k y
k=0
1 B) 1 1 1 p q pq
The sum of the first n terms of the series
A)
97.
6n n 1
B)
9n n 1
If In =
B) log 34
π
k cot 2k y= q
then
k=0
C) p + q + pq
D) p+q+pq
3 5 7 +... 2 + 2 2 + 2 1 1 +2 1 + 22 + 32
C)
12n n 1
D)
15n n 1
C) 1 - log3 4
D) log 3 0.25
1 - sin2nx dx, then I1 , I 2 , I3 . .. . are in 1 - cos2x
0
B) G.P
C) H.P
D) none of these
For any odd integer n 1 n3 - (n-1)3 + . . . .+ (-1)n-1 13 is equal to 2
A)
125
k=0
-1
A) A.P 99.
k tan 2k x= P
4 n 10 2 9
1- x + 2 and log9 (4.3x -1 ) are in A.P, then x is equal to If 1, log 9 3
A) log 4 3 98.
-1
D)
is
1 1 1 A) p q pq
96.
D) none of these
(666... . 6)2 + (88.....8) is equal to n digits
95.
D) n
1 1 1 If x > 1 and , , are in G.P, then a, b, c are in x x x
A) A.P 94.
S depends on : A
B) n, b a
93.
D)None of the above
n 1 2n 1 4
n 2 2n 1 B) 4
2
C)
n 1 2n 1 4
D)
n 1 2n 1 4
2
IIT- MATHS 1 3 7 15 + + + + . . . . is 2 4 8 16
100. The sum of first n terms of the series A) 2n - 1
B) 1 - 2-n
C) 2-n - n + 1
D) 2-n + n -1
101. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation A) x2 - 18x - 16 =0
B) x2 - 18x + 16 =0
C) x2 + 18x - 16 =0
D) x2 + 18x + 16 =0
n n +1 102. The sum of first n terms of the series 1 +2.2 + 3 + 2.4 + 5 + 2.6 + .. . . is 2 2
2
2
2
2
2
2
when n
is even. When n is odd the sum is n n 1 A) 2
2
B)
103. If in H.P and f(k) = A) A.P
C)
D)
then B) G.P
are in C) H.P
D) none of these
104. If cos (x - y), cosx, cos (x +y) are in H.P then cosx sec y/2 is equal to A) ±
B)
C) ±2
D) none of these
105. 0.2 + 0.22 + 0.222 + . . . . . to n terms is equal to A) 106. Let
B) nbe the roots of x2 - x + p = 0 and
C)
D)
be the roots of x2 - 4x + q = 0 If
,
are
in G.P. then integral values of p and q respectively are : A) -2, -32
B) -2,3
C) -6, 3
107. If a,b,c and d are in G.P., A)-1 108. If
B) 2 , b,
A) A.P.
are in A.P. then B) G.P.
D) -6, -32 is
C) 0
D) 1
are in : C) H.P.
D) None of these
C) a + d > b +c
D) none of the above
109. If a,b,c, d are in H.P. then : A) a + b > c +d
B) a +c > b +d
126
ARTHMETIC PROGESSION 110. Consider an A.P., with first term 'a' and common difference 'd'. Let terms. If
is independent of x, then
A) a = 2d
B) a =d
C) 2a = d
denote the sum of first k
D) None of these
111. A G.P. consists of an even number of terms, if the sum of all the terms is 5 times the sum of the terms occupying odd places, the common ratio will be equal to A) 2
B) 3
C)4
112. The value of x, for which
A) 113. The coefficient of
D)5 and 1 are in A.P. is
B)
C)
D)
in the polynomial given by is
A) 5511 114. If which
B) 5151
C) 1515
D) 1155
are three consecutive terms of a G.P. with common ratio r, the value of r for holds is given by
A) 1< r < 3
B) -3 < r < -1
C) r >3 or r <1
D) none of the above
115. If x1 x2 x3 as well as y1 y2 y3 are in G.P with some common ratio then (x1y1) (x2y2) (x3y3) A) lie on a straight line
B) lie on an ellipse
C) lie on circle
D) are vertices of triangle
116. Let f(x) be a polynomial function of second degree if f(1)=f(-1) and a, b, c are in A.P then f--1(a), f1(b), f1(c) are in A) Arithmetic - Geometric progression C) G.P
127
B) A. P D) H.P
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
If tan-1x, tan-1y, tan-1z are in A.P and x, y, z are also in A.P., (y being not equal to 0, 1 or -1), then A) x, y, z are in G.P. C) x = y = 1
2.
B) x, y, z are in H.P. D) (x-y)2 + (y-z)2 + (z-x)2 = 0
If d, e, f are in G.P. and the two quadratic equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root, then A)
are in H.P..
B)
C) 2dbf = aef + cde 3.
4.
A) 1 -
: -2 : 1 +
B) 1 : -
:-
C) 1 -
:
D) 1 +
:-2:1-
If a, b, c, d are distinct positive numbers then the inequality an + an > bn + cn holds for every positive integer n provided a, b, c, d are in
9.
D) increasing progression
B) Tpq = p+q
C) Tp+q > Tpq
D) Tpq > Tp+q
B) a + c = b
C) a
b
c
D) ac = b2
If a, b, c be three unequal positive quantities in H.P., then A)
8.
C) H.P
If the first and (2n-1) th terms of an A.P., a G.P. and a H.P., are equal and their nth terms are a, b and c respectively, then A) a = b = c
7.
B) G.P
The pth term Tp of an H.P is q (p+q) and qth term Tq is p (p+q) when p > 1, q >1 then A) Tp+q = pq
6.
D) b2 df = ace2
If three unequal numbers p, q, r are in H.P., and their squares are in A.P., then the ratio p : q : r is
A) A.P 5.
are in G.P..
B)
C)
The sum of n terms of the series
is
A)
B)
C)
D)
Given that 0 < x <
/4 and
/4 < y <
D)
/2 and
= a,
, then
128
ARTHMETIC PROGESSION
10.
A)
B) a+b-ab
C)
D)
If a, b, c are in H.P., then the value of
is
A)
B)
C)
D) none of these
A)
B)
C)
D)
11.
12.
If 1, logyx, logzy, - 15 logxz are in A.P., then A) z3 = x C) z -3 = y
13.
n2
B) x = y-1 D) x = y-1 = z3
I f b1, b2, b3 (b1 > 0) are three successive terms of a G.P. with common ratio r, the value of r for
which the inequality b3 > 4b2 - 3b1 holds is given by A) r > 3 C) r = 3.5 14.
B) r < 1 D) r = 5.2
If logx a, ax/2 and logb x are in GP., then x is equal to A) loga (logba) C) -loga (loga b)
15.
B) loga (logea) - loge (logeb) D) loga (loge b) - loga (logea)
If a, b, c are in H.P., then A)
are in H.P.,
B) C) aD)
129
are in G.P.. are in H.P.,
IIT- MATHS 16.
If A) a = 1/2
17.
then B) b = 8/3
C) c = 9/2
D) e = 0
I f 1, log9 (31-x+2) and log3(4.3x-1) are in A.P., then x is equal to
A) log43
B) log3 4
C) 1 - log3 4
D) log3 (0.75)
130
ARTHMETIC PROGESSION
131
IIT- MATHS
3
BASIC TRIGONOMETERY
132
BASIC TRIGONOMETRY
DEFINITION OF TRIGONOMETRIC RATIOS Let a revolving line OP starts from OM and revolves into the position OP, thus tracing out the angle MOP. Draw PM perpendicular to the initial line OM. In the triangle MOP, OP is the hypotenuse, PM is the perpendicular, and OM is the base. The trigonometrical ratios, or functions, of the angle MOP are defined as follows.
Perp. MP , i.e., , is called the Sine of the angle AOP; Hyp. OP Base OM , i.e., , is called the Cosine of the angle AOP; Hyp. OP MP Perp. , i.e., , is called the Tangent of the angle AOP; OM Base
P
O
M
Base OM , i.e., , is called the Cotangent of the angle AOP; Perp. MP OP Hyp. , i.e., , is called the Secant of the angle AOP; OM Base
Hyp. OP , i.e., , is called the Cosecant of the angle AOP; Perp. MP
FUNDAMENTAL RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS OF AN ANGLE It is clear from the definitions that if one of the trigonometric ratios of an angle is known, the numerical magnitude of each of the others is also known. Let the angle MOP (fig. of previous article) be , then
133
sin2 + cos2 = 1 or
sin2 =1 –cos2 or cos2 = 1 – sin2
1 + tan2 = sec2 or sec2 – tan2 = 1
1 + cot2 = cosec2 or cosec2 – cot2 = 1
tan =
sin . cosec = tan . cot = cos . sec = 1
It is possible to express a trigonometrical ratio in terms of any one of the other ratios:
sin cos and cot cos sin
IIT- MATHS 1
e.g. sin
2
, cos
1 cot
cos ec 1 cot 2 , sec
cot 2
, tan
1 cot
1 cot
1 cot 2 cot
i.e., all trigonometrical functions have been expressed in terms of cot . This is left as an exercise for you to derive other results of this type.
SIGNS OF TRIGONOMETRIC RATIOS Tracing of the changes in the sign and magnitude of the trigonometrical ratios of an angle, as the angle increases from 0° to 360°. Let the revolving line OP be of constant length a. When it coincides with OA, the length OM1 is equal to a; and, when it coincides with OB, the point M1 coincides with O and OM1 vanishes. Also, as the revolving line turns from OA to OB, the distance OM1 decreases from a to zero.
Whilst the revolving line is in the second quadrant and is revolving from OB to OA , the distance OM2 is negative and increases numerically from 0 to a [i.e., it decreases algebraically from 0 to –a].
B
P1
P2
A
M2
M1 M3
O
P3
M4
A
P4 B
In the third quadrant, the distance OM3 increases algebraically from –a to 0; and, in the fourth quadrant, the distance OM4 increases from 0 to a. In the first quadrant, the length M1P1 increases from 0 to a; in the second quadrant, M2P2 decreases from a to 0; in the third quadrant, M3P3 decreases algebraically from 0 to –a; whilst in the fourth quadrant M4 P4 increases algebraically from –a to 0. Therefore it is clear that
134
BASIC TRIGONOMETRY B i
i
A
A
O i
i
B
PERIODS OF THE TRIGONOMETRICAL FUNCTIONS As an angle increases from 0 to 2 radians. i.e., whilst the revolving line makes a complete revolution, its sine first increases from 0 to 1, then decreases from 1 to – 1, and finally increases from –1 to 0, and thus the sine goes through all its changes, returning to its original value. Similarly, as the angle increases from 2 radians to 4 radians, the sine goes through the same series of changes. Also, the sines of any two angles which differ by four right angles, i.e., 2 radians, are same. This is expressed by saying that the period of the sine function is 2 . Similarly, the cosine, secant, and cosecant go through all their changes as the angle increases by 2 . The tangent, however, goes through all its changes as the angle increases from 0 to radians, i.e., whilst the revolving line turns through two right angles. Similarly for the cotangent. The period of the sine, cosine, secant and cosecant is therefore 2 radians; the period of the tangent and cotangent is radians. Since the values of the trigonometrical functions repeat over and over again as the angle increases, they are called periodic functions.
GRAPHS OF THE TRIGONOMETRIC RATIOS The variations in the values of the trigonometric ratios may be graphically represented to the eye by means of curves constructed in the following manner.
135
IIT- MATHS Y
Sine-Graph:
/ 2
1 O /2
2
X
–1
Y 1 O
/ 2
Cosine-Graph :
/ 2
3 / 2 2
X
–1
Y
Tangent-Graph: / 2
O
3 / 2X
/2
Cosecant-Graph:
Y
2 X
3 2
1 O
–1
2
2
X
The secant-graph and the cotangent-graph are left as an exercise to the students.
136
BASIC TRIGONOMETRY
TRIGONOMETRIC RATIOS OF SOME ANGLES Angle
0°
30°
45°
60°
1 2
1 2 1 2
3 2 1 2
3 2 1 2
3
3
1 3 2 3
1 3 2 3
3 2 1 3 3
2 2 3
2
90°
120°
135° 1 2 1 2
150°
180°
1 2 3 2 1 3 3
2 2
2 3
Note: Later on we shall learn that infact tan(90°–) = and tan(90°+) = – etc. ]
TRIGONOMETRIC RATIOS FOR AN ANGLE OF ANY MAGNITUDE Complementary Angles Two angles are said to be complementary when their sum is equal to a right angle. Thus any angle and the angle 90° – are complementary..
Supplementary angles Two angles are said to be supplementary when their sum is equal to two right angles, i.e., the supplement of any angle is 180° – .
Allied or Related Angles 1 1 n and n , where n is any integer, are known as allied or related angles. 2 2 The trigonometric functions of these angles can be expressed as trigonometric functions of ,
The angles
with either a plus or a minus sign. The following working rules can be used in determining these functions. 1. Assuming that 0 < < 90°, note the quadrant in which the given angle lies. The result has a plus or minus sign according as the given function is positive or negative in that quadrant. 2. If n is even, the result contains the same trigonometric function as the given function. But if n is odd, the result contains the corresponding cofunction, i.e., sine becomes cosine, tangent becomes cotangent, secant becomes cosecant and vice-versa.
137
IIT- MATHS P2
P1 90°–
+ve
A –ve O
+ve
+ve A –ve
–ve
P3
P4
Here
angle AOP1 = angle AOP4 (measured clockwise) = – angle AOP2 = 180° – angle OP1A = 90° – and similarly the other angles. It is clear from the figure that when equals
sin
cos
tan
cot
sec
cosec
–
–sin
cos
–tan
–cot
sec
–cosec
90°–
cos
sin
cot
tan
cosec
sec
90° +
cos
–sin
–cot
–tan
–cosec
sec
180° –
sin
–cos
–tan
–cot
–sec
cosec
180° +
–sin
–cos
tan
cot
–sec
–cosec
360° –
–sin
cos
–tan
–cot
sec
–cosec
360° +
sin
cos
tan
cot
sec
cosec
TRIGONOMETRIC RATIOS OF COMPOUND ANGLES An angle made up of the algebraic sum of two or more angles is called compound angle. Some formulae and results regarding compound angles:
sin (A + B) = sin A cosB + cosA sinB
sin(A – B) = sinA cosB – cos A sinB
cos (A + B) = cosA cos B – sinA sin B
cos(A – B) = cosA cosB + sin A sin B.
tan A tan B 1 tan A tan(A + B) = 1 tan A tan B , tan (45° + A) = 1 tan A
tan A tan B tan(A–B) = 1 tan A tan B ,
cot A cot B 1 cot A cot B 1 cot (A + B) = cot A cot B , cot (A – B) = cot B cot A
sin(A + B) sin(A – B) = sin2A – sin2B = cos2B – cos2A
1 tan A tan (45° – A) = 1 tan A
138
BASIC TRIGONOMETRY cos(A + B) cos(A – B) = cos2A – sin2B = cos2B –sin2A.
tan A tan B tan C tan A tan B tan C tan (A + B + C) = 1 tan A tan B tan B tan C tan C tan A
.MAXIMUM AND MINIMUM VALUES OF acos + bsin Let a = rsin , b = rcos so that r =
a2 b2
also, acos + bsin = r(cos sin + sin cos ) = rsin ( ) Now the maximum and minimum values of sin( ) are 1 & – 1 respectively.. Hence – r rsin( ) r
– a 2 b 2 a cos b sin a 2 b 2
Hence the maximum value =
a2 b2
and minimum value is a 2 b 2 .
TRIGONOMETRIC RATIOS OF MULTIPLES OF AN ANGLE 2 tan A 1 tan 2 A
sin2A = 2sinA cosA =
1 tan 2 A cos2A = cos A –sin A = 1 – 2 sin A = 2 cos A–1 = , 1 tan 2 A 2
2
2
2
1 + cos2A = 2cos2A, 1 – cos2A = 2sin2A 2 tan A 1 tan 2 A
tan2A
sin3A = 3sinA – 4sin3A = 4sin(60° – A) sinA sin(60° + A)
cos3A = 4 cos3A – 3cosA = 4cos(60°–A) cosA(cos60°+A)
3 tan A tan 3 A tan 3A = tan(60°–A) tanA tan(60°+A) 1 3 tan 2 A
PRODUCT OF SINES/COSINES IN TERM OF SUMS
2 sinA cosB = sin (A + B) – sin (A – B)
2 cos A sin B = sin (A + B) – sin (A – B)
2 cos A cos B = cos (A + B) + cos (A – B)
2 sin A sin B = cos (A – B) – cos (A + B)
SUM OF SINES/COSINES IN TERM OF PRODUCTS 139
sinC + sinD = 2sin
CD CD cos 2 2
IIT- MATHS
sinC – sinD = 2 cos
CD CD sin 2 2
cosC + cosD = 2cos
CD CD cos 2 2
cosC – cosD = –2sin
CD CD sin 2 2
tanA + tanB =
sin ( A B) sin ( A B) , tanA – tanB = cos A cos B cos A cos B
TRIGONOMETRIC RATIO OF SUBMULTIPLE OF AN ANGLE
| sin
A A cos | 1 sin A 2 2
or sin
A 3 A A ve, if 2 n 2n cos 1 sin A 4 2 4 2 2 ve, otherwise
A A cos | 1 sin A 2 2
| sin
or
A A sin cos 2 2
tan
A 5 2n ve, if 2n 1 sin A 4 2 4 ve, otherwise
A tan 2 A 1 1 2 tan A
The ambiguities of signs are removed by locating the quadrant in which
A lies or one can use the 2
following figure. sin
A A cos is ve 2 2
sin sin
sin
A A cos is ve 2 2
4
A A cos is ve 2 2 sin
sin
A A cos is ve 2 2
|a cosA + bsinA| Also cosA sinA =
sin
sin
A A cos is ve 2 2
A A cos is ve 2 2
A A cos is ve 2 2
A A cos is ve 2 2
a2 b2 2 sin 4 A =
2 cos A 4 140
BASIC TRIGONOMETRY
IDENTITIES A trigonometric equation is an identity if it is true for all values of the angle or angles involved. A given identity may be established by (i) reducing either side to the other one, or (ii) reducing each side to the same expression, or (iii) any convenient modification of the methods given in (i) & (ii).
CONDITIONAL IDENTITIES When the angles A, B and C satisfy a given relation, many interesting identities can be established connecting the trigonometric functions of these angles. In providing these identities, we require the properties of complementary and supplementary angles. For example, if A + B + C = , then
sin (B + C) = sinA, cosB = –cos (C + A)
cos (A + B) = –cosC, sinC = sin(A + B)
tan (C + A) = tanB, cotA = –cot(B + C)
cos
sin
CA B A BC cos , sin cos 2 2 2 2
tan
BC A B CA cot , tan cot 2 2 2 2
AB C C AB sin , cos sin 2 2 2 2
Some important identities: If A, B, C are angles of a triangle (or A + B + C = ):
141
tanA + tanB + tanC = tanA tanB tanC
cotA cotB + cotB cotC + cotC cotA = 1
tan
A B B C C A tan + tan tan + tan tan 1 2 2 2 2 2 2
cot
A B C A B C cot cot cot cot cot 2 2 2 2 2 2
sin2A + sin2B + sin2C = 4sinA sinB sinC
cos2A + cos2B + cos2C = –1 – 4cosA cosB cosC
sinA + sinB + sinC = 4cos
cosA + cosB + cosC = 1 + 4 sin
A B C cos cos 2 2 2 A B C sin sin 2 2 2
IIT- MATHS
TWO SIMPLE TRIGONOMETRICAL SERIES
n 2 (n 1) sin sin 2 2 sin + sin( )+sin ( 2 ) + ... + sin{ ( n 1) }= sin 2
n 2 (n 1) cos sin 2 2 cos +cos( )+cos( 2 )+ ... +cos{ ( n 1) }= sin 2
142
BASIC TRIGONOMETRY
WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01
1
sin 3 cos3 is equal to sin cos sin cos
(a) sin 2
(b) cos 2
(c) sin cos
(d) sin 2
Ans : (C) Solution : We can write the given expression as
sin cos sin 2 cos 2 sin cos sin 3 cos3 1 1 sin cos sin cos =1- ( 1 sin cos ) = sin cos ILLUSTRATION : 02 If cos
(a) 1
2cos 1 0 , , then tan 2 cos 2
(b)
2
(c)
3
(d)
tan
is equal to 2
1 Z 3
Ans : (C) Solution : From the given relation we have
2cos 1 2 cos 2 cos 1 1 cos = 1 2 cos 2 cos
or
2cos 2 1 cos 2 2cos 2 2 2 cos 1 2sin 2 2
or
cos 2 2 cos 2 2 1 2sin 2 ........... (1) 2
cos 2 1 2sin 2 cos2 2 2 2 1 cos 2 1 ==> 2 1 2sin 2 1 2sin 2 2 2
143
IIT- MATHS 3sin 2 2 sin 2 or 2 1 2sin 2 2
............ (2)
From (1) and (2) we get tan 2 3 2 2 tan 3 tan ==> 2 2 tan 2 ILLUSTRATION : 03 If Pn cos n sin n , then Pn Pn 2 kPn 4 where (a) k 1
(b) k sin 2 cos 2
(c) k sin 2
(d) k cos 2
Ans : (B) Solution : We have Pn Pn 2 cos n sin n cos n 2 sin n 2 = cos n 2 cos 2 1 sin n 2 sin 2 1 = sin 2 cos n 2 cos 2 sin n 2 = sin 2 cos 2 cos n 4 sin n 4 = sin 2 cos 2 Pn 4 kPn 4 where k = sin 2 cos 2
ILLUSTRATION : 04 If
(a)
sin 4 cos4 sin 8 cos8 , then + is equal to a2 b2 a3 b3 1 a b
1 (b) a b 2
1
(c)
3
a b
(d) a b
Ans : (C) Solution :
sin 4 cos 4 k Let a2 b2
144
BASIC TRIGONOMETRY k
then 1 sin 2 cos 2 a k b k ==>
1
a b
2
sin 8 cos8 k 2 a 4 k 2b 4 1 2 so that a 3 b3 a3 b3 k a b a b 3
ILLUSTRATION : 05 sin 3 A sin 3 1200 A sin 3 240 0 A is equal to
(a) (9/4) sinA
(b) (3/4) sin3A(c) - (3/4) sin3A
(d) 0
Ans : (C) Solution : The given expression is equal to 1 0 0 0 0 3sin A sin 3 A 3sin 120 A sin 360 3 A 3sin 240 A sin 720 3 A 4 3 1 0 0 0 0 = sin A sin 180 60 A sin 180 60 A sin 3 A sin 3 A sin 3 A 4 4
=
3 3 0 0 sin A sin 60 A sin 60 A sin 3 A 4 4
3 3 0 = sin A 2 cos 60 sin A sin 3 A 4 4 3 = sin 3 A 4
ILLUSTRATION : 06 If tan , tan , tan are the roots of the equation x 3 px 2 r 0 then the value of (1 + tan?) (1 + tan2 ) (1 + tan2 ) is equal to 2
(a) p r (b) 1 p r
2
(c) 1 p r
2
Ans : (B) Solution : From the given equation we have
tan tan 0 and
tan p
tan tan tan r
so that 1 tan 2 1 tan 2 1 tan 2
145
(d) None of these
IIT- MATHS = 1 tan 2 tan 2 tan 2 tan 2 tan 2 tan 2 =1 +
tan
2
2 tan tan tan tan
2
2 tan tan tan tan tan 2 tan 2 tan 2 = 1 + p 2 2 pr r 2 = 1 p r
2
ILLUSTRATION : 07 If A and B be acute positive angles satisfying 3sin 2 A 2sin 2 B 1, 3sin 2 A 2sin 2B 0 then (a) B
A 4 2
(b) B
A 4 2
(c) B
A 2 4
(d) A
B 4 2
Ans : (A) Solution : From the given relations we have 3 Sin2B = sin 2 A 2
and 3sin 2 A 1 2sin 2 cos 2 B 3 sin 2 A so that tan 2 B 2 cot A or 1- tan2BtanA =0 3sin 2 A ==> A+2B =
A ==> B 2 4 2
ILLUSTRATION : 08 If
(a)
a c cos x cos x cos x 2 cos x 3 then is equal to b d a b c d a d
(b)
c d
(c)
b c
(d)
d a
Ans : (C) Solution :
146
BASIC TRIGONOMETRY cos x cos x 2 2 cos x cos a c b = b d cos x cos x 3 = 2 cos x 2 cos c
ILLUSTRATION : 09 sec 2
4 xy
x y
(a) x y 0
2
is true if and only if (c) x y
(b) x y , x 0
(d) x 0, y 0
Ans : (C) Solution : 2
2
Since sec 2 1 so 4 xy x y x y 0 x y Which is true if and only if x y 0 , because for x y 0 , sec 2 becomes indeterminant.
ILLUSTRATION : 10 cos A sin C If cos B sin C , then is equals
(a) tan
A B tan A B tan C
(b) tan
A B tan A B tan C
(c) tan
A B tan A B sin C
(d) tan
A B tan A B cos C
2
2
2
2
2
2
Ans : (B) Solution : cos A sin C cos B sin C cos A cos B sin C sin C ==> cos A cos B sin C sin C
==> tan
A B AB tan cot C tan 2 2
==> tan tan
147
A B A B tan tan C 2 2
2
2
2
2
2
IIT- MATHS
SECTION A SINGLE ANSWER TYPE QUESTIONS 1.
If cos20° - sin 20° = P the cos40° is equal to A) - p
2.
2 p2
B) p 2 p 2
B) 0
B) 8 cos2A
B) p2 = q (q+2)
The period of A) 2
9.
B) -1/2
D) f(x) 2
is
C) 1/2
D) 1
B) 1
C) 3
D) 1
C) /2
D) /4
sin2x is 3cos4x
B)
The period of sin (x + 4x + 9x+ ......... n2x ) is
4 C) n n 1 2n 1
3 D) n n 1 2n 1
C) 1/4 and 4
D) 1/6 and 6
The value of tan3 cot can't lie between A) 1/2 and 2
11.
C) 2 < f(x) < 1
π 3π 5π 7π 9π + cos + cos + cos + cos 11 11 11 11 11
12 6 A) n 2n 1 n 1 B) n 2n 1 n 1 10.
D) p2 =q2 (q2 - 4)
Maximum value of 2sin2x - cos2x is A) 2
8.
D) 4cos2A
C) p2 + q2 = 2q
B) f(x) = 1
The value of cos A) 0
7.
C) 1/8 cos2A
If f(x) = cos2x + sec2x its value always is A) f(x) < 1
6.
D) 1/2
If sec and cosec are the roots of x2-px+q=0 then A) p2 = q (q-2)
5.
C) 1
sin 2 3A cos 2 3A is equal to sin 2 A cos 2 A A) cos2A
4.
D) none of these
If tan2 - 2 tan2 +1 then cos2 + sin2 is equal to A) -1
3.
C) p 2 p 2
B) 1/3 and 3
The maximum value of cos 1, cos 2 cos 3.......cos n under the restriction 0 1 , 2 , 3 ,......... n
A) 1/2n/2
and cot 1, cot 2 ....... Cos n = 1 is 2
B) 1/2n
C) -1/2n
D) 1
148
BASIC TRIGONOMETRY π 5π 12. If tan , x and tan 9 18 A) 2x = y 13.
B) x > y
tan
D) ± 1/5
B) (0,1]
C) [1/3 )
D) [1 )
C) 2
D) 4
B) 2
C) 3
D) 4
If 3sin2A + 2sin2B = 1 and 3sin2A - 2sin2B = 0 where A and B are acute angles, then (A + 2B) is equal to B) /4
C) /2
D) /6
The number of solutions of the equation |sinx| = |cos3x| in [-2 , 2 ] is B) 28
1 1 2 + + 4 4 2 sec α cosec α sec α + cosec 2 α
B) 1
C) 24
D) 30
C) sin2
D) cos2
=
If sin α + cosec α = 2 then sinn α + cosecn α = B) 2n
C) 0
D) 2
If x2 = 1 - tan2 θ then tan3 θ cosec θ + sin θ = A)
149
3
If tan x - tan2x = 1 then the value of tan4x - 2tan2x - tan2x + 2tanx + 1 is
A) n 23.
C) ± 3/4
B) 1
A) 0 22.
D)
The no. of roots of equation x sinx = 1 in the interval 0 < x < 2 π is
A) 32 21.
1 2 2 2
Let ABC be a acute angled triangle such that A = π /3 and cot B cos C = P. The possible values of P will be
A) /3 20.
C) 1
B) ± 1/2
A) 1 19.
D)
If sin ( π cos θ ) = cos ( π sin θ ), then sin2 θ is equal to
A) 0 18.
C) 1/8
B) 1/ 3
A) (0, 1/3] 17.
D) x = 2y
2π π 2π π - 3 tan tan - tan is equal to 5 15 5 15
A) ± 1/4 16.
are also in A.P. then
C) x = y
B) cos /8
A) 3 15.
π 7π , y and tan 9 18
π 3π 5π 7π 1 + cos 1 + cos 1 + cos 1+ cos is equal to 8 8 8 8
A) 1/2
14.
are in A.P. and tan
2 1/ 2
2 x
B)
x
2
3/ 2
2
3/ 2
C) 2 x 2
D)
2 5/ 2
2 x
IIT- MATHS 24.
The value of the expression 1 A) 1
1- cos 2 A sinA 1 + cosA + is equal to 1 + cosA 1 - cosA sinA
B) tanA
C) cosA
D) sinA
n
25.
If
sinθ i=1
i
= n , then cos θ 1 + cos θ 2 + cos θ 3 +......+cos θ n =
A) n 26.
B) n-1
For 0 < <
π if , x = 2
A) xyz = xz + y 27.
D) zero
¥
¥
¥
cos 2n φ, y = sin 2n φ, Z =
n =0
n=0
n =0
B) xyz = xy+z
cos 2n φsin 2n φ , then
C) xyz = x+y+z
D) xy2 = y2+x
C) m [-2, 2]
1 D) m , 1 4
If sin6 α + cos6 a = m, then 1 1 A) m , 6 6
28.
C) 2n
B) m 0, 1
The smallest value of θ in the first quadrant which satisfies the equation exp {(1+cos2 θ +cos4 θ +cos6 θ +..... )loge16} = 256 is A) /4
29.
If x =
B) /6
B) cos
B) zero
C) x
D) 13
B) a2 + 2ab - b2 =0
C) a2 + 2ac + b2=0
D) a2 + 2ac - b2=0
If (1-sinA)(1-sinB)(1-sinC)=(1+sinA)(1+sinB)(1+sinC) then each equal side is equal to B) 1
C) sinAsinBsinC
D) cosAcosBcosC
If cosec θ + cot θ = 11/2, then tan θ is equal to A) 21/22
35.
D) None
If sin and cos are roots of equation ax2 + bx + c =0 then the relation-ship between a, b, c is
A) zero 34.
C) -1
B) cosx
A) a2 + 2ab + b2 =0 33.
D) None
3(sinx - cosx)4 + 6(sinx + cosx)2 + 4 sin 6 x + cos 6 x is A) sinx
32.
C) x
tan 2 θ 1 If 0 < θ < π /2, the value of the expression tanθ -1 + tanθ 1 - tanθ - secθcosecθ is
A) 1 31.
D) /2
2sinθ 1 - cosθ + sinθ then = 1 + cosθ + sinθ 1 + sinθ
A) sin 30.
C) /3
1 + tanAtanB
B) 15/16 2
+ tanA - tanB
C) 44/117 2 1/ 2
D) 117/44
is equal to
150
BASIC TRIGONOMETRY A) tan2A + tan2B 36.
B) cos2Acos2B
C) secAsecB
D) tanAtanB
If tan2 α tan2 β + tan2 β tan2 γ + tan2 γ tan2 α + 2tan2 α tan2 β tan2 γ = 1 then the value of sin2 α + sin2 β + sin2 γ is A) 0
37.
4 3 m 2 1
2
B)
4
1 3 3
151
D)
C) k <
1 9
2
4
D) k > 1/3
C) 1/2
D) 1
C) tan c tan
D) tan c cot
B) 2
C) 1/ 2
D) 2
B) 8
C) 10
D) 12
The value of sin π /14 sin 3 π /14 sin 5 π /14 sin 7 π /4 sin 9 π /14 sin11 1 π /4 sin
If
sin α + β cos α - β
B) 1/16 =
1- m 1+ m
C) 1/64
π π then tan - α tan - β 4 4
B) 1 + m
13π is equal to 14
D) 1/32 is equal to
C) 2m
D) 2 + m
If A and B are acute angles such that A+B and A-B satisfy the equation tan2 - 4tan +1=0 then (A, B) be B) ( /4, /6)
C) ( /6 /4)
D) ( /2, /6)
The equation cos4x - sin4x + cos2x + α 2 + a = 0 will have at least one solution if A) -2 2
47.
4
4 3 m 2 1
The number of integral values of k for which the equation 7cosx + 5sinx = 2k+1 has a solution is
A) ( /6, /6) 46.
2
If sec( α - β ), sec α , sec ( α + β ) are in A.P, then cos α sec β /2 be
A) m 45.
3 3
B) cot c tan
A) 1 44.
C)
3 4 m 2 1
A+B A -B If cos A/ cos B = sin (c - θ )/ sin (c + θ ) then tan tan is 2 2
A) 4 43.
1
B) 3/4
A) 2 42.
4
2
If cos6 α + sin6 α + k sin22 α = 1 then k is equal to
A) cot c cot 41.
4 3 m 2 1
B) k
A) 1/4 40.
D) 1/2
If A, B, C are acute positive angles such that A +B + C = and cot A cot B cot C = k then A) k
39.
C) -1
If sin α + cos α = m then sin6 α + cos6 α is equal to A)
38.
B) 1
B) -3 1
C) 2 1
D) 1 2
The number of points inside or on the circle x2+y2=4 satisfying tan4x + cot4x + 1 = 3sin2y
IIT- MATHS A) one 48.
B) two
C) four
D) infinite
I f tan θ 1, tan θ 2, tan θ 3, tan θ 4 are roots of the equation x4-x3 sin 2 + x2 cos2 - x cos - sin
= 0 then tan θ1 + θ 2 + θ3 + θ 4 is equal to A) sin 49.
B) cos
C) tan
D) cot
If x = α , β satisfies both the equations cos2x + acosx + b= 0 and sin2x + p sinx + q = 0 then relation between a, b, p and q is A) 1+b+a2 = p2-q-1
50.
B) 2
C) 3
B) cosec A
D) 4
C) sin A
D) tan A
(cosec θ - sin θ ) (sec θ - cos θ ) (tan θ + cot θ ) is equal to A) 4 cos sin
53.
D) a2+b2 = p+q
π sin 2π - Α cos π + A tan - A 2 is equal to π sin - A cos 2π + A sin(π - A) 2 A) cosA
52.
C) 2(b+q) = a2+p2-2
The number of solutions of the equation 1 + x2 + 2x sin (cos-1y) = 0 is A) 1
51.
B) a2+b2 = p2+q2
B) 4 sec tan
C) 4 cosec cot
D) 1
If α, β and λ are variables subject to the relation 2tan α + 3tan β + 6 tan γ = 7, then the minimum value of tan2 α + tan2 β + tan2 γ = A) zero
54.
B) 1
d a a c B) b d c b
B) a=0
If p =
a b
D) None
bc
C) a > 1
D) a > 0
a 2 cos 2 θ + b 2 sin 2 θ + a 2 sin 2 θ + b 2 cos 2 θ then the maximum value of P is
A) a - b
57.
C)
If 0 < A < /2 and sinA+cosA+tanA+cotA+secA+cosecA=7, and sinA and cosA are the roots of the equation 4x2 - 3x + a =0, then A) a=1
56.
D) 3
If a sin2 α + b cos2 β = C, bsin2 β + acos2 β = d and atan α - b tan β , then a2/b2 = A) 1
55.
C) 2
B) b + a
C)
a b 2
D)
a b 2
4 4 sin 8 θ cos 8 θ + = If sin θ + cos θ = 1 , then a3 b3 a b a+b
1
A)
1 8
a b
B)
a b
1 4
C)
3
a b
D)
1 a b 152
BASIC TRIGONOMETRY 58. If A and B are positive acute angles satisfying the equations 3cos2A + 2cos2B = 4 and 3sinA 2cosB = sinB cosA
then A + 2B is equal to
A) /4 59.
a d c a b c d b
B)
B) 2
d a c a b c d b
C)
C)
B) -1/2
2 B) n n 1
B) - 5
B)
13 A1 16
B) tan + tan
If /2 < α < π , π < β <
A)
153
C) tan A + tan B = 1
D) tan A tan B = 0
a2 is equal to b2 D)
b c b d a c a d
3
D) 1/ 3
C) 1
D) -1
C) 2 n 1
D) 2n n 1
C) 5
D) - 1/5
C)
3 13 A 4 16
D)
3 A 1 4
If + = /2 and + γ = then tan equals A) 2 (tan + tan )
69.
B) tan A tan B < 1
If A = cos2 +sin4 , then for all values of A) 1 A 2
68.
D) 12
If 3cosx = 2cos (x - 2y) then tan (x - y) tan y is equal to A) 1/5
67.
C) 6
The period of tan (x + 2x + 3x + ......... + nx) is
A) n n 1 66.
B) 9
The minimum value of cosx. cos (120-x) cos (120+x) is A) -1/4
65.
D) - 13/18
If A + B = π /4 then (tan A +1) (tan B +1) is equal to A) 1
64.
C) 13/18
If a sin2 x + b cos2x = c ; bsin2y + a cos2y = d and a tanx = b tany then
b c d b A) a d c a 63.
B) - 24/25
If A+B+C = (A, B, C > 0) and the angle C is obtuse then A) tan A tan B > 1
62.
D) /2
The minimum value of 9tan2 θ + 4cot2 θ is A) 13
61.
C) /6
If α is the root of 25cos2 θ + 5cos θ - 12 = 0, π /2 < α < π then sin 2 α is equal to A) 24/25
60.
B) /3
21 221
B)
C) tan + 2tan
D) 2tan + tan
3π 15 12 ; sin α = and tan β = the value of sin ( β - α ) is 2 17 5 21 221
C)
171 221
D)
171 221
70.
IIT- MATHS The number of values of x in the interval [0, 5 ] satisfying the equation 3sin x-7sinx+2=0 2
A) 0 71.
B) 5
76.
C) [ /2, ] U [3 /2, 2 ]
D) x [ , 2 ]
If
A π 1- sinA = sinA / 2 - cosA / 2 then 2 - 4
D) 1
B) II, III
could lie in quadrant C) III, IV
D) I, IV
B) 8
C) 10
D) 12
If cosec - sin = m; sec - cos = n then (m2n) 2/3 + (mn2) 2/3 =
If < /8 the value of
B) 1
C) 2
D) - 1
2 + 2 + 2 + 2cos4θ
B) 2cos 2
C) 2cos /2
D) 2sin /2
If sinx + sin2x = 1, then the value of cos2x + cos4x is B) 2
C) 1.5
D) None
6 sin 6 θ + cos 6 θ - 9 sin 4 θ + cos 4 θ + 15 is equal to B) 10
C) 12
D) - 12
If sin + sin2 +sin3 =1, then cos6 - 4cos4 +8cos2 = A) 3
83.
D) 2
Total number of roots of the equation 3cosx = |sinx| belongs to [-2 , 2 ] are
A) -10 82.
C) - 3
B) (0, /2
A) 1 81.
C) -1
3 A) x , , 2 2 2
A) 2 cos 80.
D)
If [cosx] + [sinx + 1] = 0, then value of x satisfying f(x) where xε 0, 2π
A) 0 79.
C) /3
B) 5
A) 6 78.
D) 2cos36°
The value of y for which the equation 4sinx+3cosx=y2 - 6y + 14 has a real solution is/are
A) I, II 77.
B) /2
B) 1
A) 3 75.
C) 2cos18°
If sin x + sin2x = 1 then the value of cos12 x + 3cos10 x + 3cos8x + cos6x - 1 is equal to A) 0
74.
B) 2sin18°
If A and B are acute positive angles satisfying the equation 3sin2A + 2sin2B = 1 and 3sin2A 2sin2B = 0 then A +2B = A) /4
73.
D) 10
If cos A = tan B, cos B = tan C, cos C = tan A, then cos2 A is equal to A) sin 18°
72.
C) 6
B) -3
C) 4
D) -4
If sinx+cosx = a, then |sinx-cosx| = 154
BASIC TRIGONOMETRY A) 84.
B)
2 a2
C)
2a
D)
2 a2
2a
If x = γ sin .cos , y = γ sin .sin , z = γ cos then x2 + y2 + z2 = A) sin2
B) sin2
D) 2
C) cos2 4xy
85.
2
For all real values of x and y, the equation sec θ = x + y 2 is possible when A) x = y
86.
B) x y
sin θ + cos θ = m, then sin6 θ +cos6 θ = A) m R
87.
B) 2 /7
D) -2 /7
B) x ? 4
C) x ? 6
D) None
B) a - b
C) a/b
D) 2 ab
If x = a cos3, y = bsin3 θ , then
a C) x
2/3
2/3
If tan θ =
y b
b y
2/3
x B) b
1
2/3
2/3
b D) x
1
y a
2/3
2/3
a y
1 2/3
1
sinα - cosα , then sinα + cosα
A) sin -cos = 2 sin C) sin = 2 (sin +cos ) 92.
C) - /7
The minimum value of a2tan2? + b2cot2 θ is
x A) a
91.
D) None
If x = sin2 θ + cos2 θ +tan2 θ +cosec2 θ +sec2 θ +cot2 θ , 0, then
A) a + b 90.
C) m
If f(x) = 3(sinx-cosx)4 + 6(sinx+cosx)2 + 4(sin6x+cos6x) and g(13) = π /7 then (gof)x =
A) x =1 89.
D) x = y = 0
2 1 4 - 3 m 2 -1 if 4
B) m R
A) /7 88.
C) x = y 0
B) sin +cos D) None
2 cos
If tan α equals to the integral solution of the inequality 4x2 - 16x + 15 < 0 and cos β equals to the slope of the bisector of the first and quadrant, then sin ( α + β ) sin ( α - β ) is equal to A) 3/5
93.
D) 4/5
B) a2 + (b - c )2
C) c2 + (a - b )2
D) b2 + (a +c)2
For what value of α lying between 0 and π is the inequality sin α cos3 α > sin3 α cos α valid A) (0, /4)
155
C) 2 / 5
1 If maximum value of |a sin2 θ + b sin θ cos θ +c cos2 θ (a+c)| is k/2 then k2 is equal to 2
A) b2 + (a -c)2 94.
B) - 3/5
B) (0, /2)
C) ( /4, /2)
D) (- /4 /4)
IIT- MATHS 95.
The values of a for which the equation cos2x + a sin x = 2a - 7 has a solution A) 2 a 6
96.
B) 1 a 5
B) 7/16
C) (
B) ( / 2, 2 / 3]
D) 4/15
2 ,5 / 6] 3
D) (
5 , ] 6
If ABCD is a cyclic quadrilateral such that 12 tan A - 5 = 0 and 5cos B + 3 = 0, then the quadratic equation whose root are cosC, tanD is A) 39x2 - 16x - 48 = 0 C) 39x2 - 88x + 48 = 0
99.
C) 1/16
Given both θ and are acute angles and sin = 1/2 cos = 1/3 then the value of + belongs to A) ( / 3 / 2]
98.
D) 1 a 6
If sinx cos y = 1/4 and 3 tanx = 4 tany then sin(x+y) A) 5/16
97.
C) a 5
B) 39x2 + 88x + 48 = 0 D) 39x2 + 16x + 48 = 01
If sinA, cos A, tan A are in G.P. then cot6A - cot2A is equal to A) 1
B) -1
C) 1/2
D) -1/2
C) -1/2
D) 0
100. Minimum value of 4x2 - 4x |sin | - cos2 is A) -2
B) -1
101. The number of solution of equation e cos 2 x - e -cos2 x + 4 = 0 in [0, 2 π ] A) 1
B) 2
C) 3
D) none of these
102. sinx + siny = y2 - y + a will have no solution in x and y if a belongs to A)
0, 3
B) 3, 0
C)
1 D) 2 , 4
, 3
103. In an isosceles right angled triangle, a straight line drawn from the mid point of one equal sides to the opposite angle. It divides the angle into two parts and ( π /4 - ), Then tan and tan ( π /4 - ), then tan and tan ( π /4- ) are equal to A)
1 1 , 3 4
B)
1 1 , 4 5
C)
1 1 , 5 6
D)
1 1 , 2 3
104. The equation of sinx (sinx + cosx) = k has real solutions then A) 0 k
1 2 2
C) 0 k 2 3
B) 2 3 k 2 3
D)
1 2 1 2 k 2 2
π π π 105. cosec + θ + x cos θ cot + θ = sin + θ then x = 2 2 2 156
BASIC TRIGONOMETRY A) cot B) sin
C) tan
D) cos
4 3π π - α + sin 4 3π + α - 2 sin 6 + α + sin 6 5π - α = 106. 3 sin 2 2 A) 0
B) 1
C) 3
D) 2
107. (sin α + cosec α )2 + (cos α + sec α )2 = k + tan2 α + cot2 α then k = A) 9
B) 7
C) 5
D) 3
π 108. If x cos α + y sin β = x cos β + y sin β = 2a , 0 < α, β < then 2
A) cos + cos =
C) sin + sin =
4ax 2 x y2
B) cos cos =
4ay 2 x y2
D) sin sin =
109. If x = acos2 θ sin θ , y = a sin2 θ cos θ , then A) a sin
B) a2sin
x
2
B) -2cos
4a 2 x 2 x2 y 2
3
x 2 y2
=
C) a2sin2
3π 110. If π < θ < , then the expression is equal to 2
A) +2sin
+ y2
4a 2 y 2 x2 y 2
1 - cosθ 1 + cosθ + 1+ cosθ 1- cosθ
C) +2sec
111. If cosecA = cosecBcosecC + cotBcotC, then cosecB = A) cosecAcosecC+cotAcotC C) cosecAcosecC ? cotAcotC
157
D) a2(sin2 +cos2 )
B) cosecAcosecC - cotAcotC D) None
D) -2cosec
IIT- MATHS
SECTION - B SINGLE ANSWER TYPE QUESTIONS 1.
For any real x one has A) cos (cos x) > sin (sin x) C) cos (sin (cos x)) > sin (cos (sin x))
B) cos (sin x) > sin (cos x) D) cos (cos (cosx)) > sin (sin (sinx)) 7
2.
2
0
2
0
2
0
I f sec 20 + sec 40 + sec 80 = a and
tan k =1
2
kπ = b then the value of a-b can be expressed as 16
A) sec2 - tan2 B) cos2 - sin2 C) 3(sin2 + cos4 ) - 2(sin6 + cos6 ) D) none of these 3.
If 3 sin β = sin (2 α + β ) then A) [cot + cot ( + )] [cot - 3 cot (2 )] = 6 B) sin = cos ( + ) sin C) 2 sin = sin ( + ) co s D) tan ( + ) = 2 sin
4.
5.
Let Pn (u) be a polynomial in u of degree n. Then, for every positive integer n, sin 2nx is expressible is A) P2n (sinx)
B) P2n (cos x)
C) cos x P2n-1 (sinx)
D) sin x P2n-1 (cos x)
If cos α =
3 5 and cos β = , then 5 13
A) cos ( + ) =
33 65
1 C) sin2 2 65
6.
56 65
D) cos ( - ) =
63 65
The equation sin6x + cos6x = a2 has real solutions if A) a (-1, 1)
7.
B) sin ( + ) =
1 B) a 1, 2
1 1 C) a , 2 2
1 D) a ,1 2
If tan α and tan β are the roots of the equation x2 + px + q = 0 (p 0), then A) sin2 ( + ) + p sin ( + ) cos ( + ) + q cos2 ( + ) = q B) tan ( + ) = p/q -1 C) cos ( + ) = 1-q D) sin ( + )=-p
8.
If sin + sin = a and cos θ + cos = b, then 158
BASIC TRIGONOMETRY 1 A) cos 2 2
a
2
B) cos 2
b2
4 a2 b2 2 2 C) tan 2 a b 9.
11.
B) (a2+b2 ) cos = 2ab D) sin = 2.375
θ (1+sec θ ) (1+sec2 θ ) (1+sec 4 θ ) .... (1+sec2n θ ). Then 2
Let fn ( θ ) = tan A) f2 =1 16
B) f3 =1 32
C) f4 =1 64
D) f5 =1 128
If (secA + tanA) (secB + tanB) (secC + tanC) = (secA-tanA) (secB-tanB) (secC-tanC) B)-1
C) 0
¥
For 0 < /2 if A) xyz = xz + y If tanx =
D) none of these
¥
2n x = cos φ, y = n =0
13.
b2
Which of the following statements are possible a, b, m and n being non-zero real numbers ?
A) 1 12.
2
a2 b2 2 D) cos 2
A) 4 sin2 = 5 C) m 2 n 2 cos ec m 2 n 2 10.
a
¥
sin 2n φ
z=
n =0
B) xyz = xy + z
cos
2n
φsin 2n φ , then
n =0
C) xyz = x+y+z
D) xyz = yz+x
C)y-z = a-c
D) y-z = (a-c)2 + 4b2
2b , (a c) a -c
y = a cos2x + 2b sinx cos x + c sin2x z = a sin2x - 2b sinx cosx + c cos2x, then A) y = z
B) y + z = a+c n
14.
cosA + cosB sinA + sinB + sinA - sinB cosA - cosB
AB A) 2tann 2
15.
AB B) 2cotn 2
C) 0
D) none of these
B) cot 760
C) tan 460
D) cot 440
In a triangle tanA + tanB + tanC = 6 and tanA tanB = 2, then the values of tanA.tanB and tanC are A) 1, 2, 3
159
(n, even or odd) =
3 + cot76 0 cot16 0 = cot76 0 + cot16 0 A) tan 160
16.
n
B) 2, 1, 3
C) 1, 2, 0
D) none of these
IIT- MATHS 17.
If cot θ + tan θ = x and sec θ - cos θ = y, then 1 x 2 2/3 2 2/3 C) (x y) - (xy ) = 1
A) sinq cosq =
18.
19.
If
D) (x2y)1/3 + (xy2)1/3 = 1
x cosA = where A B then y cosB
A B x tan A y tan B A) tan x y 2
A B x tan A y tan B B) tan x y 2
sin A B y sin A x sin B C) sin A B y sin A x sin B
D) x cosA + y cosB = 0
If tan θ =
sinα - cosα then sinα + cosα
A) sin -cos = C) cos2 = sin2 20.
B) sin tan = y
2 sin
B) sin + cos = 2 cos D) sin2 + cos2 = 0
Let 0 /2 and x = X cos + Y sin , y = X sin -Y cos such that x2 + 4xy + y2 = aX2 + bY2, where a, b are constants. Then A) a = -1, b = 3
B) = /4
C) x = 3, b = -1
D) θ = π /3
160
BASIC TRIGONOMETRY
SECTION - A SINGLE ANSWER TYPE QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
B
B
B
B
D
C
C
B
A
B
A
16.
17. 18. 19.
20. 21. 22. 23
24. 25.
26.
12. 13.
A
C
27. 28.
14. 15.
D
C
29. 30.
A
C
D
C
C
B
D
C
C
D
B,C
D
A
C
A
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
D
D
D
C
C
B
A
A
B
B
B
B
C
A
B
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
C
B
D
C
A
B
D
B
C
C
B
C
D
B
D
61
62
63
64
65
66
67
68
69
70
71
72 73
74
75
B
B
B
A
B
A
D
C
D
C
B
B
A
A
161
A
IIT- MATHS
76
77
78
79
80
81
82
83
84
85
86
87 88
89
90
A
B
B
C
A
C
C
C
D
A
C
A
D
A
91
92
93
94
95
96
97
98
99 100 101 102 103 104 105
A,B,C D
A
A
A
B
B
A
A
106 107 108 109
B
B A,B,C A
B
D
D
C
D
D
C
110 111
D
C
SECTION - B SINGLE ANSWER TYPE QUESTIONS 1.
2.
3.
AB
AD ABC CD BCD BD AB ACD BD
16
17
18
4.
5.
19
20
AB ABC ABC AB CD
BC
6.
7.
8.
9.
10.
11.
12. 13.
14. 15.
A B AB CD
BC BC
BC CD
162
BASIC TRIGONOMETRY
163
IIT- MATHS
4
TRIGONOMETRIC EQUATIONS
164
TRIGONOMETRY EQUATIONS An equation involving one or more trigonometrical ratios of unknown angle is called trigonometric equation e.g. cos2x – 4 sinx = 1 It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle whereas, trigonometric equation is satisfied only for some values (finite or infinite in number) of unknown angle. e.g. sin2x + cos2x = 1 is a trigonometrical identity as it is satisfied for every value of x R.
SOLUTION OF A TRIGONOMETRIC EQUATION A value of the unknown angle which satisfies the given equation is called a solution of the equation e.g. / 6 is a solution of sin =
1 . 2
GENERAL SOLUTION Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. We use the following formulae for solving the trigonometric equations: ( n I) sin = 0 = n, cos = 0 = (2n + 1)
, 2
tan = 0 = n, sin = sin = n + (– 1)n cos = cos = 2n tan = tan = n + sin2 = sin2 or cos2 = cos2 or tan2 = tan2 = n , sin = 1 = (4n + 1) sin = –1 = (4n – 1)
2 2
cos = 1 = 2n cos = –1 = (2n + 1) sin = sin and cos = cos = 2n +
Note: Everywhere in this chapter n is taken as an integer, if not stated otherwise. The general solution should be given unless the solution is required in a specified nterval or range.
SOME IMPORTANT POINTS TO REMEMBER While solving a trigonometric equation, squaring the equation at any step should be avoided as far as possible. If squaring is necessary, check the solution for extraneous values. 165
IIT- MATHS Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of genuine solution. The answer should not contain such values of angles, which make any of the terms undefined. Domain should not be changed. If it is changed, necessary corrections must be incorporated. Check that the denominator is not zero at any stage while solving equations. Some times you may find that your answers differ from those in the package in their notations. This may be due to the different methods of solving the same problem. Whenever you come across such situation, you must check their authenticity. This will ensure that your answer is correct. While solving trigonometric equations you may get same set of solution repeated in your answer. It is necessary for you to exclude these repetitions, e.g. n + of
, ( n I) forms a part 2
k , k I the second part of the second set of solution (you can check by putting 5 10
k = 5 m + 2 (mI). Hence the final answer is
k ,k I . 5 10
Some times the two solution set consist partly of common values. In all such cases the common part must be presented only once. Now we present some illustrations for solving the different forms of trigonometric equations. Which will highlight the importance of above mentioned points.
SOLVING SIMULTANEOUS EQUATIONS Here we will discuss problems related to the solution of two equations satisfied simultaneously. We may divide the problems in two categories. (i)
Two equations in one unknown
(ii)
Two equations in two unknowns.
TRIGNOMETRIC INEQUATIONS
y
/6
5/6
y = 1/2 x
166
TRIGONOMETRY EQUATIONS From, the graph of y = sin x, it is obvous that, between 0 and 2p sinx > 1/2 for
5 x . 6 6
Hence sin x > 1/ 2 Þ
2np + p/6 < x < 2np+ 5p/6
5 2n , 2 n The required solution set = n I 6 6
PROBLEMS BASED ON BOUNDARY CONDITIONS If the problem involves only one equation consisting of more than one variable or equation involves variable of different natures then the boundary conditions of trigonometric functions is generally used. It must be noted that |sinx| 1 ; | cosx | 1; |sec x| 1; | cosec x | 1; |tan x| 0; |cot x| 0
167
IIT- MATHS
WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 Number of solutions of the equation tan x sec x 2cos x lying in the interval [0. 2 ] is (a) 0
(b) 1
(c) 2
(d) 3
Ans : (A) Solution : The given equation can be written as
sin x 1 2 cos x cos x
== sin x 1 2cos 2 x cos x 0
==> 2sin 2 x sin x 1 0
==>
2sin x 1 sin x 1 0
==> x
==> sin x
2
sin x 1 0
as cosx 0
5 , in 0, 2 so that required number of solutions is 2. 6 6
ILLUSTRATION : 02 2 If 6 cos 2 2 cos
(a)
3
2 sin 2 0, , then = 2
(b)
3 , cos 1 3 5
1 3 (c) cos 5
(d)
3 , cos 1 3 5
Ans : (D) Solution : The given equation can be written as ==> 10 cos 2 cos 3 0 ==> 5cos 3 2cos 1 0 ==> cos
1 3 or cos 2 5
==>
1 3 or cos as 3 5
==>
3 , cos 1 3 5
ILLUSTRATION : 03 The set of values of x for which
tan 3x tan 2 x 1 is 1 tan 3x tan 2 x 168
TRIGONOMETRY EQUATIONS (b) 4
(a)
(c) n / n 1, 2,3,..... 4
(d) 2n / n 1, 2,3...... 4
Ans : (C) Solution : The given equation can be written as tan 3x 2 x 1 ==> tanx = 1 ==> x n
4
But for these values of x, tan2x is not defined so the given equation has no solutions. ILLUSTRATION : 04 The number of all possible triplets such than a1 a2 cos 2 x a3 sin 2 x 0 for all is (a) 0
(b) 1
(c) 3
(d) infinite
Ans : (D) Solution : The given equation can be written as a1 a2 cos 2 x
a3 1 cos 2 x 0 2
a3 a3 ==> a1 a2 cos 2 x 0 2 2
which is zero for all values of x. If a1
a3 a2 2
k k or a1 , a2 , a3 k for any 2 2
Hence the required number of triplets is infinite. ILLUSTRATION : 05 In a triangle ABC, the angle A is greater than the angle B. If the values of the angles A and B satisfy the equation 3sinx - 4sin3x - k =0, 0 < k < 1, then the measure of angle C is (a)
3
(b)
2
(c)
2 3
(d)
5 6
Ans :(C) Solution : The given equation can be written as sin 3 x k , 0 k 1 Since A and B satisfy this equation 0 3 A, 3B < as 0 < k < 1
169
IIT- MATHS Also sin 3 A k sin 3B ==> sin 3 A sin 3B 0 ==> 2cos
3 A B 3 A B sin 0 2 2
==> either cos
But sin
3 A B 3 A B 0 or sin 0 2 2
3 A B 3 A B 0 as A > B and 0 < 3A, 3B < so cos 0 2 2
==> cos
3 C 0 2
==> sin
3C 2 0 ==> C . 2 3
ILLUSTRATION : 06 The equation sinx + cosx =1 has a solution in the open interval (a) 0, 2
(b) , 2
3 (c) , 2
(d) None of these
Solution :
1 sin x cos x 1 ==> sin x 4 2 sin 4 n n ==> x n 1 ==> x n 1 , n I 4 4 4 4
==> x 2n or x 2n 2
so that x does not belong to the intervals given by (a), (b), or (c) for any value of x. ILLUSTRATION : 07 3 7 1 1 sin The principal value of 2 cos cos 6 is
(a)
5 6
(b)
2
(C)
3 2
(d) None of these
Ans : (B) Solution : 3 3 7 1 1 sin 1 sin and cos cos 3 6 2 2 170
TRIGONOMETRY EQUATIONS 5 5 1 1 = cos cos 2 cos cos 6 6 3 7 1 sin 1 cos cos 6 2
so that
5 6
5 a 3 6 2
ILLUSTRATION : 08 If sin 4 x cos 4 y 2 4sin x cos y, and 0 x, y (a) -2
(b) 0
(c) 2
2
then sinx + cosy is equal to
(d) none of these
Ans : (C) Solution : The given equation can be written as sin 4 x cos 4 y 2 4sin x cos y 0 2
2
2
2
==> sin 2 x 1 cos 2 y 1 2sin 2 x 2 cos2 y 4sin x cos y 0 ==> sin 2 x 1 cos 2 y 1 2(sin x cos y )2 0
which is true if sin 2 x 1 , cos 2 y 1 and sin x cos y as 0 x, y we get sinx=cosy=1 2 ==> sin x cos y 2 ILLUSTRATION : 09 n
1 sin x ....... 1 sin n x ...... 1 cos 2 x The general solution of the equation 1 sin x ....... sin n x ........ 1 cos 2 x (a)
1
(c) 1
n
n 3
n 1
n (b) 1 n 6
n 6
(d)
1
n 1
n , n I 3
Ans : Solution : n
1 sin x ....... 1 sin n x ...... 1 cos 2 x The equation 1 sin x ....... sin n x ........ 1 cos 2 x ==>
171
1 1-sinx 2sin 2 x X 1 sin x 1 2 cos 2 x
is
IIT- MATHS ==> cos 2 x cos 2 x sin x sin 2 x sin 3 x
(sinx +1 0)
==> cos 2 x sin 2 x sin x cos 2 x sin 2 x ==> 2sin 2 x sin x 1 0 ==> sin x
1 1 1 8 1 3 ==> or sin x 2 4 4
ILLUSTRATION : 10 5 5 The number of solutions of the equation sin x cos x
(a) 0
(b) 1
(c) infinite
1 1 sin x cos x is cos x sin x
(d) none of these
Ans : Solution : The given equation can be written as
sin 5 x cos5 x sin x cos x sin x cos x sin x cos x ==> sin x cos x 1 sin x cos x 5
5
1 4 3 2 2 3 4 ==> sin 2 x sin x sin x cos x sin x cos x sin x cos x cos x =1 2 2 2 2 2 2 2 2 2 2 ==> sin 2 x sin x cos x 2sin x cos x sin x cos x sin x cos x sin x cos x 2
==> sin 2 x 1 sin 2 x cos 2 x sin x cos x 2 1 1 2 ==> sin 2 x 1 sin 2 x sin 2 x 2 2 4
==> sin 3 2 x 2 sin 2 2 x 4 sin 2 x 8 0 2
==> sin 2 x 2 sin 2 x 2 0 ==> sin 2 x 2 , which is not possible for any . ILLUSTRATION : 11 cos 3 1 if 2 cos 2 1 2
(a) n
3
(b) 2n
3
(c) 2n
6
(d) 2n
6
Ans :(B)
172
TRIGONOMETRY EQUATIONS Solution : 4 cos3 3cos 1 The given equality can be written as 2 2 cos 2 1 1 2
4cos 3 cos 1 2
==>
4 cos 2 3
==> cos
2
1 2
==> 2n
3
3 , as for this value of L.H.S. of the given equation is not defined. cos 2
173
IIT- MATHS
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
sin2x - 2cos x +
1 = 0 then x equal to 4
A) 2n ± /3 2.
B) 2n ± /4
The set of values of x for which
C) 2n ± /6
tan3x - tan2x = 1 is 1 + tan3xtan2x
A) C) {n /4 n = 1,2......} 3.
B) /4 D) 2n /4 ; n = 1, 2, 3 .......
The most general value of which satisfies sin = -1/2 and tan = 1/ 3 A) 2n /6
4.
B) 2n + 11 1 /6
C) 2n + 7 /6
D) ? = n + 3 /6
If 1 + cos α + cos2 α +....... = 2 - 2 then α (0 < α < π ) is A) 3 /4
B) / 4
B) x = 6 (n - 1)
D) x = 5 (n+
1 ) 2
C) 13/18
D) -13/18
If |K| = 5 and 0 £ θ £3 6 0° , then the number of different solutions of 3cos + 4sin = K is
sin 4 x + cos 4 x =
A) x =
B) two
C) one
D) infinite
7 sinxcosx then x equal to 2
n n 1 2 2
C) x = n 1 10.
1 ) 2 2
C) x = 5 (n +
B) -24/25
A) zero 9.
D) / 8
If α is the root of 25cos2 θ + 5cos θ - 12 = 0 π /2 < α < π A) 24/25
8.
C) / 6
6x x If sin = 0 and cos = 0 then 5 5
A) x = (n - 5) 7.
D) 2n + /4
B) = n + (-1)n (3 /10)
C) = 2n ± /6
6.
is
Solution of the equation 4cos2 sin - 2sin2 = 3sin is 3 A) n + (-1)n 10
5.
D) n ± /3
n
B) x = n 1
24
n
D) x = n 1
6 n
12
If tan x + tan 4x + tan 7x = tan x. tan 4x tan 7x then x is equal to A) n /6
B) n /12
C) /12
D) n /2 174
TRIGONOMETRY EQUATIONS 11.
12.
If sin 4 θ = cos θ - cos 7 θ A)
n n 1 / 18 3
B)
n n 1 / 9 3
C)
n n 1 / 6 3
D)
n n 1 / 12 2
If tan 7 θ = cot 5 θ then general solution for θ is A) 2n ± /24
13.
B) (2n-1) /24
B) 5
C) 8
D) 9
The most general solutions of the equation secx - 1 = ( 2 1) tanx are given by A) n /8
15.
D) n + /12
The number of distinct solutions of sin 5 θ cos 3 θ = sin9 θ . cos7 θ in [0,2 ] is A) 4
14.
C) n + /24
B) 2n , 2n + /4
D) n /4
C) 2n
The most general values of ? satisfying the equation (1+2sin )2 +
3sinθ -1
2
= 0 are given
by B) n + (-1)n
A) n /6
16.
sinx.sin(600-x) sin(600+x) = A) x = n + (-(A)n C) x n 1
17.
n
D) 2n + 11 1 /6
6
B) x = D) x
n n 1 3 18 n n 1 3 9
cos3θ 1 = , if 2cos2θ -1 2
B) = 2n /3
C) =2n /6
D) =2n /6
General solution of the equation 1+sin2x = (sin3x-cos3x)2 is/are given by A) n 2
19.
C) 2n + 7 /6
1 , n I, then 8
3
A) = n + /3 18.
7 6
B) n , n C) n , n 4 4 4
D) None
If [x] denotes the greatest integer less than or equal to x and let f(x) = sinx +
3 cosx, then the
π most general solution of f(x) = f are 10 A) n /4 20.
175
B) 2n /3
C) n + (-1)n /6 - /3 D) None
The smallest positive value of x for which tanxtannx = 1(n N) is
IIT- MATHS
A) 2 n 1 21.
B)
2 n 1
C)
3 2 n 1
D) None
If 5 cos2 θ + 2cos2 θ /2 +1 = 0 - then is equal to A) = /4
B) = /3
C) = /2
D) /6
22.
If the system of equations (sin3 ) x - y + z = 0 (cos2 ) x + 4y +3z = 0 and 2x + 7y + 7z = 0 have non - trivial solution, then is equal to A) = n B) = (2n+1) C) 2n 1 D) 4n 1 / 2 2 2
23.
The equation 3sin2x+2cos2 x + 31-sin2x+2sin 2x = 28 is satisfied for the values of x given by A) tan x = 1
24.
B) cosx = 0
B) 1
If sin =
B) x =
n 2
C) tan x = 1 / 4
D) x n / 4
B) /3, /2
C) /6,/4
D) /4, /3
B) {x|x = 2np ± 2p/3 n e z} D) [p/3, p/2]
B) [0, 5/6]
C) [5/6, 2]
D) [/6, 5/6]
The equation cos4x – (a+2) cos2x – (a+3)=0 possesses a solution if B) a < -2
C) –3 a 2
D)
If sin8x + cos8y + 2 = y sin2x cos2y and 0 x, y /2 then sinx + cosy is equal to A) –2
32.
D) 2
If sin 4sin2x – 8sinx + 3 0 0 / 2 then the solution set for x is
A) a >-3 31.
C) infinite
If max {5sin + 3sin () = 7 then the set of possible values of is
A) [0, /6] 30.
D) {/4, /3}
3 , + 4sin = 2 ( 3 1) 0 / 2 then is equal to
A) {x|x = 2np ± p/3 n e z } C) [x/3, 2p/3 ] 29.
C) {/3}
2sin 2 cos 2 x = 1-cos sin 2x if 2
A) /6, /3 28.
1 is 2
x 1 2 2 .sin x x The number of solutions of 2cos 2 x 2 0 x / 2 is
A) x = 2n 1 / 2 27.
2
2
A) 0 26.
B) {/3, /4}
D) tan x =
2
0
A) {/4, /6} 25.
C) tan x =
B) 0
C) 2
D) None
{2, e, , 2 4 7} are given by The most general values of x for which 3 sin x cos x min R
176
TRIGONOMETRY EQUATIONS A) 2n B) 2n + 2/3 33.
If
2 cos 2 X 1 2 sin 2 X 1 2 2 then x =
A) (2n+1) 34.
4
B) (2n-1)
38.
B) 2n ± /4
The general solution of the equation
42.
D) n + (-1)n /4
C) 2n + 7/4
C) n - /4
3 1 sin
D) 2n ± /4
3 1 cos 2
n A) n 1
/ 12 4
B) 2n / 4 / 12
n C) n 1
/ 12 4
D) 2n / 4 / 12
sin x cos x The number of distinct real roots of cos x
cos x sin x
cos x cos x 0 in the interval - /4 x / 4 is sin x
cos x
B) 2
C) 1
D) 3
The smallest positive root of the equation tan x – x = 0 lies in B) (/2, )
C) (, 3/2)
D) (3/2, 2)
7 cos3x + sin 2 x = -2 then x is equal to 6 A)
41.
5 , cos-1 (-3/2} D) {/3, /6} 3
If sin cot cos tan , then is equal to 2 2
A) (0,/2) 40.
C) {/3,
B) n + (-1)n 7/4
A) 0 39.
D) All
The most general value of which satisfies both the equations tan = -1 and cos = 1 / 2 will be
A) n +/4 37.
C) (2n±1)
B) {/3, 5/3}
A) n +7 /4 36.
4
The solution set of (2cosx –1) (3+2 cosx) = 0 in the interval 0 x 2 is A) {/3}
35.
C) n + (-1)n/4 +/6 D) None
6k 1 3
B)
6k 1 3
The general solution of sin2 sec +
C)
6k 1 3
D)
6k 1 2
3 tan = 0 is
A) = n + (-1)n+1 /3, = n n z
B) = n n z
C) = n + (-1)n+1 /3 n z
D) =
n nz 2
The value of x between 0 and 2 which satisfy the equation sin x 8 cos 2 x 1 are in A.P. with common difference A) /4
177
B) /8
C) 3/8
D) 5/8
IIT- MATHS 43.
The inequality 2sin2x – 3sinx + 1 > 0 holds for all values of x 5 , 6
B) (0, /6)
5 , 6
3 2 4
D) (0, /3) ,
A) (0, /3)
3
C) (0, /2) , 44.
1 + sin x + sin2x +………..¥ = 4 + 2 3 , where x e [0,] A) /3 or 2 /3
45.
D) n - /3
C) 8
D) 10
B) 2n + /3 n z
C) 2n ± n z
D) n + (-1)n /3 n z
B) 4
C) 2
D) 1
B) 1
B) n + /4
C) 2
D) 3
sin x i cos x is purely imaginary are given by 1 i C) n
D) 2n + /2
The equation (cosp-1)x2 + (cosp)x+sinp=0 where x is a variable has real roots if p lies in the interval A) (0, 2)
52.
B) 6
The values of x [-2, 2] such that A) n - /4
51.
C) 2n- - /3
The number of values of x for which sin2x + cos4x = 2 is A) 0
50.
B) 2n + /3
The number of solutions of |cosx|= sinx 4 is A) 8
49.
D) /2
3 = 2 are The most general values of satisfying tan + tan 4
A) n ± /3 n z 48.
C) /4
The number of values of x in [0,5] satisfying the equation 3cos2x – 10cosx + 7 = 0 is A) 5
47.
B) /6
If sec + tan = 2 3 then general solution for is A) n + /3
46.
B) (-, 0)
C) (-/2, /(B)
D) (0, )
n If 2tan2x –3secx is equal to 0 for exactly 7 distinct values of x 0, , n N then the greatest 2 value of n A) 4
53.
B) 10
C) 13
D) 15
If [x] denotes the greatest integer less than or equal to x and sinx = [1+sinx]+[1-cosx] has A) no solution in [-/2, /2]
B) no solution in [/2, ]
C) no solution in [, 3/2]
D) no solution for x
178
TRIGONOMETRY EQUATIONS 54.
If cos2 =
1 2 1 cos then = 2
A) 2n + /4 55.
B) 2n ± /3
B)
n /8 2
n C) 1
If cos6 + cos4 + cos2 + 1 = 0 0 A) {/6, /4, /2}
57.
D) None
The general solution of sinx - 3sin2x + sin3x = cosx - 3cos2x + cos3x is A) np + p/8
56.
C) both
B) {/6, /4}
3 2
n /8 2
1 D) 2n cos
then q is equal to 2 D) {/3, /6}
C) {/4}
The value of lying between = 0 and = /2 and satisfying the equation 1 cos 2 sin 2 4 sin 4 2 2 4 sin 4 0 cos 1 sin cos 2 sin 2 1 4 sin 4
A) 10/24 58.
B) 7/24
The value of ‘a’ for which the equation a2 – 2a + sec2 { (a+ x) } = 0 has solution is A) a = 2
59.
B) a = 1
B) 1/3
B) 2 a 2
D) 1/ 2 a 1/ 2
C) 110
1 x e [0, 6p] is 4
D) 15/3
If max {5 sin+ 3sin(-a)}=7, then the set of possible values of is .
3
2
C) , 3 3
B) x : x 2n
2 , n I 3
D) None
1 1 The number of solutions of the equation sin5x – cos5x = cos x sin x (sinx cosx) is
A) zero
179
D) ½
C) 1 a 1
B) 30
A) x : x 2n , n I
63.
C) 2/3
The sum of all the solutions of the equation cosx. cos ( / 3 x) cos ( / 3 x ) A) 15
62.
D) –1
The equations sin4x – 2cos2x + a2 = 0 is solvable if A) 3 a 3
61.
C) a = ½
2 19 3 sinpx + cos px = x2 - 3 x + 9 , then x is equal to
If
A) –1/3 60.
D) /24
C) 5/24
B) 5
C) ¥
D) None
IIT- MATHS 64.
If the system of equations (sin3)x – y +z=0, (cos2)x +4y+3z=0, 2x+7y+7z=0 have non-trivial solutions, then =
A) n, n+ (-1)n 6 C) both
B) n ± a, n + (-1)n 3 D) None
180
TRIGONOMETRY EQUATIONS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
2.
2 sin x cos2x = sinx if A) x = n + /6 (n l)
B) x = n - /6 (nÎ l)
C) x = n (n Î l)
D) x = n + /2 (n Î l)
X X cos 2 X 2sin sin 2 X cos2 X sin 2 X has a root for which 2 2
The equation 2 sin A) sin 2x=1
3.
B) sin2x=-1
1
B) sin (x - p/4) =
2 1
C) cos (x + p/4) =
5.
D) cos (x-p/4) =
2
2
C) two values of x and two values of
D) two pairs of values of (x, )
The equation sinx = [1+sinx] + [1-cosx] has (where [x] is the greatest integer less than or equal to x)
3 2
B) no solution in , 2 D) no solution for x
Solutions of the equation sin7x + cos2x = -2 are A) x =
2 k 3 , n, k I 7 14
C) x = n + /2, n I
,nI 4 D) none of these
B) x = np +
The solutions of the system of equation sinx siny =
3 /4, cosx cosy =
A) x1 =
2n k 3 2
B) y1 =
k 2n 6 2
C) x2 =
2n k 6 2
D) y2 =
k 2n 3 2
3 /4 are .........
2sin2x + sin22x = 2, x, then x = A) ±/2
181
1
B) one value of x and two values of
C) no solution in ,
8.
2
A) no value of x and
7.
1
3 cos = 6x - x2-11, 4, holds for
sin +
A) no solution in , 2 2
6.
D) cos2x = -1/2
sinx + cosx = 1 + sinx cosx, if A) sin (x+p/4) =
4.
C) cosx = 1/2
B) ± /4
C) ± 3/4
D) none of these
IIT- MATHS
9.
1 tan x If 1 tan x = tany and x-y , then x, y are respectively 6
A) 10.
5 , 24 24
If cosx = A) p
11.
B) -
7 11 , 24 24
C)
115 119 , 24 24
D) none of these
1 sin 2 x , 0 x, then a value of x is B) 0
C) tan-12
D) none of these
The solution of the equation cos103x - sin103x = 1 are A) -
2
B) 0
C)
2
D) p
182
TRIGONOMETRY EQUATIONS
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A
A
A
A
A
D
B
B
A
B
A
16.
17. 18. 19.
20. 21. 22. 23
24. 25.
26.
12. 13.
B
D
27. 28.
14. 15.
B
C
29. 30.
B
B
A
C
A
B
A
B
A
A
A
A
A
D
C
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
C
B
C
B
C
A
D
C
C
A
B
C
B
A
B
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
C
A
B
A
A
D
D A,B, A,B C,D
B
A
B
B
B
61
62
63
64
B
A
A
A
183
B
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
AB C
2.
3.
4.
A B AD BD CD
5.
6.
7.
8.
A B AC AB AB CD C C
9.
AB C
10.
11.
BC AB
184
TRIGONOMETRY EQUATIONS
185
IIT- MATHS
5
INVERSE TRIGNOMETRIC FUNCTIONS
186
INVERSE TRIGONOMETRY FUNCTIONS
DEFINITION If a function is one to one and onto from A to B, then function g which associates each element y B to one and only one element x A, such that y = f(x), then g is called the inverse function of f denoted by x = g(y). Usually, we denote g = f-1 {Read as f inverse} –1 x = f (y).
If cos = x, then may be any angle whose cosine is x, and we write = cos–1 x. It means that is an angle whose cosine is x. Thus, sin–1 where,
is an angle, whose sine is , i.e. = sin–1 = n + (–1)n 2 2 2 6
is the least positive value of . 6
The functions sin–1 x, cos–1 x, tan–1 x, cot–1 x, cosec–1 and sec–1 x are called inverse circular or inverse trigonometric functions. Each of the inverse circular function is multivalued. To make each inverse circular function single valued, we define principal value. If x is positive, the principal values of all the inverse circular functions lie between 0 and . If x is negative, the principal values of cos–1x, sec–1 x and cot–1x lie 2 between and , while that of sin–1x, tan–1x and cosec–1 x lie between and 0. 2 2
Function
Domain
sin–1x
[ –1, 1]
2 , 2
cos–1 x
[–1, 1]
[0, ]
tan–1x
R
, 2 2
cot–1 x
R
(0, )
sec–1 x
R – (–1, 1).
[0, ] – { /2}
cosec–1 x
R – (–1, 1)
2 , 2 – {0}
Note. sin–1 x is not to be interpreted as
Range (Principal Values)
1 . The sin–1 x is merely a symbol denoting a certain angle sin x
whose sine is x. The ‘–1’ used in sin–1 x is not an exponent. Similar argument also works for cos– 1 x, tan–1x etc. Remark 1. The inverse trigonometric functions are also written arc sin x, arc cos x etc.
187
IIT- MATHS
GRAPHS OF INVERSE TRIGONOMETRIC FUNCTIONS sin = x = sin–1 x, where , 2 2
y = sin–1x
y
y=x
/2
y = sin x
–1
x 1
and x [– 1, 1] – /2
y y = cosec x y=x
2
cosec = x = cosec–1 x
1
–1
y = cosec x
- 2 –1
x
where , 0 0, 2 2
2
1 1 – 2
and x (– , –1] [1, )
y
–1
y = cos x
cos = x = cos–1 x where [0, ] and x [–1, 1]
2
y=x
1
2
1
x
y = cos x
y
y=x
sec = x = sec–1x
where 0, , 2 2 and x (– , –1] [1, )
–1
1 1
y = sec–1x x
–1
y = sec x
188
INVERSE TRIGONOMETRY FUNCTIONS y y = tan x
y=x
2 –1
y = tan x – 2
tan = x = tan–1 x
x
2
where , and x (– , ) 2 2
– 2
y y = cot x y=x –1
y = cot x
–1
cot = x = cot x where (0, ) and x (– , )
x
PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS Property I: (i) (ii) (iii) (iv) (v) (vi)
sin–1 (sin ) = ; for all [– /2, /2] cos–1 (cos ) = ; for all [0, ] tan–1 (tan ) = ; for all (– /2, /2) cosec–1 (cosec ) = ; for all [– /2, /2], 0 sec–1 (sec ) = ; for all [0, ], /2 cot–1 (cot ) = ; for all (0, )
Property II: (i) (ii) (iii) (iv) (v) (vi)
sin (sin–1x) = x, cos (cos–1 x) = x, tan (tan–1 x) = x, cosec (cosec–1 x) = x, sec (sec–1 x) = x, cot (cot–1 x) = x,
for all x [–1, 1] for all x [–1, 1] for all x R for all x [– , –1] [1, ) for all x (– , –1] [1, ) for all x R
Property III: (i) (ii) (iii) (iv) 189
sin–1 (– x) = – sin–1 ( x), cos–1 (– x) = – cos–1 (x), tan–1 (– x) = – tan–1 x, cosec–1 (– x) = – cosec–1 x,
for all x [–1, 1] for all x [–1, 1] for all x R for all x [– , –1] [1, )
IIT- MATHS (v) (vi)
sec–1 (– x) = – sec–1 x, cot–1 (– x) = – cot–1 x,
for all x (– , –1] [1, ) for all x R
Property IV: 1 x
for all x (– , 1] [1, )
1 x
for all x (– , 1] [1, )
1 1 cot x, = 1 x cot x ,
for x 0 for x 0
(i)
sin–1 = cosec–1 x,
(ii)
cos–1 = sec–1 x,
(iii)
tan
–1
Property V: (i)
sin–1 x + cos–1x =
, 2
for all x [–1, 1]
(ii)
tan–1 x + cot–1 x =
, 2
for all x R
(iii)
sec–1 x + cosec–1 x =
, 2
for all x (– , – 1] [1, )
Property VI: If x, y 0, then
(i)
1 x y tan , if xy 1 1 xy π , if xy 1 tan–1x + tan–1 y = 2 1 x y π tan , if xy 1 1 xy
(ii)
1 tan–1x – tan–1y = tan 1 xy .
xy
Property VII: (i)
sin–1 x + sin–1y =
sin 1 x 1 y2 y 1 x 2 , if 1 x, y 1 and x 2 y2 1 or if xy 0 and x 2 y 2 1. 1 x 1 y 2 y 1 x 2 , if 0 x, y 1 and x 2 y 2 1. π sin π sin 1 x 1 y2 y 1 x 2 , if 1 x, y 0 and x 2 y 2 1.
190
INVERSE TRIGONOMETRY FUNCTIONS (ii) sin –1 x – sin–1 y =
sin 1 x 1 y 2 y 1 x 2 , if or if 1 x 1 y 2 y 1 x 2 , if π sin 1 x 1 y 2 y 1 x 2 , if π sin
1 x, y 1 and x 2 y 2 1
xy 0 and x 2 y2 1. 0 x, y 1, 1 y 0 and x 2 y 2 1. 1 x 0,0 x y 1 and x 2 y 2 1.
Property VIII: (i)
cos–1x + cos–1y =
cos1 xy 1 x 2 1 y2 , 2π cos 1 xy – 1 x 2 1 y2
(ii)
if 1 x, y 1
,
if
and
xy0
1 x, y 1 and x y 0
cos–1 x – cos–1 y =
cos1 xy 1 x 2 1 y 2 , if 1 x, y 1 and x y. cos1 xy 1 x 2 1 y 2 , if 1 y 0, 0 x 1 and x y.
Property IX: (i)
x 2 1 x
sin–1 x = cos–1 1 x 2 tan 1
= cot
(ii)
1 x2 x sec 1 2 x 1 x
–1
cos ec 1 1 , x (0, 1) x
1 x2 cos–1 x = sin–1 1 x 2 tan 1
x
1 1 1 cosec 1 sec 2 2 x 1 x 1 x
= cot –1
(iii)
1 cos 2 1 x
tan–1 x = sin–1
= cot
191
x
–1
x
1 2 1 x
, x (0, 1)
1 x2 1 1 2 1 sec 1 x cos ec x x
, x>0
IIT- MATHS
Property X: 1 2x 1 x 2 , π sin 1 2 2sin–1 x = sin 2x 1 x , 1 2x 1 x 2 , π sin
(i)
if
2 1 2
x
2 1 2
x 1
if 1 x
if
1
1
1 2 1 1 if x 2 2 1 if x 1 2
1 3x 4x 3 , π sin –1 3sin x = sin 1 3x 4x 3 , 1 3x 4x 3 , π sin
(ii)
if 1 x
Property XI: (i)
2π cos 1 2x 2 1 , 2 cos x = cos1 2x 2 1 , –1
if
0 x 1
1 1 4x 3 3x , if 1 x 2π cos 2 1 1 1 3 3 cos–1 x = 2π cos 4x 3x , if x 2 2 1 1 4x 3 3x , if x 1 cos 2
(ii)
if 1 x 0
Property XII:
(i)
1 2x , π tan 2 1 x 1 2x 2tan–1x = tan 1 x 2 , 1 2x π tan , 1 x2
if
x 1
if 1 x 1 if x 1
192
INVERSE TRIGONOMETRY FUNCTIONS
(ii)
3x x 3 1 , π tan 2 1 3x 3 1 3x x –1 tan , 3 tan x = 1 3x 2 3 π tan 1 3x x , 1 3x 2
if
x
if
1 3
1 1 x 3 3
if x
1 3
Property XIII
(i)
(ii)
193
1 2x , π sin 2 1 x 1 2x 2 tan–1 x = sin , 1 x2 1 2x π sin , 1 x2
1 1 x 2 cos , 1 x 2 2 tan–1 x = 1 x2 cos , 2 1 x
if
x 1
if 1 x 1 if x 1
if 0 x
if x 0
IIT- MATHS
WORKED OUT ILLUSTRATIONS 1ILLUSTRATION : 01 If 0 x 1 and sin 1 x cos 1 x tan 1 x , then (a) /2
(b) /4
(c) = /4
(d) /4 /2
Ans : (D)
Solution :
sin 1 x cos 1 x tan 1 x
tan 1 x and 0 tan 1 x since 0 x 1 when we fined 2 4 4 2
ILLUSTRATION : 02 If x
(a)
1 1 1 , the value of cos cos x 2sin x is 5
24 25
(b)
24 25
(c)
1 5
(d)
1 5
Ans :(C)
Solution : The given expression is equal to
cos cos 1 x sin 1 x cos sin 1 x 2
1 = sin sin x x
1 5
ILLUSTRATION : 03
2 tan 1 cos ec tan 1 3 tan cot 1 3 is equal too (a) /16
(b) /6
(c) /3
(d) /2
Ans :(C)
Solution : The given expression is equal to
2 tan 1 cos ec tan 3 6 194
INVERSE TRIGONOMETRY FUNCTIONS 1
1 2 1 1 2x 2 tan 6 3 3 3 3
= 2 tan
ILLUSTRATION : 04 2 tan 1 1 tan 1 2 tan 1 3 is equal too (a) /4
(b) /2
(c)
(d) 2
Ans : (D)
Solution : 1 2
1 tan 1 3 The given expression is equal to 2 tan 1 2
1 1 = 2 tan 3 tan 3 2
ILLUSTRATION : 05
1 x2 1 4 , then x
1 If tan
(a) x tan 2
(b) x tan 4
(c)x = tan(1/4)
Ans : (D)
Solution : Taking x tan , tan
1 = tan
1
1 x2 1 sec 1 tan 1 x tan
1 cos 1 1 tan 1 tan tan 1 x sin 2 2 2
So that according to the given condition
1 1 tan x 4 tan 1 x 8 or x tan 8 2
ILLUSTRATION : 06 sec 2 tan 1 2 cos ec 2 cot 1 3 is equal too (a) 1 Ans :
195
(b) 5 (D)
(c) 10
(d) 15
(d) x tan 8
IIT- MATHS Solution : The given expression is equal to 2
1 tan tan 1 2 1 cot cot 1 3
2
= 1+ 4 +1 9 = 15
ILLUSTRATION : 07
1 1 2 The equation 2 cos x sin 2 x 1 x
(a) 1 x 1
(b) 0 x 1
is valid for all values of satisfying (c) 0 x
1 2
(d)
1 x 1 2
Ans : (D)
Solution : If we denote cos 1 x by y, then Since 0 cos 1 x 0 2 y 2 Also since
………(1)
sin 1 2 x 1 x 2 2 2
sin 1 sin 2 y 2 2
2y 2 2
………..(2)
From (1) and (2) we find 0 2 y
0 y
2
4
1 0 cos x
which holds if
4 1 x 1 2
ILLUSTRATION : 08
u
If u cot 1 tan tan 1 tan , then tan is equal to o 4 2 196
INVERSE TRIGONOMETRY FUNCTIONS (a)
(b) cot
tan
(c) tan
(d) cot
Ans :
Solution : Let
1 1 tan tan x, then u cot tan x tan tan x
x x 2x 2 2
=
u 2
2x
u 4 2
x
u 4 2
tan x tan
tan x tan 4 2
ILLUSTRATION : 09 1
1 2
1
1 2
1
The value of cos 2sin 3cos (a) 7/4
(b) 11/4
(c) /12
1 1 4 tan 1 is equal too 2
(d) 25/12
Ans : (D)
Solution : 1
1 2
1
1 2
1
We have - cos sin 3cos
=
1 2
2 3 25 2 x 3 x 4 3 6 4 4 12
ILLUSTRATION : 10 3 5
1 1 1 If cos ec x 2 cos 7 cos then the value of is
(a) 44/117 Ans : (B) 197
(b) 125/117
(c) 24/7
(d) 5/3
IIT- MATHS Solution : 2 3 3 1 7 1 2cot 7 cos cot cot 1 2x7 4 5 1
1 = cot
1
24 3 cot 1 7 4
24 3 x 1 cot 1 7 4 = 24 3 7 4
cot 1
44 117
cos ec 1
125 117
198
INVERSE TRIGONOMETRY FUNCTIONS
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
cos 1 (cos 5 / 4) is given by A) 5/4
2.
B) 3/4
cos 1
B) /2
140 221
tan
1 1 If cot x sin
A) 2 9.
199
C) /2
B) 0
D)
B) 0, /4
C) - /4, /4
D) /4, /2
B) /2
C)
D) 3/2
B) 1/ 3
C) 1
D)
C) –2
D) ½
1 / 4 then x equal to 5 B) 3
Number of solutions of sin -1 x + sin -1 2x = /3 is A) 2
10.
D)
2 2 tan 3 tan tan is equal to 4 15 5 15
A) - 3 8.
2
If x + 1 / x = 2, the principle value of sin 1 x is A) /4
7.
C)
1 1 x The smallest and the largest values of tan ; 0 x 1 are 1 x
A) 0, 6.
171
B) cos 1 221
2x If x 1 then 2tan-1 x + sin-1 1 x 2 is equal to
A) 4tan-1 x 5.
D) /4
C) 0
15 1 2 tan 1 is equal to 17 5
1 A) cos
4.
D) - 5/4
yz xz 1 xy tan 1 tan 1 If x 2 y 2 z 2 y 2 then tan is equal to zr xr yr A)
3.
C) -/4
B) 3
C) 1
4 x2 x3 x6 -1 -1 2 x sin x + ........ + cos x + ....... = for 0 <|x| < If 2 4 2 4
D) 4
2
3
IIT- MATHS A) 11.
1 2
B) 1
C) – ½
If a and b are roots of the equation 6X 2 +11X + 3= 0 then A) both cos 1 and cos 1 are real C) both cot 1 and cot 1 are real 2n
12.
i 1
X i 1
B) x
tan
3
3
D) x 2 , 2
C) x 1,0
C)
D) 2
B)
cot
C) tan2a
D) cot a
C) 16/7
D) 17/6
-1 3 -1 3 The value of tan sin + cot is 5 2 B) 7/16
3 If sin-1x + sin-1y + sin-1z = then the value of 2 B) 1
x x
101
y 101
303
303
y
x x
202
y 202
404
404
y
is equal to
C) 2
D) 3
C) a = x = b
D) a > b, x
ax x b 1 cos-1 a b sin a b is possible if B) a < x < b
If sin-1x = cot--1x , then A) x2 =
20.
2
1 2
u 1 tan a then tan is equal to If u = cot-1 tan a tan 4 2
A) a > x > b 19.
,
D) 2n-1
holds for
B) /2
A) zero 18.
1
n n 1 2
2(tan-11 + tan-12+tan-13) is equal to
A) 6/17
17.
C)
A)
16.
is equal to
1 1 2 The formula 2 sin x sin 2 x. 1 x
A) /4 15.
i
B) 2n
A) x0,1 14.
B) both cos ec 1 and cos ec 1 are real D) none of these
2n
If X i n then A) n
13.
D) –1
5 1 2
B) x2 =
5 1 2
C) sin(cos-1x)=
5 1 2
D) x =
5 1 2
If x1, x2, x3, x4 are roots of the equation x 4 x 3 sin 2 x 2 cos 2 x cos sin 0 then tan 1 x 1 tan 1 x 2 tan 1 x 3 tan 1 x 4 is equal to
200
INVERSE TRIGONOMETRY FUNCTIONS A) B) /2 - 21.
x2 2 x2 1
If minimum value of
-1 If cos
201
D) 1, ½
C) 0
D) none of these
C) (a + b) (a2 + b2)
D) (a - b) (a2 - b2)
C) ¼ 2
sin x sin x y
1
2
D) 3/2
2 then the value of K is K
C) 8
D) 16
B) 5/4
C) 1
D) ½
C) 2 / 8
D) 2 / 32
Minimum value of (sec-1 x)2 + (cosec-1 x)2 is B) /32
The number of real solutions of tan 1 x( x 1 ) sin 1 x 2 x 1 / 2 is B) one
C) two
D) infinite
1 3 1 1 1 3 sin 1 sin 1 =‡ is then The two angles are A = 2 tan 2 2 1 and B = 3 sin 3 3 5
B) A < B
C) A = B
D) A + B = 0
An integral solution of the equation tan-1x + tan-1y = tan-13 is equal to A) (2, 7)
32.
C) 0, 1/2
x y 5π y x 2 y2 1 x + cot -1 = sin 1 and sin then value of 2 2 is 2 b 12 2 b 12 a b
A) A > B 31.
B) (a + b) (a2 - b2)
B) 6
A) zero 30.
B) 0,1/2
B) ½
A) /8 29.
x2 2
x 2 xy y 2 x 1 y cos / 6 then the value of If cos is 4 2 3 9 2 3
A) ¾ 28.
x2 2
1
D)
1
A) 4 27.
x2 2
x2
1 3 21 1 cos ec 2 tan 1 sec tan 2 2 á is equal to 2 2
A) ¾ 26.
C)
If sin 1 (1 x ) 2 sin 1 x / 2, then x equals
A) (a - b) (a2 + b2) 25.
x2 1
B) 1, 0
A) 0, -1/2
24.
B)
If sin 1 x sin 1 (1 x ) cos 1 x , then x equals A) 1, -1
23.
D)
The value of sin (cot-1 (cos (tan-1x)) is A)
22.
C)
B) (4, -13)
The number of real solutions of (x, y) where
C) (5, -8)
D) (1, 2)
cos 1 (cos X ) tan x ,0 x 2 is 2
A) 2 33.
B) 1
IIT- MATHS D) None
C) 4
Sum of infinite series
3 3 2 3 1 cot 1 22 cot 1 32 .... cot is 4 4 4 -1
B) tan-12
A) /4
34.
tan
1
m 1
B) -/4
1 2 x2 2 x 1 x 1 cos-1 2 4 A) |x| 1
36.
B) x R
2x
C) 0 x 1
1 x2
if 0 x 1
2a The solution of sin 1 a 2 ab 1 ab
B) f(x) =
1 x2
if x < 1
B)
2 1 1 b cos 2 1 b
1 2x tan 2 is 1 x
1 ab ab
C)
ab 1 ab
D)
ab 1 ab
B) /2
C) 1
D) none of these
C) p/3
D) 4p/3
2 1 The principal value of sin sin is 3 A) 2p/3
B) - 2p/3
3 1 1 1 The equation sin x cos x cos 2 has
A) no solution C) infinite no.of solution 41.
2x
If xy + yz + zx = 1 then tan-1x + tan-1 y + tan-1z is equal to A)
40.
D) –1 x < 0
D) None
-1
A)
39.
D) None
x cos 1 2 - cos--1 x holds for
C) not finite if x > 1
38.
C) ± /4
2 1 1 1 2x 1 1 x sin cos The value of f(x) = tan 2 1 x2 2 1 x 2 is
A) f(x) =
37.
D) None
2m 4 is equal to 2 m m 2
zA) /4
35.
C) tan-13
B) unique solution D) none of the these
If tan + tan ( + /3) + tan ( - /3) = k tan 3 then the value of k is 202
INVERSE TRIGONOMETRY FUNCTIONS A) 1 B) 1/3 42.
B) 1
1 3 2 7
1 If tan
D)
1 7 2 3
D) 3/2
C) - 3tan-1 x
D) -2tan-1 x
4 3 5 5
4 3 3 8
B) ,
4 3 5 7
C) ,
D) ,
B) xR
C) x(--1] [1]
D) x(-1, 1)
x 1 x2
sin 1 x holds is C) [0,1]
D) [-1,0]
C) –1/2
D) 3 / 2
C) 25
D) 27
If sin 1 x cos 1 x / 6 then x is equal to B)
3 /2
sec2(tan-12) + cosec2(cot-13)+2 is equal to B) 15
x y 2 2 -1 If cos a +cos b = a, then x 2xy cos y a 2 ab b2 -1
A) sin2
B) cos2
C) tan2
D) cot2
1 1 1 If tan-1 ( x 1 ) tan x tan ( x 1 ) tan ( 3 ) then x=
A) zero 203
B) + 3 tan-1 x
B) (-1,1)
A) 17
52.
C) ¼
1 The set of values of x for which tan
A) 1/2
51.
1 7 2 3
Let f(x) = sec 1 x tan 1 x then f(x) is real for
A) R
50.
C)
x 1 2x 1 23 tan 1 tan 1 then x equal to x 1 2x 1 36
A) x[-1, 1]
49.
3 7
3 1 1 3x x tan , 3 If x e the value of 2 is 3 1 3x
4 3 5 8
48.
1 2
B) ½
A) , 47.
B)
x 2 xy y 2 x -1 y + + sin = π / 6 hen value of If sin is 9 4 3 16 3 4
A) - 3tan-1 x 46.
D) 3
-1
A) ¾
45.
C) 2
If sin -1x + sin -1 2x = π / 3 then x equal to A)
44.
D) 1/6
9 100 100 100 If sin -1x + sin -1 y + sin -1z = 3π / 2 the value of x y z x101 y101 z101 is
A) 0 43.
C) 3
B) –1
C) 3
D) 13
IIT- MATHS 53.
1 If l is a root of x2 + 3x+1=x0, then tan-1l + tan1 is equal to A) /2
54.
B) - /2
The equation 2cos-1x = sin-1(2x A) –1 x 1
55.
If cos tan
1
sin cot
1
2
C) y = -1
D) None
C) 60°
D) 30°
C) 2
D) infinite
C) 0
D) –1/2
2
B) 1
If 2tan-1 (cosq) = tan-2 (2 cosecq) then q equal to B) n + /4
C) - /4
D) 2n - /4
3 5 cos x cos 1 is equal to 5 3 cos x x 2
x 2
1 B) 2 tan 2 tan
C)
1 x tan 1 2 tan 2 2
1 2
x 2
1 D) 2 tan tan
Minimum value of (tan–1 x)2 + (cot–1 x)2 is A)2/4
B) 2/8
C) 2/2
D) 2/32
1 1 1 x 3 3x 2 then If f ( x ) cos x cos 2 2 2 3
1 B) f 2 cos
1 3
1 /3 D) f 2 cos
A) f / 3 C) f / 3 63.
x1
5 2 If (tan x ) (cot x ) , then x equals 8 1
1 2
62.
2
B) 1
1 A) tan tan
61.
2
1
If < 1/32 then the number of solutions of (sin-1x)3 + (cot-1 x)3 = ap3 is
A) n - /4 60.
D)
2 y then
1
B) 45°
A) –1 59.
1
2x k x 3 1 -1 tan and B = tan k 3 then the value of A – B is If A = 2k x
A) 0 58.
C) 0 x
B) y = 3 / 2
A) 0° 57.
D) None of these
1 x 2 ) is valid for all values of x satisfying
B) 0 x 1
A) y = 4/5
56.
C) /3
2 3
2 /3 3
1 3
1 3
If a, b, g are the roots of the equation x 3 mx 2 3x m 0, then the general value of 204
INVERSE TRIGONOMETRY FUNCTIONS tan -1α + tan -1β + tan -1 γ is A) (2n+1) /2 64.
D) /4 + n
B) 1
C) 3
D) 4
1 x 2 x 1 /2 is The number of real solutions of tan-1 x( x 1 ) sin
A) zero 66.
C) n/2
The number of positive integral solution of the equation tan–1x + tan–1 1/y = tan–1 3 A) 2
65.
B) n
B) 1
C) 2
D)
The sum of the infinite series 2 1 3 2 n n 1 1 1 1 sin 1 sin ...... sin .... sin 2 6 12 n ( n 1 ) -1
A) /8
67.
D)
B) 4
C) 6
D) 8
If sin1x + sin-1y = 2/3, cos-1x – cos-1y = /3, then the number of values of (x, y) is A) two
205
C) /2
n 2 10n 21.6 , , n N, then the maximum value of n is If cot-1 6 A) 2
68.
B) /4
B) four
C) zero
D) None
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTION 1.
The x satisfying sin-1x + sin-1 (1-x) = cos-1 x are A) 0
2.
If
B) 1/2
D) cos-1x
B) x (0, 1/ 2 )
C) x (1/ 2 , 1)
D) x = 0.75
C) x = 3
D) x = 4
B) x = 2
5 2 Let q = tan-1 tan and f = tan-1 tan then 4 3 A) >
6.
C) sec-1x
6 sin-1 (x2-6x+8.5) = , if A) x = 1
5.
B) tan-1x
sin-1x > cos-1x holds for A) all values of x
4.
1
8 = 5 / 18 9 e
B) f
7 = /12 4 e
D) f
8 = 13 / 8 9 e 7 = 11 / 12 9 e
If a £ sin-1 x + cos-1x + sin-1 x £ b, then A) =0
B) = /2
C) = /4
D) =
The greatest and least values of (sin-1x)3 + (cos-1x)3 are A) 3/32
9.
D) none of these
Let f(x) = e cos sin( x / 3 ) then
C) f
8.
C) + = 7/12
B) 4 - 3 = 0
A) f
7.
D) 2
1 X 1 then which of the following are real ? 2
A) sin-1x 3.
C) 1
tan-1 A) /4
B) - 3/8
C) 73/8
D) /2
a( a b c ) b( a b c ) c( a b c ) tan 1 tan 1 is bc ca ab B) /2
C)
D) 0
206
INVERSE TRIGONOMETRY FUNCTIONS
SECTION - A SINGLE ANSWER TYPE QUESTION
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
B
B
A
D
B
B
D
B
C
B
C
16.
17. 18. 19.
20.
21. 22. 23
24. 25.
26.
12. 13.
B
B
27. 28.
14. 15.
D
A
29. 30.
D
D
B
A,C
B
B
C
C
C
C
C
B
C
C
A
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
A,B C,D
C
D
B
C
A,C
D
D
C
B
D
A
A
C
C
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
B
C
B
B
A
A
B
B
D
B
D
A
B
D
61
62
63
64
65
66
67
68
B
A
B
A
C
C
C
D
207
A
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS
1.
2.
AB
AB D
3.
4.
CD BD
5.
6.
7.
8.
BC BC AD AC
9.
CD
208
INVERSE TRIGONOMETRY FUNCTIONS
209
IIT- MATHS
6
PROPERTIES OF TRIANGLE
210
PROPERTIES OF TRIANGLE In a triangle ABC the angles are denoted by capital letters A, B and C and the length of the sides opposite to these angles are denoted by small letters a, b and c. Semi perimeter of the triangle is given by s =
abc and its area is denoted by . 2
SINE RULE In a triangle ABC, a b c sin A sin B sin C
COSINE RULE In a triangle ABC, b2 c2 a 2 (i) cosA = 2bc
(iii)cosC =
c2 a 2 b2 (ii) cosB = 2ca
a 2 b2 c2 2ab
PROJECTION FORMULAE (i) a = b cos C + c cos B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos C
NAPIER’S ANALOGY (TANGENT RULE) A BC bc cot 2 2 bc
(i) tan
B CA ca cot 2 2 ca
(ii) tan
C A B a b cot 2 2 ab
(iii) tan
HALF ANGLE FORMULAE (a)
(b)
(i) sin
A 2
(iii) sin
C 2
(i) cos
A = 2
(iii) cos
211
s bs c bc
(ii) sin
B 2
s c s a ca
s a s b ab
ss a bc
s s c C 2 ab
(ii) cos
B = 2
ss b ca
IIT- MATHS
(c)
(i) tan
s b s c s s a
A = 2
2 2Δ s s a s b s c bc bc
(iii) sin C =
B = 2
s cs a ss b
s a s b ss c
C
(iii) tan 2 (i) sin A =
(ii) tan
(ii)
sinB=
2 2Δ s s a s b s c ca ca
2 2Δ s s a s b s c . ab ab
m-n THEOREM Let D be a point on the side BC of a ABC such that BD : DC = m : n and ADC = , BAD = and DAC = . Then (i)
(m + n) cot = m cot – n cot
(ii)
(m + n) cot = n cot B – m cot C
CENTROID AND MEDIANS OF A TRIANGLE The line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called the median of the triangle. The three medians of a triangle are concurrent and the point of concurrency of the medians of any triangle is called the centroid of the triangle. The centroid divides the median in the ratio 2 : 1 a sin B
sin =
2b 2 2c 2 a 2
CIRCUM CIRCLE The circle which passes through the angular points of a ABC, is called its circumcircle. The centre of this circle i.e., the point of concurrency of the perpendicular bisectors of the sides of the ABC, is called the circumcentre.
A
E
F
O A
A
B a/2
D
a/2
C
212
PROPERTIES OF TRIANGLE Radius of the circumcircle is given by the following formulae R=
a b c abc 2sin A 2 sin B 2 sin C 4
BD In BDM, = tan A or MD
b
a 2 = tan A, i.e., a = tan A, 2x x
c
A
F
Similarly, 2 y = tan B, = tan C 2z
E
z M A
a b c tan A + tan B + tan C = 2x 2y 2z
and
tan A. tan B. tan C = 2x . 2 y . 2z
a
b
B
y x
D
C
c
But in a triangle ABC,tan A + tan B + tan C = tan A. tan B. tan C
a b c abc x y z 4xyz .
ORTHOCENTRE AND PEDAL TRIANGLE OF A TRIANGLE. In a triangle the altitudes drawn from the three vertices to the opposite sides are concurrent and the point of cuncurrency of the altitudes of the triangle is called the orthocentre of the triangle. The triangle formed by joining the feet of these perpendiculars is called the pedal triangle i.e. DEF is the pedal triangle of ABC. A
F E P 0
90 – C
B
C
D
INCIRCLE The circle which can be inscribed within the triangle so as to touch each of the sides of the triangle is called its incircle. The centre of this circle i.e., the point of concurrency of angle bisectors of the triangle is called the incentre of the ABC. A
E
r I
900– B/2
B/2
B
213
r D
F r
C/2
C
IIT- MATHS Radius of the Incircle is given by the following formulae r=
A B C A B C = (s – a) tan = (s – b) tan = (s – c) tan = 4R sin sin sin . s 2 2 2 2 2 2
ESCRIBED CIRCLES The circle which touches the side BC and the two sides AB and AC produced is called the escribed circle opposite the angle A. Its centre and radius will be denoted by I1 and r1 respectively. Radii of the excircles are given by the following formuale A
(i) (ii)
r1 =
A A B C s tan 4R sin cos cos sa 2 2 2 2
B
C
F1
B A B C s tan 4R cos sin cos r2 = sb 2 2 2 2
E1 L
(iii)
D1
I1
C A B C s tan 4R cos cos sin . r3 = sc 2 2 2 2
M
BISECTORS OF THE ANGLES If AD bisects the angle A and divide the base into portions x and y, we have, by Geometry, x AB c y AC b
x y xy a c b bc bc
ac ab and y = bc bc Also let be the length of AD we have ABD + ACD = ABC
x=
1 A 1 A 1 c sin b sin bc sin A, 2 2 2 2 2
i.e.,
bc sin A 2bc A cos b c sin A b c 2 2
A
B
x
D
y
C
SOLUTION OF TRIANGLES When any three of the six elements (except all the three angles) of a triangle are given, the triangle is known completely. This process is called the solution of triangles. (i) (ii)
b 2 c2 a 2 If the sides a, b and c are given, then cos A = . B and C can be obtained in 2bc the similar way. If two sides b and c and the included angle A are given, then using
tan
BC b c A BC cot , we get . 2 bc 2 2 214
PROPERTIES OF TRIANGLE BC A Also = 900 – , so that B and C can be evaluated. 2 2 The third side is given by a = (iii)
b sin A . sin B
c If two sides b and c and the angle B (opposite to side b) are given, then sin C = sin B, b b sin A give the remaining elements. If b < c sin B, there is sin B no triangle possible (fig 1). If b = c sin B and B is an acute angle, then there is only one
A = 1800 – (B + C) and a =
triangle possible (fig 2). If c sin B < b < c and B is an acute angle, then there are two values of angle C (fig 3). If c < b and B is an acute angle, then there is only one triangle (fig 4). A A
c
c b
B
B
D
(Fig 1)
b c sinB
c sinB
D (Fig 2) A
A b
c b
b D
B
C2
c sinB
b c
c sinB
C2
C1
B
C1
(Fig 4)
(Fig 3)
This case is, sometimes, called an ambiguous case. Let I be the in-centre of O be the circumcentre of the triangle ABC. Let OL be parallel to BC. Let IOL . IM = r OC = R, NOC A
tan
IL IM LM IM ON OL BM BN BM NC
r R cos A B r cot R sin A 2
A B C sin sin R cos A 2 2 2 A B C B 4R sin sin sin .cot R sin A 2 2 2 2
A
4R sin
215
cos A cos B cos C 1 cos A cos B cos C 1 sin A sin C sin B sin A sin C sin B
cos B cos C 1 tan 1 . sin C sin B
I O B
L
NM
C
IIT- MATHS
WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 1
If two sides of a triangle are
(a)
(b)
3
1
6 2
3 2
(c)
and
6 2
1 3
, and the included angle is 600, then the third side is
(d) 2 3
Ans : (B)
Solution : If a represents the third side then 2
2
1 1 a 2 6 2 6 2 cos 60 0 1 1 2x x 6 2 6 2
1 26 2 a 2 6 22 2 26 2
2 a
3 3 a 4 2
ILLUSTRATION : 02 If the angles A, B, C of a triangle ABC are in arithmetical progression then (a) tan A tan C 3 tan A tan C 3
(b) tan A tan C 3 tan A tan C 3
(c) tan A tan C 3 tan A tan C 3
(d) tan A tan C 3 tan A tan C 3
Ans : (C)
Solution : Since A, B, C are in Arithmetical progression
2 B A C also A + B +C = 1800 so that B = 600 In a triangle ABC, we know TanA + tanB + tanC = tanAtanBtanC tanA+tanC =
3 (tanAtanC-(A)
tanA+tanC - 3 tanAtanC = 3
ILLUSTRATION : 03 216
PROPERTIES OF TRIANGLE If the angles of a triangle are in the ratio 1: 3: 5 and denotes the smallest angle, then the ratio of the largest side to the smallest side of the triangle is 3 sin cos 2 sin
(a)
(b)
3 cos sin 2 sin
(c)
cos 3 sin 2 sin
(d)
3 cos sin 2 sin
Ans : (D)
Solution : Since the angles of the triangle are in the ration 1: 3: 5 Let A 20 0 , B 60 0 , C 100 0
We have 3 5 180 0 20 0 ; a b c then from sin A sin B sin C
c sin C sin 120 0 We get the required ratio = a sin A sin
3 cos 1 sin 2 2 = sin
ILLUSTRATION : 04 The expression
a b cb c a c a b a b c is equal to
2 A (a) cos 2
2 A (b) sin 2
2 A (c) cot 2
2 A (d) tan 2
Ans : (C)
Solution : The given expression is equal to
b c2 a 2 a 2 b c 2
b 2 c 2 a 2 2bc
2bc b 2 c 2 a 2
2b cos A 2bc 1 cos A A cot 2 2bc 2bc cos A 1 cos A 2
ILLUSTRATION : 05 In a cyclic quadrilateral ABCD; a, b, c, d denote the length of the sides AB, BC, CD and DA respectively, then cosA is equal to
(a)
a 2 b2 c2 d 2 2ab cd
217
(b)
b2 c2 d 2 a2 2bc da
(c)
c2 d 2 a 2 b 2 2cd ab
(d)
d 2 a 2 b 2 c2 2da bc
IIT- MATHS Ans : (D)
Solution :
D
We have from ABD
c
d
BD2 = a 2 d 2 2ad cos A
C
from BDC
A a
BD2 = b 2 c 2 2bc cos C
B
= b 2 c 2 2bc cos A
Equating the two values we get
b
cos A
d a 2 b2 c 2 2 da bc
ILLUSTRATION : 06 2 p q 1 pq A B tan p , tan q , In a triangle ABC, if then 1 p 2 1 q 2 is equal to 2 2 Ans : (C)
Solution : The given expression is equal to A B A B 2 tan tan 1 tan tan 2 2 2 2 A B 1 tan 2 1 tan 2 2 2 C C A B A B = 2 sin cos 2 cos sin sin C 2 2 2 2 2 2
ILLUSTRATION : 07 In a triangle ABC if (a) an acute angle
cos A cos B cos C a , then A is a b c bc
(b) an obtuse angle
(c) a right angle
(d) equal to B-C
Ans :
Solution : We have
cos A cos B cos C a b c
218
PROPERTIES OF TRIANGLE
b 2 c 2 a 2 c 2 a 2 b 2 a 2 b 2 c 2 2abc a 2 b2 c2 a (given) 2abc bc
a 2 b 2 c 2 2a 2 b 2 c 2 a 2 b 2 c 2 a 2 0 cosA = 0 A = 2
ILLUSTRATION : 08 If A, B, C, D are the angles of a quadrilateral, then
tan A tan B tan C tan D is equal to cot A cot B cot C cot D
(a) tanAtanBtanCtanD
(b) cotAcotBcotCcotD
(c) tan 2 A tan 2 B tan 2 C tan 2 D
(d)
Ans:
tan A tan B tan C
(A)
Solution :
We have tanA B tan 360 0 C D tanC D
tan A tan B tan C tan D 1 tan A tan B 1 tan C tan D
tan A tan B1 tan C tan D 1 tan A tan Btan C tan D 0 tan A tan B tan C tamD
tan A tan B tan C
tan A tan B tan C tan D 1 1 1 1 tan A tan B tan C tan D tan A tan B tan C tan D
=
tan A tan B tan C tan D tan A tan B tan C tan D cot A cot B cot C cot D
ILLUSTRATION : 09 A B acb cos k, then 2 2 2c B C (b) k sin (c) k sin 2 2
In a triangle ABC, if sin A (a) k cos 2
Ans : (D)
Solution : sin
219
A B acb cos k 2 2 2c
C (d) k cos 2
IIT- MATHS
s b s c x ss - b a c b k bc
ca
2c
s b ss c s b k c ab c
cos
C k 2
220
PROPERTIES OF TRIANGLE
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
In any DABC if cot A/2, cot B/2, cot C/2 are in A.P. then a,b,c are in A) A.P
2.
C) H.P
D) none of these
If the triangles A,B, C of a triangle are in A.P and sides a,b,c are in G.P. then a2, b2, c2 are in A) A.P
3.
B) G.P
B) H.P
C) G.P
D) none of these
If twice the squares of the diameter of a circle is equal to half the sum of the squares of the sides of inscribed triangle ABC then sin 2 A sin 2 B sin 2 C is equal to A) 1
4.
B) 2
C) 4
sin A sin( A B ) If in a triangle ABC sin C sin( B C ) then
B) a2, b2, c2 are in A.P D) a2, b2, c2 are in H.P
A), a, b, c in A.P C) a,b,c are in H.P 5.
a 2 b 2 sin( A B ) If in a triangle ABC, 2 then the triangle is a b 2 sin ( A B )
A) right angled or isosceles C) equilateral 6.
1 A
C) G.P
D) none of these
B) G.P
C) H.P
D) none of these
B)
1 A
C) A
1 1 1 A1 A2 A3 D)
1 A2
B) 12 cm
C) 16 cm
D) 18 cm
If orthocentre H of a DABC bisect the altitude AD of the triangle ABC, then value of tanB tan C is A) 3
221
D) none of these
If length of the side BC of a DABC is 6cm and BAC = 120° then the distance between in centre and excentre of the circle touching the side BC internally is A) 10 cm
11.
B) H.P
If A, A1, A2, A3 are the areas of incircle and the ex-circles of a triangle, then
A) 10.
C) G.P
If r 1 2r 2 3r 3 then a, b, c are in A) A.P
9.
B) H.P
In a triangle ABC if tan A/2, tan B/2, tan C/2 are in H.P then a,b,c are in A) A.P
8.
B) right angled and isosceles D) none of these
If in a DABC, cosA +2cosB + cosC = 2 then a, b, c are in A) A.P
7.
D) 8
B) 2
C) 1
D) 0
12.
IIT- MATHS Which of the following pieces of data do not uniquely determine an acute angle triangle ABC (R being the radius of the circumcircle) A) a, sinA, sin B
13.
B) a, b, c
C) a, sin B, R
D) a, sin A, R
In a triangle ABC. Let C / 2 .If r is the inradius and R is the circumradius of the triangle, then 2(r + R) is equal to A) a + b
14.
D) l > 4
B) 3a2 = b2 – 3c2
C) b2 = a2 – c2
D) a2+b2 = 5c2
B) tanB = b/a
D) sin2A+sin2B+sin2C=0
C) cosC =0
B) H.P
C) G.P.
D) None of these
B) 16/9
C) 25/27
D) 27/25
B) 20 3 3
C) 20 3 3
D) 3- 3
If the sides of a triangle are in A.P., and the greatest angle of the triangle exceeds the least by 900 then the sine of the third angle is 7 /4
B)
7 /2
C)
7 /8
D)
7
If one angle of a triangle is 300 and the length of the sides adjacent to it are 40 and 40 3 then the triangle is A) right angled
24.
C) 0 < l < 4
If ABC is a triangle in which B=450, C=1200 and a=40, the length of the perpendicular from A on BC produced is
A) 23.
B) l > 6
In a triangle ABC a : b : c = 4 : 5 : 6 then the ratio of the circumcircle to that of incircle is
A) 3+ 3 22.
D) 4
In a triangle ABC, if tan A/2 = 5/6, and tan B/2 = 20/37 and the sides a, b, c are in
A) 16/7 21.
C) 6
If the tangents of the angles A and B of a triangle ABC, satisfying the equation abx2 -c2x+ab=0 then
A) A.P. 20.
D) 120°
If D is the midpoint BC of a triangle ABC and AD is perpendicular to AC, then
A) tanA = a/b 19.
C) 90°
In a triangle ABC, (a+b+c) (b+c-a) = l bc if
A) 3b2 = a2 – c2 18.
B) 60°
B) 7
A) l < 0 17.
D) a + b + c
Let Tn denotes the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn +1 – Tn = 21 then n equals A) 5
16.
C) c + a
In a triangle ABC 3sinA = 6sinB = 2 3 sin C then angle A is A) 30°
15.
B) b + c
B) isosceles
C) obtuse angled
D) None of these
Points D,E are taken on the side BC of a triangle ABC, such that BD = DE = EC. If
222
PROPERTIES OF TRIANGLE BAD X , DAE Y EAC X then value of
A) 1 25.
B) 2
sin( x y ) sin ( y z ) is equal to sin x sin z
C) 4
D) None of these
If A, B, C are angles of a triangle, then the minimum value of tan a 2 A / 2 tan 2 B / 2 tan 2 C / 2 is equal to A) 0
26.
B) 1
If in a triangle ABC, A) 1/5
27.
C) ½
D) none of these
bc ca a b then cosA is equal to 11 12 13 B) 5/7
C) 19/35
D) 20/35
If P1, P2, P3 are the altitudes of a triangle ABC from vertices A,B, C and D is the area of the triangle then p1 1 p 2 1 p 3 1 is equal to A)
28.
sa
B)
B) G.P
B) /4
2
B)
2 :1:1
B) 5 cm
B) 3 abc
C) 5/12
D) /2
C) 1 : 2 : 1
D) 1 : 1 : 1
C) 5.5 cm
D) 6 cm
C) abc
D) 4 abc
B) sin /n : cos /n D) tan /n : cos /n
P Q In a triangle PQR R / 2 If tan and tan are roots of the equation ax 2 bx c 0 2 2
a 0
then
A) a + b = c
223
b 3 then angle A is equal to c 2
The ratio of the radius of circumcircle and incircle of a regular polygon of side n is A) cosec /n : cot /n C) tan /n : cot /n
34.
D) none of these
The minimum value of bc (b+c) cos A + ca(c+a) cos B + ab (a +b) cos C is equal to A) 2abc
33.
2s
Radius of the circumcircle of ABC is 3cm. If I1, I2, I3 are the centres of the excircles of the triangle ABC then radius of the circumcircle of I1 I2 I3 is A) 4.5 cm
32.
D)
If in a DABC cos B cos C + sin A sin B sin C = 1 then a : b : c is equal to A) 1 : 1 :
31.
s
C) H.P
Angles A, B and C of a triangle ABC are in A.P. If A) /6
30.
C)
If A1, A2, A3 denote the respectively the areas of an inscribed polygon of 2n sides, inscribed polygon of n sides and circumscribed polygon of n sides then A2, A1, A3 are in A) A.P
29.
sb
B) b + c = a
C) a + c = b
D) b = c
IIT- MATHS 35.
There exists a triangle ABC satisfying A) tanA + tanB + tanC = 0 B)
sin A sin B sin C 2 3 7 2 sin A cos A 3
C) (a+b)2 = c2+ab and D) sinA + sinB = 36.
3 1 3 sin A sin B , cosAcosB 2 4
Given an isosceles triangle with equal side of length b, base angle a < p/4, R, r the radii and 0, I the centres of the circumcircle and incircle respectively, then A) R =
1 b cos ec 2
B) D = 2b2 sin2a b cos
b sin 2
C) r = 21 cos
37.
If in a triangle ABC,
D) OI =
B) /4
2 sin
B)
A)
abc 2
C)
sin 2
D) None
bc ca a b r1 r2 r3 r
B) s
C) 2s
D) 3s
B) 2
C) 3
D) 4
In a triangle the length of the two larger sides are 24 and 22 respectively. If the angles are in A.P. then the third side is A) 12 2 3
B) 12 2 3
C) 2 3 2
D) 2 3 2
If A + B + C = , n = z then tan nA + tan nB +tan nC is equal to A) 0 C) tan nA tan nB tan nC
43.
D) 2/3
In any ABC b2 sin2C +C2 sin2B is equal to A)
42.
C) /6
2 sin
39.s In a triangle ABC the value of
41.
2
If the area and an angle of of a triangle are given, then the side opposite to the given angles is minimum when the triangle is isosceles with the length of the equal sides equal to A)
40.
2 sin cos
2CosA CosB 2CosC a b , then ÐA is a b c bc ca
A) /2 38.
3 2
B) 1 D) none of these
AB If in a ABC, a tanA + a tan B = (a+b) tan then 2 224
PROPERTIES OF TRIANGLE A) A = B B) A = -B 44.
B) b
The value of
C) 2b
D) –b
1 1 1 1 2 2 2 is 2 r1 r2 r3 r
A) 0 46.
D) B = 2A
Let in a ABC such that A 45 , A 75 then a + c 2 is equal to A) 0
45.
C) A = 2B
B)
a 2 b2 c2 2
C)
2 a 2 b2 c2
D)
a 2 b2 c2
If the angles of a triangle are 30° and 45° and the included side is ( 3 1) cm then the area of the triangle is A)
47.
1 3 1
B) 6
3 /2
225
B)
5 /2
C) ½
D) ¼
B) tanA+2tanB = 0 C) tanA–2tanB = 0
D) 2tanA–tanB = 0
B) a 2 b 2 3c 2
C) a 2 c 2 2 b 2
D) a 2 b 2 2 c 2
B) obtuse angled
C) isosceles
D) equilateral
B) c 2 a 2 b 2
C) b 2 c 2 a 2
D) c 2 a 2 b 2
The sides of a triangle are in the ratio 1 : 3 : 2 then the angles of the triangle are in the ratio B) 2 : 3 : 4
C) 3 : 2 : 1
D) 1 : 2 : 3
In a triangle, the lengths of the two larger sides are 10 and 9 respectively, if the angles are in A.P., then the length of the third side be A) 5- 6
56.
D) ½
1 In a triangle ABC 2ac sin ( A B C ) is equal to 2
A) 1 : 3 : 5 55.
C) –1
In a ABC if a2 sin (B-C) + b2 sin (C – A) + c2 Sin (A – B) = 0 then triangle is
A) a 2 b 2 c 2 54.
D) 4
In a ABC if median AD is perpendicular to the side AB, then which of the following is true
A) right angled 53.
D) 4
If in a ABC the median AD is perpendicular to the side AB then
A) a 2 c 2 3b 2 52.
3 1
C) 1
B) 0
A) 2tanA+tanB = 0 51.
1
If length of the sides of a DABC are 3,4 and 5 then distance between its incentre and circumcentre is A)
50.
C)
In a if a = t 2 1, b t 2 1c c = 2t then value of r r1 r2 r3 is A) 1
49.
3 1
If in a triangle ABC a = 5, b = 4 and cos (A-B) = 31/32 then the third side ‘c’ is equal to A) 4
48.
B)
B) 3 3
C) 5
If H is the orthocentre of the triangle ABC, then AH is equal to
D) 5+ 6
IIT- MATHS
A) 2R cosA 57.
B) 2R sinA
B) 1 : 3
C) 3 : 5
D) None
The cosine of the obtuse angle formed by the medians drawn from the vertices of the acute angles of an isosceles right angled triangle is A) – 4/5
60.
B) the altitudes are in H.P. D) None of these
In an isosceles right – angled triangle, a straight line is drawn from the midpoint of one of equal sides to the opposite angle. Then the ratio of the tangents of the two parts in two which it is divided by the line is A) 3 : 2
59.
C) a cotA
If in a triangle ABC, sinA, sinB, sinC are in A.P., then A) altitudes are in A.P. C) altitudes are in G.P.
58.
2abc D) cosA
B) 4/5
C) 3/5
D) – 3/5
If P1, P2, P3 are the altitudes of a triangle ABC from the vertices A, B, C and D the area of the triangle then P1 2 P 2 2 P 3 2 is equal to A)
61.
abc
4
a 2 b2 c2 2
D)
abc 2
B) 1/R
C) 1/
D) r/
If P is the product of the sines of angles of a triangle and q the product of their cosines, the tangents of the angles are roots of the equation A) qx 2 px 3 1 q x p 0
B) px 3 qx 2 1 p x q 0
C) 1 q x 3 px 2 qx p 0
D) qx 3 px 2 1 q x p 0
If length of the sides AB, BC and AC of a triangle are 8cm, 15cm, 17 cm respectively, then length of the angular bisector of ABC is A)
64.
2
C)
cosA cos B cos C is equal to P1 P2 P3
A) 1/r
63.
a 2 b2 c2
If P1, P2, P3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides, then
62.
B)
120 2 cm 23
B)
60 2 cm 23
C)
30 2 cm 23
D)
30 23
In a right angle ABC, right angled at B,D and F are the points on BC such that ADB 2ACB and AEB 3 ACB , then ratio of DE and CD will lie in the interval 1 2
A) ,1 65.
In a
1 3
B) ,1
1 2 3 3
C) ,
1 1 3 2
D) ,
ABC , B / 3 and C / 4 Let D divide BC internally in the ratio 1 : 3 then
sin BAD equals sin CAD 226
PROPERTIES OF TRIANGLE A) 1/ 6 B) 1/3 66.
B) 3 3
2bc sin
A 2
B)
2bc cos
bc
A 2
bc
abc
C) 2R b C cosec A/2
D)
4 A cos ec bc 2
B) 3
C) 6
D) 12
B) 1130
C) 930
abc then B = b c2 2
D) None
B) A.P.
C) H.P.
D) None of these
If cos2A + cos2B +cos2C = 1, then the triangle is A) isosceles
227
3 3 2
If the angles A, B and C of a triangle ABC are the sides a, b and c opposite these angles are in G.P. then a2 , b2 , c2 are in A) G.P.
71.
D)
In a triangle ABC, AD is the altitude from A, given b > c, C = 230 and AD= A) 130
70.
C) 3
If in a triangle ABC, a=5, b=4 and cos(A-B) = 31/32 then the third side c is equal to A) 2
69.
2 /3
In a triangle ABC, the length of the bisector of angle A is
A)
68.
D)
Let A 0 A1A 2A 3 A 4 A 5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A 0 A1 , A0A 2 and A0A 4 is A) ¾
67.
C) 1/ 3
B) right angled
C) obtuse angled
D) acute angled
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
If in a triangle ABC, B = 60 then A) (a-b)2 = c2 - ab
2.
1 b cosec a 2
B) 15cm
If ABC : a = 5, b = 4, A =
B) tan-1 (9/40) D) 2tan-1 (1/9)
E) none of these
B) tanB = b/a C) cos C = 0 2 2 2 E) sin A + sin B + sin C = 2
These exists a triangle ABC satisfying
D) sinA + sinB =
B)
sin A sin B sin C 2 2 7
2 (sinA + cosA ) =
3
3 +1 3 , cosA cosB = = sinA sinB 2 4
a c AC In a DABC, 2 cos 2 . Then 2 ( a c2 ac A) B = /3
B) B = C
C) A, B, C are in A.P.
D) B + C = A
In a DABC, tan C < 0. Then A) tanA tan B < 1 C) tanA + tanB + tanC < 0
9.
D) 5 3 1 cm
If tanA, tanB are the roots of the quadratic abx2 - c2x + ab = 0, where a, b, c are the sides of a triangle, then
C) (a+b)2 = c2 + ab and
8.
+ B for the value of angle C 2
A) tanA + tanB + tanC = 0
7.
D) OI = 2 sin cos / 2
C) 5 3 1 cm
A) tanA = a/b D) tan A + tan B = c2/ab 6.
b cos3 / 2
C) r = 21 cos
In DABC, A = 150, b = 10 2 cm the value of ‘a’ for which these will be a unique triangle meeting theese requirement is
A) can not be evaluated C) tan-1 (1/40) 5.
D) a2+b2+c2 = 2b2 + ac
b sin 2
B) D = 2b2 sin 2a
A) 10 2 cm 4.
C) (c-a)2 = b2 - ac
Given an isosceles triangle with equal sides of length b, base angle < /4, r the radii and O, I the centres of the circumcircle and incircle, respectively. Then A) R =
3.
B) (b-c)2 = a2 - bc
B) tanA tanB > 1 D) tanA + tanB + tanC > 0
If the sines of the angles A and B of a triangle ABC satisfy the equation c2x2-c(a+b)x+ab= 0, then the triangle 228
PROPERTIES OF TRIANGLE A) is acute - angled C) is obtuse angled 10.
12.
B) a2 + b2 - ab < c2
3 < 0) then
C) a2+b2 > c2
D) none of these
For a triangle ABC, which of the following is true ? A)
cos A cos B cos C a b c
B)
C)
sin A sin B sin C 3 a b c 2R
D)
cos A cos B cos C a 2 b 2 c 2 a b c 2abc
sin 2A a
2
sin 2B b
2
sin 2C c2
If H is the orthocentre of triangle ABC, then AH is equal to 2abc cos A If the angles of a triangle are in the ratio 2 : 3 : 7, then the sides opposite these angles are in the
A) 2R cosA 13.
3 x2-4x +
In a ABC tanA and tanB satisfy the inequation A) a2+b2 + ab > c2
11.
B) is right-angled D) satisfies sinA + cosA = (a+b)/c
B) 2R sinA
C) a cotA
D)
ratio A) 14.
2 : 2: 3 1
B) 2 :
2 :
3 +1
C)
4
2
:1 :
3 1 2
D) 2
2
4
C) 2cos2
4
D) 2 sin2
If l is the medium from the vertex A to the side BC of a DABC, then B) 4l2 = 2b2+2bc cosA D) 4l2 = (2s-a)2-4bc sin2A/2
If in a ABC, r1 = 2r2 = 3r3, then A) a/b = 4/5
B) a/b = 5/4
If two sides of a triangle are 12 and third side is A) 2 2 - 6
229
B) 2 cos2
A) 4l2 = 2b2+2c2-a2 C) 4l2 = a2+4bc cosA
19.
1
If sinb is the G.M. between sina and cosa, then cos2b is equal to 4
18.
3 1
D)
B) b sinA > a, A > /2 D) b sinA < a, A < /2, b > a
B) - 2
A) 2 sin2 17.
2
If cos (), cos, cos () are in H.P., then cosq sec /2 is equal to A) -1
16.
2 :
These exists a triangle ABC satisfying the conditions A) b sinA = a, A < /2 C) b sinA > a, A < /2
15.
C) 1 :
B)
2 6
C) a/c = 3/5
D) a/c = 5/3
8 ,the angle opposite to the shorter side is 450, then the
C)
6 2
D) none of these
IIT- MATHS
SECTION - A SINGLE ANSWER TYPE QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A
A
C
B
A
A
A
A
B
B
B
16.
17. 18. 19.
20.
21. 22. 23
24. 25.
26.
12. 13.
D
A
27. 28.
14. 15.
C
B
29. 30.
C
A
B,C A,C D
A
B
D
A,B D
C
B
A
A
B
C
B
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
B
A
A
C
AB CD
A
B
A,B
D
A
C
A
B
B
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
A
B
B
B
B
B
C
B
D
A
A
B
C
C
61
62
63
64
65
66
67
68
69
70
71
B
A
A
D
A
C
B
C
B
A
B
A
230
PROPERTIES OF TRIANGLE
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS
1.
2.
3.
4.
CD ACD AD BD
16.
17. 18. 19.
AC
AB BC BC CD
231
5.
6.
7.
8.
9.
10.
11.
12. 13.
14. 15.
ABC CD AC AC BD AB DE
BC
AC ACD AD BC
IIT- MATHS
7
CO-ORDINATE GEOMETRY
232
CO-ORDINATE GEOMENTRY The co-ordination of algebra and geometry is called co-ordinate geometry. Historically, co-ordinates were introduced to help geometry. And so well did they do this job, that the very identity of geometry was changed. The word ‘geometry’ today generally means coordinate geometry. In co-ordinate geometry all the properties of geometrical figures are studied with the help of algebraic equations. Students should note that the object of coordinate geometry is to use some known facts about a curve in order to obtain its equation and then deduce other properties of the curve from the equation so obtained. For this purpose we require a co-ordinate system. There are various types of coordinate systems present in two dimension e.g. rectangular, oblique, polar, triangular system etc. Here we will only discuss rectangular co-ordinate system in detail.
CARTESIAN CO-ORDINATES Let XOX and YOY be two fixed straight lines at right angles. XOX is called axis of x and YOY is called axis of y and O is named as origin. From any point ‘P’ a line is drawn parallel to OY. The directed line OM = x and MP = y. Here OM is abscissa and MP is ordinate of the point ‘P’. The abscissa OM and the ordinate MP together written as (x, y) are called co-ordinates of point ‘P’. Here (x, y) is an ordered pair of real numbers x and y, which determine the position of point ‘P’. Since XOX YOY, this system of representation is called rectangular (or orthogonal) co-ordinate system. When the axes of co-ordinates XOX and YOY are not at right angles, they are said to be oblique axes.
Remarks : Y nd
st
II quadrant
I quadrant
X O th IV quadrant IIIrd quadrant
x
y Lattice Point (w.r.t. co-ordinate geometry) : A point whose abscissa and ordinate both are integers.
233
IIT- MATHS
Distance between two points : The distance between two points P(x1, y1) and Q(x2, y2 ) is given by
Q(x2, y2)
y P (x1,y1)
PQ =
y2
x1 x 2 2 y1 y 2 2
y1
O
x
x1 x2
SECTION FORMULA 1.
If P(x, y) divides the line joining A(x1, y1 ) & B(x2 , y2 ) in the ratio m : n, then (i)
Internal division: x = y=
(ii)
2.
my 2 ny1 mn
External division: x = y=
mx 2 nx1 mn
mx 2 nx1 mn
my 2 ny1 mn
The coordinates of the mid-point of the line-segment joining (x1, y1) and (x2, y2) are x1 x 2 y 1 y 2 , 2 2
Remarks: If the ratio, in which a given line segment is divided, is to be determined, then sometimes, for convenience (instead of taking the ratio m : n), we take the ratio : 1 and apply the formula for internal division. If the value of turns out to be positive, it is an internal division otherwise it is an external division.
Remarks: Points P and Q are said to be harmonic conjugate of each other w.r.t. OA. Incentre and Excentre of a triangle are harmonic conjugate of each other w.r.t. to the angle bisector on which they lie.
Centres connected with a Triangle : (w.r.t. ABC, where A (x1, y1), B (x2, y2), C (x3, y3), BC = a, CA = b & AB = c). Centroid :
The point of concurrency of the medians of a triangle is called the centroid of the triangle. The centroid of a triangle divides each median in the ratio 2 : 1. The coordinates of centroid are given by
234
CO-ORDINATE GEOMENTRY x1 x 2 x 3 y1 y 2 y 3 , . G 3 3
Orthocentre : The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle. The triangle formed by joining the feet of altitudes in a is called the orthic triangle. Here DEF is the orthic triangle of ABC. Incentre :
The point of concurrency of the internal bisectors of the angles of a triangle is called the incentre of the triangle. The coordinates of the incentre are given by ax 1 bx 2 cx 3 ay1 by 2 cy 3 , . I abc abc
Excentre :
Co-ordinate
of
excentre
opposit e
to
ÐA
is
given
by
ax1 bx 2 cx 3 ay1 by 2 cy3 , and similarly for excentres (I2 & I3) opposite to I1 º a bc a bc ax1 bx 2 cx 3 ay1 by 2 cy 3 , B and C are given by I2 º abc a bc A c B
b L C
I1
BL c AI bc , also 1 LC b I1L a ax1 bx 2 cx 3 ay1 by 2 cy 3 , I3 abc a bc
Circumcentre : The point of concurrency of the perpendicular bisectors of the sides of a triangle is called circumcentre of the triangle.
Remarks : 1. Circumcentre O, Centroid G and Orthocentre H of a ABC are collinear. G Divides OH in the ratio 1 : 2, i.e. OG : GH = 1 : 2 235
IIT- MATHS 2. In an isosceles triangle centroid, orthocenter, incentre and circumcentre lie on the same line and in an equilateral triangle all these four points coincide.
Area of a triangle : Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is 1 [x (y – y )+ x2 (y3 – y1) + x3 (y1 – y2)] 2 1 2 3
=
x1
y1 1
1 x2 2 x 3
y2 1 y3 1
.......(1)
.......(2)
While using formula (1) or (2), order of the points (x1, y1), (x2, y2) and (x3, y3) has not been taken into account. If we plot the points A(x1, y1), B(x2, y2) and C(x3, y3), then the area of the triangle as obtained by using formula (1) or (2) will be positive or negative as the point A, B, C are in anti-clockwise or clockwise directions,
So, while finding the area of triangle ABC, we take modulus.
Remarks : In case of polygon with vertices (x1, y1), (x2, y2), ....... (xn, yn) in order, then area of polygon is given by
1 |(x1y2 – y1x2) + (x2y3 – y2x3) + .....+ (xn – 1yn – yn – 1xn) ) + (xny1 – ynx1)| 2
LOCUS When a point moves in a plane under certain geometrical conditions, the point traces out a path. This path of the moving point is called its locus.
Equation of locus The equation to a locus is the relation which exists between the coordinates of any point on the path, and which holds for no other point except those lying on the path. In other words equation to a curve (or locus) is merely the equation connecting the x and the y coordinates of every point on the curve.
Procedure for finding the equation of the locus of a point : (i) If we are finding the equation of the locus of a point P, assign coordinates (h, k) or (x1, y1) to P. (ii) Express the given conditions in terms of the known quantities to facilitate calculations. We sometimes include some unknown quantities known as parameters. 236
CO-ORDINATE GEOMENTRY (iii)Eliminate the parameter. So that the eliminant contains only h, k and known quantities. If h and k coordinates of the moving point are obtained in terms of a third variable ‘t’ called the parameter, eliminate ‘t’ to obtain the relation in h and k and simplify this relation. (iv)Replace h by x, and k by y, in the eliminant. The resulting equation would be the equation of the locus of P.
x=
0 3X 0 3Y ,y= 1 3 1 3
from which X =
4 4 x, Y = y . 3 3
Substitute these values, then the locus of P is
8 x + 4y + 4 = 0 3
2x + 3y + 3 = 0.
STRAIGHT LINE Any equation of first degree of the form ax + by + c = 0, where a, b, c are constants always represents a straight line (at least one out of a and b is non zero)
Slope If a straight line makes an angle ‘’in anticlockwise direction with the positive direction of x-axis, 0º < 180º, ¹90º, then the slope of the line, denoted by ‘m’ is tan. i.e. m = tan`. If A(x1, y1 ) and B(x2, y2 ), x1 x2 are any two points, then slope of the line passing through y 2 y1
A and B is given by m = x x . 2 1 Remark : (i)
If 900 , m does not exist and line is parallel to y - axis.
(ii)
If = 0°, m = 0 and the line is parallel to x-axis. Let m1 and m2 be slopes of two given lines. (a) If lines are parallel, m1 = m2 and vice versa. (b) if lines are perpendicular, m1.m2 = -1 and vice versa.
(iii)
Position of a given point relative to a given line : The fig. Shows a point P(x1, y1) lying above a given line. If an ordinate is dropped from P to meet the line L at N, then the x coordinate of N will be x1. Putting x = x1 in the equation ax + by + c = 0 gives y coordinate of
237
N=–
(ax1 c) b
IIT- MATHS
If P(x1, y1) lies above the line, then we have y1 > –
(ax1 c) b
i.e. y1 +
i.e.
(ax1 by1 c) >0 b
i.e.
L ( x 1 , y1 ) > 0 .......(1) b
(ax1 c) >0 b
Hence, if P(x1, y1) satisfies equation (1), it would mean that P lies above the line ax + by + c = 0, and if
L ( x 1 , y1 ) < 0, it would mean that P lies below the line ax + by + c = 0. b
Remark : If (ax1 + by1 + c) and (ax2 + by2 + c) have same signs, it implies that (x1, y1) and (x2, y2) both lie on the same side of the line ax + by + c = 0. If the quantities ax1 + by1 + c and ax2 + by2 + c have opposite signs, then they lie on the opposite sides of the line.
Intercept of a straight line on the axis : If a line AB cuts the x-axis and y-axis at A and B respectively and O be the origin then OA and OB with proper sign are called the intercepts of the line AB on x and y axes respectively.
Standard equations of straight lines : 1 Slope-intercept form : y = mx + c, where m = slope of the line = tanq c = y intercept
2 Intercept form : x/a + y/b = 1
y (0, b)
x y 1 a b
x intercept = a, length of x intercept = |a| x
y intercept = b, length of y intercept = |b|
(0, 0)
(a, 0)
238
CO-ORDINATE GEOMENTRY
3 Normal form : x cos + y sin = p, where , is the angle which the perpendicular to the line makes with the axis of x and p is the length of the perpendicular from the origin to the line. 0 2 and p is always positive.
(0,0)
X
p+ ve
L
Y
4.4.4 Slope point form : Equation : y – y1 = m(x – x1), where (a) One point on the straight line is (x1, y1) and (b) The direction of the straight line i.e., the slope of the line = m
4.4.5 Two point form : y 2 y1 Equation : y – y1 = x x (x – x1), where (x1, y1) and (x2, y2) are the two given points. Here 2 1 y 2 y1 m= x x . 2 1
Parametric equations of a straight line : In figure given below let BAP be a straight line through a given point A (x1, y1), the angle of slope being q. The positive direction of the line is in the sense BAP. (Direction of increasing ordinate is called the positive direction of the line). For the points P (x,y) and Q (X, Y) Shown in the figure AP is regarded as a positive vector and AQ as a negative vector, as indicated by the arrows. From the general definitions of cosq and sinq we have cosq =
x x1 y y1 , sinq = AP AP
or x – x1 = AP cosq, y – y1 = AP sinq.
x x1 y y1 r cos sin
239
IIT- MATHS or r (4 +
21 ) + 71 = 0 5
or r = –
355 . The distance between A and P is thus 355/41 units, 41
the vector AP being in the negative direction of the line.
Length of the perpendicular from a point on a line : The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is
ax1 by1 c
.
a 2 b2 c
The length of the perpendicular from origin on ax + by + c = 0 is
2
a b2
.
The distance between two parallel lines : The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is
| c1 c 2 | a 2 b2
.
Reflection of a point about a line : The image of a point (x1, y1) about the line ax + by + c = 0 is x x1 y y1 ax by c = = – 2 1 2 12 a b a b
and the foot of perpendicular from a point (x1, y1) on the line ax + by + c = 0 is x x1 y y1 ax by c = = – 1 2 12 . a b a b
FAMILY OF LINES: (Equation of any straight line through the point of intersecton of two given straight lines). The equation of any straight line passing through the intersection of the two lines ax + by + c = 0, Ax + By + C = 0 has the general form ax + by + c + (Ax + By + C) = 0 In which can have any real value ; here, is parameter which can be evaluated specifically if some further condition is imposed. Hence the general equation of the family of lines through the point of intersection of two given lines is L + L = 0 where L = 0 and L = 0 are the two given lines, and is a parameter. Conversely, any line of the form L1 + L2 = 0 passes through a fixed point which is the point of intersection of the lines L1 = 0 and L2 = 0. In other words if a linear expression L contains an unknown coefficient, then the line L = 0 can not be a fixed line. Rather it represents a family of straight lines.
Remarks : 1. If L1 = 0 and L2 = 0 are parallel lines, they will meet at infinity. 240
CO-ORDINATE GEOMENTRY 2. The family of lines perpendicular to a given line ax + by + c = 0 is given by bx - ay + k = 0, where k is a parameter. 3. The family of lines parallel to a given line ax + by + c = 0 is given by ax + by + k = 0, where k is a parameter.
CONCURRENCY OF STRAIGHT LINES : The condition for three lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is -
(i)
a1
b1
c1
a2 a3
b2 b3
c2 = 0. c3
(ii) There exist three constants , m, n (not all zero the same time) such thatL1 + mL2 + nL3 = 0, where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines. (iii) The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines.
THE ANGLE BETWEEN TWO STRAIGHT LINES : In fig. given below, GAH and LAM are two straight lines meeting the Y
K
M H A
2
1 O
G
L
X
x-axis at G and L and intersecting at A. The angles of slope are 1 and 2; the corresponding gradients are given by Let,
m1 = tan 1 , m2 = tanq2
= 2 – 1
.......(1)
........(2)
Thus, f is the angle through which GA has to be rotated about G in the counter-clockwise direction to be parallel to, and in the same sense as, LA. From (2), tan = tan(2 – 1).
tan 2 tan 1 = 1 tan tan , 2 1
m 2 m1 or, by means of (1), tan = 1 m m 1 2
.........(3)
This is the formula required ; from it we can calculate from the given-or deducible-values of the gradients of the two given lines. In numerical examples the value of the right-hand side of (3) may be positive or negative ; if the value is positive, the angle is acute ; if the value is negative, the angle is obtuse.. 241
IIT- MATHS It is a convention to tell acute angle for the angle between the two lines. For this purpose tan
m 2 m1 = 1 m m , where is the acute angle. 1 2 Remarks : 1. If the lines are parallel then 2 = 1 and, by (2), = 0 that tan = 0; thus, from (3), m2 = m1, which is otherwise obvious from (1). 1 m1m 2 2. If the lines are parallel then = 90º so that cot = 0 ; from (3), cot = m m and it 2 1
follows that, since m1 and m2 are unequal, then 1 + m1m2 = 0 or m1m2 = – 1, which is the condition that the two lines should be perpendicular.
BISECTORS OF THE ANGLES BETWEEN TWO GIVEN LINES : Angle bisector is the locus of a point which moves in such a way so that its distance from two intersecting lines remains same. The equations of the two bisectors of the angles between the lines a1x + b1y + c1 = 0 and a 1x b1 y c1
a2x + b2y + c2 = 0 are
a12 b12
a 2 x b2 y c2
=±
a 22 b 22
If the two given lines are not perpendicular i.e. a1 a2 + b1b2 0 and not parallel i.e. a1 b2 a2b1 then one of these equations is the equation of the bisector of the acute angle between two given lines and the other that of the obtuse angle between two given lines.
Remarks: Whether both given lines are perpendicular or not but the angular bisectors of these lines will always be mutually perpendicular.
The bisectors of the acute and the obtuse angles Take one of the lines and let its slope be m1 and take one of the bisectors and let its slope be m2. If
m1 m 2 be the acute angle between them, then find tan = 1 m m 1 2
242
CO-ORDINATE GEOMENTRY If tan > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle. If 0 < tan < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angles. If two lines are a1 x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then a 1x b1 y c1 a12 b12
C N
a 2x b2 y c2 A
a 22 b 22
P(x, y)
M B
will represent the equation of the bisector of the acute or obtuse angle between the lines according as a1a2 + b1b2 is negative or positive.
The equation of the bisector of the angle which contains a given point : The equation of the bisector of the angle between the two lines containing the point (a,) is a 1x b1 y c1 2 1
2 1
a b
=
a 2 x b2 y c2 2 2
a b
2 2
or
a 1x b1 y c1 2 1
2 1
a b
=–
a 2 x b2 y c2 a 22 b 22
according as a1 + b1 + c1 and a2
+ b2 + c2 are of the same signs or of opposite signs. For example the equation of the bisector of the angle containing the origin is given by a 1x b1 y c1 2 1
2 1
a b
a 2 x b2 y c2
=+
a 22 b 22
for same sign of c1 and c2 (for opposite sign take –ve sign in
place of +ve sign)
Remarks: (i)
If c1c2 (a1a2 + b1b2) < 0, then the origin will lie in the acute angle and if c1c2 (a1a2 + b1b2) > 0, then origin will lie inthe obtuse angle.
(ii)
Equation of straight lines passing through P(x1, y1) and equally inclined with the lines a1x + b1y + c1= 0 and a2x + b2y + c2 = 0 are those which are parallel to the bisectors between these two lines and passing through the point P.
The equation of reflected ray : Let L1 a1x + b1y + c1 = 0 be the incident ray in the line mirror L2 a2x + b2y + c2 = 0. Let L3 be the reflected ray from the line L2. Clearly L2 will be one of the bisectors of the angles between L1 and L3. Since L3 passes through A, so L3 L1 + L2 = 0 Let (h, k) be a point on L2. Then,
243
IIT- MATHS | a 1h b1k c1 | 2 1
2 1
a b
=
| a 1h b1k c1 (a 2 h b 2 k c 2 ) | (a 1 a 2 ) 2 (b1 b 2 ) 2
.
Since (h, k) lies on L2, a2h + b2k + c2 = 0
a12 + a22 2 + 2a1a2 + b12 + b222 + 2b1b2 = a12 + b12
= 0 or =
2(a1a 2 b1b 2 ) a 22 b 22
But = 0 given L3 = L1. Hence L3 L1 –
2(a 1a 2 b1b 2 ) L2 = 0. a 22 b 22
Remarks : Some times the reflected ray L3 is also called the mirror image of L1 in L2.
ROTATION OF THE AXES (To change the direction of the axes of co-ordinates, without changing the origin, both systems of co-ordinates being rectangular.) Let OX, OY be given rectangular axes with respect to which the coordinates of a point P are (x, y). Suppose that OU, OV are the two perpendicular lines obtained by rotating OX, OY respectively through an angle a in the counter-clockwise sense. We take OU, OV as a new pair of coordinate axes, with respect to which the coordinates of P are (x, y), then x x cos y sin y x sin y cos
x cos sin y = sin cos
or,
x y
(in matrix form)
PAIR OF STRAIGHT LINES The general equation of degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight a
h g
lines if h b f 0 g
f
c
abc + 2fgh – af 2 – bg2 – ch2 = 0 and h2 ³ ab. The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin If lines through the origin whose joint equation is ax2 + 2hxy + by2 = 0, are y = m1x and y = m2x, then y2 – (m1 + m2)xy + m1m2x2 = 0 and y2 +
2h a xy + x2 = 0 are identical. If is the angle between the b b
244
CO-ORDINATE GEOMENTRY
m1 m 2 2 4m1m 2
two lines, then tan = ±
1 m1m 2
2 h 2 ab ab
The lines are perpendicular if a + b = 0 and coincident if h2 = ab. Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Line and a Curve A
B
O
If the lines x + my + n = 0, ((n 0) i.e. the line not passing through origin) cuts the curve ax + 2hxy + by2 + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line i.e. 2
2
x my x my ax + 2hxy + by + (2gx + 2fy) c 0 is the equation of the lines n n 2
OA and OB
245
2
IIT- MATHS
WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 If , , are real roots of the equation x 3 3px 2 3qx 1 0 . Find the centroid of the triangle
1 1
1
whose vertices are , , and Solution : Since , , are the roots of the equation x 3 3px 2 3qx 1 0
+ + = 3p, + + = 3q, = 1
1
1
1
Let A = , , B , and C , Let G(x,y) the centroid of ABC, then X=
3p p 3 3
1 1 1 3q And y q 3 3
Hence co-ordinates of the centroid of ABC are (p,q) ILLUSTRATION : 02 The four points A (,0), B (,0), C (,0) and D( ,0 ) are such that , are the roots of equation ax 2 2 hx b 0 and , are those of equation a' x 2 2 h' x b' = 0. Show that the sum of the ratios in which C and D divide AB is zero if ab’ + a’b = 2hh’. Solution : Since , are the roots of x 2 2hx b 0 + =
2h b and a a
……. (1)
and , are the roots of a ' x 2 2h ' x b' 0 then
2h ' b' and a' a'
……...(2)
Let C divides AB in the ratio : 1
246
CO-ORDINATE GEOMENTRY 1. Then 1
and let D divides AB in the ratio : 1 then
. 1. 1
but given + = 0
0
2 2 0
2h 2h ' 2h 2b ' 0z a' a a ' a
or
ab’ + a’b = 2hh’
ILLUSTRATION : 03 Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0. Solution : Let x = t, then y = 4-t. Let P (t, 4 – t) be an arbitrary point on the line x +y = 4 Distance of P from 4x + 3y – 10 = 0 is unity
4t 3 4 t 10 | 42 32
|t + 2| = 5
t+2 ±5
or
t = -2 ± 5
t = 3, -7
points are (3,1) & (-7,11)
247
1
IIT- MATHS ILLUSTRATION : 04 Find a point P on the line 3x + 2y + 10 = 0 such that |PA - PB| is maximum where A is (4,2) and B is (2,4) Solution : Let P be x1 , y1 And APB 2
3
PA PB AB
Then cos =
2
2 PA.PB
Since cos 1 2
2
PA PB AB
2
2 PA.PB 2
2
1
2
PA PB AB 2 PA.PB 2
PA PB AB
2
PA PB AB | PA PB | 2 2 Maximum value of |PA - PB| is 2 2 when = 0 i.e, P lies on the line AB as well as on the given line. equation of AB is y2
42 x 4 24
y – 2 = -x + 4
x+y=6
and given line 3x +2y + 10 = 0
…… (1) …….(2)
Solving (1) and (2), we get P (-22, 28) ILLUSTRATION : 05 Find the co-ordinates of the orthocentre of the triangle formed by the lines y = 0, (1+t)x – ty + t (1+t) = 0 and (1 +u) x – uy + u (1+u) = 0 t 0 , and show that for all values of t and u, the orthocentre lies on the line x +y = 0. Solution : Let equations of BC, CA and AB are y=0, (1+t)x–ty+t (1+t)=0 and (1+u)x–uy+u (1+u) = 0 respectively. Let (h,k) be the orthocentre then Slope of OB x slope of AC = -1
248
CO-ORDINATE GEOMENTRY k 0 1 t 1 hu t
k (1+t) = -th – tu
k=-
th tu k tk th tu 1 t
……. (1)
and slope of OC x slope of AB = -1 k0
h t k
1 u 1 u
u h t k ku uh ut 1 u
……(2)
Subtracting (2) from (1) , we get K (t -u) = -h (t – u) h = k = 0 locus of orthocentre is x + y = 0 Putting h = -k in (1) Then we get k = -t u h = tu then orthocentre is (t,u – tu) ILLUSTRATION : 06 If m 1 and m 2 are the roots of the equation x 2
32x
3 1 0 . Show that the area of
33 11 2 c the triangle formed by the liens y = m1x, y m 2 x and y c is 4 Solution Since m 1 and m 2 are the roots of the equation x2
32x
3 1 0
then m1 m 2 3 2
249
m1 m 2
=
3 4 4
=
11
3 1 0
m1 m 2 2 4m1m 2 34 34
IIT- MATHS and co-ordinates of the vertices of the given triangle are (0,0), (c/m1,c) and (c/m2,c). Hence the required area of triangle
=
0 1 c 2 m1 c m2
0 1 c 1 c 1
=
1 2 1 1 c 2 m1 m 2
=
1 2 m 2 m1 c 2 m 1m 2
=
1 2 c 2
=
1 2 c . 2
11
3 1
33 11 2 c } 4 3 1
11 3 1
3 1
ILLUSTRATION : 07 One side of a rectangle lies on the line 4x + 7y + 5 = 0. Two of its vertices are (-3,1) and (1,1). Find the equations of other three sides Solution : Since (-3,1) lies on 4x + 7y + 5 = 0 And (1,1) does not lie on 4x + 7y + 5 = 0 Equation of AD, which is to 4x +7y + 5 = 0 And passing through (-3,1) is 7x – 4y + = 0
- 21 – 4 + = 0
= 25 Therefore, equation of AD is 7x – 4y + 25 = 0 (1,1) does not lie on AD
Co-ordinate of C is (1,1) Equation of BC which is parallel to AD and Passing through (1,1) is 7x – 4y + = 0 250
CO-ORDINATE GEOMENTRY 7x 1 – 4 x 1 + = 0 = -3
equation of AD is 7x – 4y –3 = 0 Equation of DC which is parallel to AB passing through (1,1) is 4x + 7y + = 0 4 x 1 + 7 + 1 + = 0 = -11
Therefore equation of DC is 4x + 7y –11 = 0 ILLUSTRATION : 08 A line through the variable point A (k+1, 2k) meets the lines 7x +y – 16 = 0, 5x – y – 8 = 0 at B,C,D respectively. Prove that AC, AB, AD are in H.P. Solution : Given lines are 7x – y – 16 = 0
…… (1)
5x – y – 8 = 0
…… (2)
x – 5y +8 = 0
…… (3)
Let the equation of line passing through A (k+1, 2k) making an angle with the + v e direction of x – axis be x k 1 y 2k r1 , r2 , r3 cos sin
(if AB = r1 , r2 , r3 ) (if AB = r1 , AC = r2 , AD = r3 )
B [(k+1) + cos , 2k + sin ] C [(k+1) + cos , 2k + sin ] D [(k+1) + cos , 2k + sin ]
Points B, C, D satisfying (1), (2) and (3) respectively Then r1 r2 =
r3
251
91 k 7 cos sin
31 k 5 cos sin
91 k 5 sin cos
IIT- MATHS
1 1 5 cos sin 5 sin cos r2 r3 31 k 91 k
=
15 cos 3 sin 5 sin cos 91 k
=
14 cos 2 sin 91 k
2
= r 1
Hence r2 , r1 , r3 are in H.P. ILLUSTRATION : 09 The three sides of a triangle are L r x cos r y sin r p r 0 , where r = 1,2,3. Show that the orthocentre is given by L1 cos 2 3 L 3 cos 3 1 L 3 cos 1 2 Solution : The given lines are L1 x cos 1 y sin 1 p1 0 L 2 x cos 2 y sin 2 p 2 0 L 3 x cos 3 y sin 3 p 3 0
Now equation of AD is …. (1)
L 2 L 3 0
(xcos 2 y sin 2 p 2 ) + (x cos 3 y sin 3 p 3 ) 0 (xcos 2 cos 3 ) + (x sin 2 sin 3 )p 2 p 3 0
cos
cos
2 3 Slope of AD = sin sin 2 3
cos
1 and slope of BC = sin 1
Since AD BC
Slope of BC x slope of AD = -1
cos 2 cos 3 cos 1 1 sin 2 sin 3 sin 1
cos 1 cos 2 cos 3 cos 1 sin 1 sin 2 sin 3 sin 1
cos 1 2 cos 3 1 0
cos1 2 cos3 1
Now from (1), 252
CO-ORDINATE GEOMENTRY L2
cos 1 2 L3 0 cos 3 1
L 2 cos 3 1 L 3 cos 1 2
. ….....(2)
Similarly, we can obtain equation of altitude BE as L 3 cos1 2 L1 cos 2 3
.…….(3)
From (2) and (3), we get L1 cos 2 3 L 2 cos 3 1 L 3 cos 1 2
ILLUSTRATION : 10 A (3,0) and B (6,0) are two fixed points and U (, ) is a variable point on the plane. AU and BU meet the y – axis at C and D respectively and AD meets OU at V. Prove that CV passes through (2,0) for any position of U in the plane. Solution : The equation of BU is Y-=
0 x 6
So that the coordinates of D are 0,
6 6
Similarly the coordinates of C are 0,
3 3
Now, the equation of AD is x 6 y 1 3 6
and the equation of OU is x = y Solving (1) and (2), we get x
6 6 ,y 6 6
6 6 , Hence coordinates of V are 6 6 Then the equation of CV is 6 3 3 y 6 3 x 0 6 3 0 6
253
IIT- MATHS
y
3 9 x 3 63
y=
3 x 1 3 2
Which pass through the point (2,0) for all values of (, ). ILLUSTRATION : 11 A variable line is drawn through O to cut two fixed straight lines L 1 and L 2 in R and S. A point S is chosen on the variable line such that
mn m n . Show that the locus P is a OP OR OS
straight line passing through the point of intersection L 1 and L 2 . Solution : Let the equation of the variable line through ‘O’ be
x y and let cos sin
OR r1 .OS r2 and OP r3
Then co-ordinates of R, S and P are R r1 cos , r1 sin , Sr2 cos , r2 sin , P r3 cos , r3 cos .
R lies on and lies on L 1 and S lies on L 2 r1 sin c and ar2 cos br2 sin 1 r1
c 1 and r2 sin a cos b sin
From the given condition mn m n r3 r1 r2
From the given condition mn m n r3 r1 r2
mn m n r3 r1 r2
mn m sin n a cos b sin = r3 c
…… (1)
Let co-ordinates of P be (h,k) then h = r3 cos , k r3 sin from (1),
mn
mr3 sin n ar3 cos br3 sin c
mn
mk n ah bk c
locus of P is 254
CO-ORDINATE GEOMENTRY n ax by
my m n c
y n ax by 1 m 1 0 c
ax by 1 m y c 0 nc
L1
m L2 0 nc m where nc
L1 L 2 0
Locus of P is point of intersection of L 1 and L 2
255
IIT- MATHS
SECTION A SINGLE ANSWER TYPE QUESTION 1.
If A and B are two points having co-ordinates (3,4) and (5,-2) respectively and P is a point such that PA = PB and area of triangle PAB = 10 sq units then co-ordinates of P are A) (7,4) or (13,2)
2.
B) (7,2) or (1,0)
C) (2,7) or (4,13)
D) (1,2) or (2,1)
The position of a moving point in the x – y plane at time is given by (ucosa.t, u sina. t -
1 2 9t ) 2
where u, a, g are constants. The locus of the moving point is A) a circle 3.
B) an ellipse
B) x + 2y = 5
D) x + 2y = 1
C) (-2,-2)
D) (1,1)
B) circle
C) pair of st lines
D) parabola
The equations of the lines through (-1,-1) and making an angle 45° with the line x + y = 0 are given by B) xy + x – y – y2 = 0 D) xy + x +y +1 = 0
Let the algebraic sum of the perpendicular distances from the points (2,0), (0,2) and (1,1) to a variable line is zero, then the line passing through a fixed point whose co-ordinates are A) (1,2)
10.
C) 2x + y = 5
B) (2,2)
A) x2 – xy + x – y = 0 C) xy +x + y = 0 9.
B) xsin + ycos = 2a sin2 D) xsin -y cos =2a sin2
The locus of a point P which divides the line joining (1,0) and (2 cos, 2sin) internally in the ratio 2 : 3 for all is a A) straight line
8.
D) hyperbola
One vertex of an equilateral triangle with centroid at the origin and one side as x+y–2=0 is A) (-1, -1)
7.
C) a parabola
The equation of the line passing through the intersection of the line x – 3y + 1 = 0 and 2x + 5y – 9 = 0 and 0 at distance 5 from the origin is A) 2x – y = 5
6.
D) hyperbola
The equation of a line which passes through (acos3, a sin3) and perpendicular to the line xsec + ycosec = a A) xcos - ysin =2a cos2 C) xsin + ycos = 2a cos2
5.
C) an ellipse
If A and B are fixed point then the locus of a point which moves in such a way that the angle APB is a right angle is A) a circle
4.
B) a parabola
B) (1,1)
C) (2,1)
The product of the perpendiculars from the points
D) (2,2)
5 ,0 5 ,0
to the straight line
2x cos - 3y sin = 6 A) 5
B) independent of q
C) cos2q
D) 7sin2q 256
CO-ORDINATE GEOMENTRY 11. Circum center of the triangle whose vertices are (2,-1) (3,2) and (0,3) is A) (1,-1) 12.
B) (-1,1)
B) - 54
B) –2
C) 2
B) 0
B) 2
B) 5 units
C) 1
C) triangle not possible
B) 2x + y – 7 = 0; 3x – 5y + 2 = 0 D) 2x – y + 5 = 0
B) equilateral
C) right angled
D) none of these
B) 3 : 4
C) 2 : 1
D) 4 : 3
B) 0
C) 4
D) 3
Let PS be the median of the triangle with vertices P(2,2) Q(6,-1) R (7,3). The equation of the line passing through (1, -1) and parallel PS is A) 2x – 9y – 7 = 0 C) 2x + 9y – 11 = 0
257
5 are
The number of integer values of m for which the x –coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is A) 2
23.
D) 10 units
A straight line through the origin 0 meets the parallel lines 4x + 2y = 9 and 2x + y +6 = 0 at points P and Q respectively then the point O divides the segment PQ in the ratio A) 1 : 2
22.
D) ½
The straight line x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is A) isosceles
21.
D) –1
The equation of the line through the point of intersection of the lines x – 3y + 1 = 0 and
A) 3x + 2y – 7 = 0, 5x – 7y + 12 = 0 C) 2x + y – 5 = 0
20.
D) 1
C) ½
2x + 5y – 9 = 0 and whose distance from the origin is
19.
D) 24
If co-ordinates of the vertices B and C of a ABC are (0,0) and (5,0) respectively and co-ordinates of incentre of ABC is (3,4) then length of the side AC is A) 25 units
18.
C) a and b both true
Number of lines passing through (3,4) and whose difference of the intercept is 2 A) 4
17.
D) 1 : 2 externally
Let equation of the BC of a ABC is 2x + 3y –1 = 0 co-ordinates of the incentre and circum centre of ABC are (2,1) and (-1,3) respectively then value of cosB + cosC is A) 1
16.
C) 2 : 1 externally
2 2 If the points at 1 2at 1 at 2 at 2 and (a,0) are collinear then value of t 1t 2 is
A) –1 15.
B) 1 : 1 externally
Distance between the lines 5x + 12y – 1 = 0 and 10x + 24y + k = 0 is 2 then the value of K is A) 50
14.
D) (-1,-1)
Ratio in which the join of (2,1) and (-1,2) is divided by the line x + 3y + 5 = 0 A) 1 : 1 internally
13.
C) (1,1)
B) 2x – 9y – 11 = 0 D) 2x + 9y + 7 = 0
The orthocentre of the triangle formed by the lines xy = 0 and x +y = 1 is
IIT- MATHS 1 1 2 2
1 1 3 3
A) , 24.
B) ,
1 1 4 4
D) ,
C) (0,0)
The area of the triangle formed by joining the origin to the points of intersection of the line 5x 2y 3 5 and circle x2 + y2 = 10 is
A) 6 25.
B) 5
B) 1/2
C) 2
D) 3
The vertices of a triangle are A(-1,-7) , B (5,1) and C(1,4). The equation of the bisector of ABC is A) x – 7y + 2 = 0
27.
D) 3
The area bounded by the curves x + 2|y| = 1 and x = 0 A) 1/3
26.
C) 4
B) x + 7y – 2 = 0
C) x + 6y + 2 = 0
D) x – 7y – 2 = 0
If a, c, b are in G.P. then the line ax + by + c = 0 A) has a fixed direction B) always passes through a fixed point C) form a triangle with the axes whose area is constant D) always cuts intercepts on the axes such that their sum is zero
28.
Area of the triangle formed by the lines y2 – 9xy +18x2 = 0 and y = 9 is A) 27/4
29.
30.
B) 0
1
1
1
1
1
1
1
1
A) a 2 b 2 p 2 q 2
B) a 2 b 2 p 2 q 2
C) a 2 p 2 b 2 q 2
D) a 2 p 2 b 2 q 2
Two points A and B move on the x–axis and the y–axis respectively such that the distance between the two points is always the same. The locus of the middle point of AB is B) a circle
C) a parabola
D) an ellipse
The range of values of the ordinate of a point moving on the line x=1, and always remaining in the interior of the triangle formed by the lines y=x, the x–axis and x+y=4, is A)(0,1)
32.
D) 27
Line L has intercepts a and b on the co–ordinate axes, when the axes are rotated through a given angle; keeping the origin fixed, the same line has intercepts p and q, then
A) a straight line 31.
C) 9/3
B)(0,1)
C)(0,4)
D) None of these
Let A=(1,0) and B(2,1). The line AB turns about A through an angle /6 in the clockwise sense, and the new position of B is B’. The B’ has the coordinates. 3 3 3 1 A) 2 , 2
3 3 3 1 , 2 2
B)
1 3 1 3 , 2 2
C)
D) None of these
258
CO-ORDINATE GEOMENTRY 33.
The bisector of the acute angle formed between the lines 4x–3y+7=0 and 3x–4y+14=0 has the equation A) x+y–7=0
34.
B)x–y+3=0
B) (2,–1)
B) =–3
3 x+y=4
5 2
5 1 2 2
1 5 2 2
B) ,
C) ,
B) x2 = 9y2
D) None of these
D) None of these
C) x2 – 9y2 = 0
D) y2 – 4x2 = 0
The point P (1,1) is translated parallel to y = 2x in the first quadrant through a unit distance. The co-ordinates of the new position of P are
2
5
,1
1 5
B) 1
1
,1
5
5
1
C)
5
,
2 5
2
D)
5
,
1 5
If P and P’ be the perpendiculars from the origin upon straight lines xsec + ysec = a and xcos - y sin = a cos2 respectively, then the value of expression 4p2 +P’2 B) 3a2
C) 2a2
D) 4a2
The line 3x + 2y = 24 meets y – axis at A and x – axis at B. The perpendicular bisector of AB meets the line through (0,-1) parallel to x – axis at C. Then area of D ABC is A) 182 sq units
259
C) x+ 3 y=4
Let O be the origin and A, B be the two points having co-ordinates (0,4) and (6,0) respectively. If a point P moves in such a way that the area of the OPA is always twice the area of POB then P lies on
A) a2
45.
D) (a+b, b)
If P(1+t/ 2 ,2+t/) be any point on a line then the range of values of t for which the point P lies between the parallel lines x+2y=1 and 2x+4y=15 is
A) 1
44.
D) =4
C) (0,–a)
B) x+ 3 y+4=0
A) y2 = 9x2
43.
2
The coordinates of two consecutive vertices A and B of a regular hexagon ABCDEF are (1,0) and (2,0) respectively. The equation of the diagonal CE is
3 2
42.
C) =4
B) (a,0)
A) , 41.
t m 2lm
2
If (a,b) be an end of a diagonal of a square and the other diagonal has the equation x–y=a then another vertex of the square can be
A) 39.
1 D) tan
If the lines x - 2y - 6 = 0, 3x + y - 4 = 0 and lx + 4y + l2 = 0 are concurrent, then
A) (a–b,a) 38.
D) None of these
2 2 1 t m tan C) t 2 m2
B) /2
A) =2 37.
C) (0,4)
The diagonals of the parallelogram whose sides are lx+my+n=0, lx+my+ n =0, mx+ly+n=0, include an angle A) /3
36.
D) x=2y–12=0
The equations of the three sides of a triangle are x=2, y+1=0 and x+2y=4. The co–ordinates of the circumcentre of the triangle are A) (4,0)
35.
C) 3x+y–11=0
B) 91 sq units
C) 48 sq units
D) 100 sq units
The line PQ whose equation is x – y = 2 cuts the x – axis at P and Q is (4,2). The line PQ is rotated about P through 45° in the anticlockwise direction. The equation of the line PQ in the new posi-
IIT- MATHS tion is A) y = 2 46.
B) y = 2
C) x = 2
D) x = -2
I f the co-ordinates of the vertex A of a ABC is (1,2) and equation of the perpendicular bisectors
of AB and AC are 3x + 4y – 1 = 0 and 4x + 3y – 5 = 0 then the equation of median AD is
47.
A) 11x – 10 y + 9 = 0
B) 10 x – 11y + 12 = 0
C) 3x + 4y –11 = 0
D) 3x + 4y + 12 = 0
Let the equation of the side BC of ABC is x + y + 2 = 0. If co-ordinates of its orthocentre and circum centre are (1,1) and (2,0) respectively, then radius of the circum circle of ABC is A) 3
48.
B) 10
C) 2 2
D)
2
Let co-ordinates of the vertices A and B of a triangle ABC are (6,0) and (0,6) respectively and co-
21 21 ordinates of its orthocentre is , then co-ordinates of its circum center is 4 4 13 13 , 16 16
11 11 , 16 16
A) 49.
B)
2
2 2 ,2 2 3
B) 2 3 x y 1
54.
2 and
1
D) 1 and
2
3
C) y 2 3 x 2
D) x y 4
B) y – 2 = 0, 4x – 3y = 6 D) none of these
Let P (-1,0) Q (0,0) and R (3, 3 3 ) be three pts then the equation of the bisector of the angle PQR is A)
53.
C)
The equation of the lines through the point (2,3) and making an intercept of length 2 units between the lines y +2x =3 and y + 2x = 5 are A) x + 3 = 0, 3x + 4y = 12 C) x – 2 = 0, 3x +4y = 18
52.
31 31 , 8 8
D)
A ray of light traveling along the line x + y = 1 is incident on the x – axis and after refraction is incident on the x-axis and after refraction it enters the other side of the x-axis by turning /6 away from the x-axis. The equation of the line along which the refracted ray travels is A) x + 2 3 y 1
51.
15 15 , 16 16
C)
Let L be the lines 2x + y = 2. If the axes are rotated by 45° without transforming the origin, then the intercepts made by the line L on the new axes are respectively A) 1 and
50.
B)
3x x y0 2
B) x 3y 0
C)
3x y 0
D) x
3 y0 2
Let PQR be a right angled isosceles triangle, right angled at P(2,1). If the equation of the line QR is 2x +y = 3 then the equation representing the pair of lines PQ and PR is A) 3x 2 3y 2 8xy 20 x 10y 25 0
B) 3x 2 3y 2 8xy 20x 10 y 25 0
C) 3x 2 3y 2 8xy 10x 15y 20 0
D) 3x 2 3y 2 8xy 10x 15y 20 0
In an isosceles triangle ABC, the coordinates of the points B and C on the base BC are respec260
CO-ORDINATE GEOMENTRY tively (2,1) and (1,2). If the equation of the line AB is y =
A) 2y = x + 3 55.
B) y = 2x
B) 4
6 2 2
60.
D) 3x + 4y = 2
B) x – y = 1
C) x = 0
5 2
D) x =1
B) c2 – a2 = m2
C) 2c2(1+m2) = a2
D) c2 + a2 = m2
C) y–3x+9=0, 3y–x=3=0
D) y–3x+3=0, 3y+x+9=0
On the portion of the straight line x+y=2 which is intercepted between the axes, a square is constructed away from the origin, with this portion as one of its side. If p denote the perpendicular distance of a side of this square from the origin, then the maximum value of p is
2
B) 2 2
C) 3 2
D) 4 2
Two vertices of a triangle are (3,–2) and (–2, 3) and its orthocentre is (–6, 1). Then its third vertex is B) (–1,6)
C) (1,–6)
D) None of these
A ray of light coming from the point (1,2) is reflected at a point A on the x–axis and then passes through the point (5,3). The co–ordinates of the point A is 3 ,0 5
261
C) 3x + 4y = 7
B) y+3x+9=0, 3y+x–3=0
A) 63
62 2
A) y– 3x+9=–, 3y+x–3=0
A) (1,6) 62.
B) 3x + 4y =
If one of the diagonal of a square is along the line x=2y and one of its vertices is (3,0) then its sides through this vertex are given by the equation
A) 61.
D) 5
The pair of straight lines joining the origin to the common points of x2 +y2 = a2 and y = mx +c are perpendicular to each other if A) 2c2 =a2 (1+m2)
59.
C) 8
If a ray traveling along the line x = 1 gets reflected from the line x + y = 1 then the equation of the line along which the reflected ray travels is A) y = 0
58.
D) y = x –1
There are two parallel lines, one of which has the equation 3x + 4y = 2. If the lines cut an intercept of length 5 on the line x +y = 1 then the equation of the other line is A) 3x 4 y
57.
1 x 1 2
Let A = (1,2) B (3,4) and let C (x,y) be a point such that (x-1) (x-3) + (y-2) (y-4) = 0. If ar D (ABC) =1 then the maximum number of position of C in the x – y plane is A) 2
56.
C) y =
1 X then the equation of the line AC is 2
5 ,0 13
B)
C) 7,0
D) None of these
In a ABC, side AB has the equation 2x+3y=29 and the side AC has the equation x+2y=16. If the mid point of BC is (5,6) then the equation of BC is
A) 2x+y=7 64.
88 3
B)
C)
1
B)
x
2
1
y
2
4 p
69.
x
2
1 y
2
2 p2
B) x2 – y2 = a2 – b2 D) x2 + y2 = a2 – b2
B) (1,-3)
C) (-1,1)
D) (3,3)
Two sides of an isosceles triangle are given by the equation 7x – y + 3 = 0 and x+y–3=0. If its third side passes through the point (1,-10) then its equation are A) x – 3y – 7 = 0 ; 3x + y – 31 = 0
B) x – 3y – 31 = 0 ; 3x + y – 7 = 0
C) x – 3y – 31 = 0 ; 3x + y + 7 = 0
D) x + 3y – 31 = 0 ; 3x + y + 7 = 0
In a rhombus ABCD the diagonals AC and BD intersect at the point (3,4). If the point A is (1,2) the diagonal BD has the equation B) x + y – 1 = 0
C) x + y + 1 = 0
D) x + y – 7 = 0
If each of the points (x, 4) and (-2,y) lies on the line joining the points (2, -1) and (5,-3) then the point P(x1y1) lies on the line A) x = 3y
71.
1
D)
The image of the point (-1,3) by the line x – y = 0 is
A) x - y + 1 = 0 70.
4
The locus of the point of intersection of lines x cos + y sin = a and xsin - ycos = b is ( is a variable)
A) (3, -1) 68.
D) None of these
3
2 2 C) x y p 2
2
A) 2(x2 + y2) = a2 + b2 C) x2 + y2 = a2 + b2 67.
4 7
The locus of the midpoint of the position intercepted between the axes by the line x cos + y sin = P where P is a constant is A) x2 + y2 = 4p2
66.
C) 2x–y=17
ABC is an equilateral triangle such that the vertices B and C lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is A) 8
65.
B) x+y=1
IIT- MATHS D) None of these
B) x = -3y
C) y = 2x + 1
D) 2x + 6y +1 = 0
The ratio in which the line 3x + 4y+2 = 0 divides the distance between 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is A) 7 : 3
72.
C) 2 :3
D) 1 : 2
If the foot of the perpendicular form the origin to the straight line is at the point (3,-4) then the equation of the line is A) 3x – 4y = 25
73.
B) 3 : 7
B) 3x – 4y +25 = 0
C) 4x + 3y – 25 = 0
D) 4x – 3y +25 =0
Point Q is symmetric to P(4,-1) with respect to the bisector of the first quadrant. Then length of PQ is 262
CO-ORDINATE GEOMENTRY A) 5 2 B) 5 74.
B) (1,-1)
9 9 2 2
C) (1,1)
7 7 2 2
B) ,
B) 6 sq units
D) 4 sq units
B) x2 + 2x – 4y + 5 = 0 D) x2 + 2x + 4y – 5 = 0
B) identical
B)
C) parallel
D) none
1 10 2
C)
5 2
D) 5
a2 2
D) 3a 2
Area of the quadrilateral formed by |x| + |y| = a is A) a 2
81.
C) 11/2 sq units
Distance between orthocentre and circumcentre of the triangle whose vertices are (3,-1) (2,1) (0,0) is A) 10
80.
D) 1,1
If the points a 1b1 a 2 b 2 a 3b 3 are collinear then the lines aix biy 1 0 i = 1, 2, 3 are A) concurrent
79.
11 11 , 2 2
C)
Locus of all such points which is equidistant from (1,2) and x – axis is A) x2 – 2x – 4y + 5 = 0 C) x2 – 2x + 4y + 5 = 0
78.
D) (3,5)
If the Co-ordinates of the midpoints of the sides of a triangle are (1,-1) (2,3) (3,2) then area of triangle is A) 5 sq units
77.
5
In center of the triangle whose vertices are (6,0), (0,6) and (7,7) is A) ,
76.
D)
Orthocentre of the triangle whose vertices are (1,1) (3,5) (3,0) is A) (-1,1)
75.
C) 2 5
B) 2a 2
C)
If one vertex of an equilateral triangle of side 2 is the origin and another vertex lies on the line x = x 3 y then the third vertex can be A) (0,2)
82.
B)
B) y = ¼
B) 2
mn
263
2 ,1
x at an angle 45° is
C) y = ½
C) 2 2
2 2
m n
D) y =1
D) 4
B) m n
1
C) m n
1
D) m n
If the vertices P,Q,R of a triangle PQR are rational points, which of the following points of the triangle PQR is always irrational A) Centroid
86.
D)
Area of the parallelogram formed by the lines y = mx, y = mx +1, y = nx and y = nx+1 equals A)
85.
C) (0,2)
The area a bounded by the curves y = |x| -1 and y = - |x| +1 is A) 1
84.
2 ,1
The line which is parallel to x-axis and crosses the curve y = A) x = ¼
83.
B) Incentre
C) circumcentre
D) orthocentre
The equation of the straight line passing through the point (-2,3) and making intercepts of equal
IIT- MATHS length on the axes is A) 2x + y + 1= 0 87.
2
96.
D) none of these
B) ellipse
C) hyperbola
D) circle
B) b, 2a, c are in G.P
C) b, a/2, c are in A.P.D) b, –2a,c are in G.P
B) (2,1)
C) (1,2)
D) None of these
B) 2x 1 y 1 1
C) x 1 2 y 1 1
D) 2x 1 2 y 1 1
B) ay–bx+2b=0
C) ax+by+2b=0
D) None of these
If a ray traveling the line x=1 gets reflected the line x+y=1 then the equation of the line along which the reflected ray travels is A) y=0
95.
2
1 2h C) ab 1 n
The straight line y=x–2 rotates about a point where it cuts x–axis and becomes perpendicular on the straight on the line ax+by+c=0 then its equation is A)ax+by+2a=0
94.
2
n 2h B) ab 1 n
A line passing through P(4,2) meets the x and y–axis at A and B respectively. If O is the origin, then locus of the center of the circumcircle DOAB is A) x 1 y 1 2
93.
D) (a - b)2 = 4h2
L is a variable line such that the algebraic sum of the distances of the points (1,1),(2,0) and (0,2) from the line is equal to zero. The line L will always pass through. A) (1,1)
92.
C) a – b = 2h
If bx +cy=a, where a,b,c are the same sign, be a line such that the area enclosed by the line and the axes of reference is 1/8 unit2 then A) b,a,c are in G. P
91.
B) a + b = -2h
Let AB be a line segment of length 4 with the point A on the line y=2x and B on the line y=x. Then locus of middle point of all such line segment is a A) parabola
90.
D) x +y –2 =0
The slope of one of the lines represented by ax2 + 2hxy +by2 = 0 be n times the other than n 1 n A) ab 2h
89.
C) x – y + 5 = 0
If one of the lines of the pair ax2 + 2hxy + by2 = 0 bisects the angle between positive direction of the axes, then a, b, h satisfy the relation A) a + b = 2h
88.
B) x – y = 5
B) x–y=1
C) x=0
D) None of these
A line passing through the point (2,2) and the axes enclose an area . The intercepts on the axes made by the line are given by the two roots of A) x 2 2 | | x | | 0
B) x 2 | | x 2 | | 0
C) x 2 | | x 2 | | 0
D) None of these
Let A=(1,2), B=(3,4) and let C=(x,y) be a point such that (x–1)(x–3)+(y–2)(y–4)=0. If ar(DABC) = 1 then maximum number of positions of C in the x–y plane is. A) 2
B) 4
C) 8
D) None of these
264
CO-ORDINATE GEOMENTRY 97. The limiting position of the point of intersection of the lines 3x+4y=1 and 91+c)x+3c2y=2 as c tends to 1 is A) (–5,4) 98.
C) (4,–5)
D) None of these
If the point (a,a) falls between the lines x + y = 2 then A) |a| = 2
99.
B) (5,–4)
B) |a| = 1
C) |a| < 1
D) |a| < 1/2
If A (cos, sin) B (sin, cos) C(1,2) are vertices of a ABC then as varies the locus of its centroid is A) x2 + y2 – 2x - 4y + 1 = 0
B) 3 (x2+ y2) – 2x – 4y + 1 = 0
C) x2 + y2 – 2x – 4y + 3 = 0
D) x2 + y2 +2x + 4y – 3 = 0
100. If a line joining points A(2,0) and B(3,1) is rotated through A in anticlockwise direction through an angle 15°, then the equation of the line in the new position is A)
3x y 2 3
B) 3x y 2 3
C) x 3y 2 3
D) 3x y 3
101. If a straight line L perpendicular to the line 5x – y = 1 such that the axes of the D formed by the line L and the co-ordinate axes is 5, then the equation of the line L is A) x + 5y + 5 = 0
B) x + 5y ±
2 0
C) x + 5y ± 5 0
D) x 5y 5 2 0
102. The equations of the lines on which the perpendicular form the origin make 30° angle with x – axis and which form a triangle of area 50/ 3 with axes are A) x +
3 y ± 10 = 0
B) 3x y 10 0
C) x 3y 10 0
D) 3x y 10 0
103. If the extremities of the base of an isosceles triangle are the points (2a,0) and (0,a) and the equation of one of the sides is x = 2a then the area of the triangle is A) 5 sq units
B) 5/2 sq units
C) 25/2 sq units
D) 2 sq units
104. If one vertex of an equilateral triangle is at (2,-1) and the base is x + y – 2 = 0, then the length of each side is A)
3 2
B)
2 3
C)
2 3
D)
3 2
105. If co-ordinates of orthocentre and centroid of a triangle are (4,-1) and (2,1) then co-ordinates of a point which is equidistant from the vertices of the triangle is A) (2,2)
B) (3,2)
C) (2,3)
D) (1,2)
106. Let co-ordinates of the two fixed points A and B are (a,0) and (0,b) respectively. A variable line meet the axes at P and Q so that BP is always perpendicular to AQ. Then locus of the point of intersection of BP and AQ is B) x 2 y 2 ax by 0 D) x 2 y 2 ax by 0
A) ax + by + a + b = 0 C) y 2 4a x b
107. Locus of the centres of the circles touching the line 3x – 4y + 1 = 0 and 12x + 5y – 1 = 0 are A) 21x+77y –18 = 0 265
B) 99x–27y+8=0
C) (A) and (b) both
D) none of these
IIT- MATHS 108. Equation of the line passing through (1,1) and give an intercept between the lines 5x+12y+7=0 and 5x + 12y – 32 = 0 of length 3 unit is A) 12x – 5y – 7 = 0
B) 12x+5y+7 = 0
C) 12x + 5y – 7 = 0
D) 12y + 5x – 7 = 0
109. Equations of the straight lines, inclined at 30° to the axis of x such that the length of its (each of their) lines segments between the co-ordinates axes is 10 units is A) x 3y 5 3 0 C) x 3y 5 3 0
B) x 3y 5 3 0 D) x 3y 5 3 0
110. P(3,1) Q(6,5) and R(x,y) are three points such that the angle PRQ is a right angle and the area of PQR = 7 then the number of such points R is A) 0
B) 1
C) 2
D) 4
111. The vertices of a triangle ABC are (1,1) (4,-2) (5,5) respectively. The equation of perpendicular dropped from C to the internal bisector of angle A is A) y – 5 = 0
B) x – 5 = 0
C) 2x + 3y – 7 = 0
D) x + 5 = 0
112. Let 0 < < /2 be a fixed angle. If P (cos, sin) and Q (cos ( - ), sin ( - ) then Q is obtained from P by A) clockwise rotation around origin through an angle B) anticlockwise rotation around origin through an angle C) reflection in the line through origin with slope tan D) reflection in the line through origin with slope tan /2 113. The angle between a pair of tangent s drawn from point P to the circle x 2 y 2 4x 6 y 9 sin 2 cos 2 0 is 2. The equations of locus of point ‘p’ is A) x 2 y 2 4x 6 y 4 0
B) x 2 y 2 4x 6 y 9 0
C) x 2 y 2 4x 6 y 4 0
D) x 2 y 2 4x 6 y 9 0
114. In the ABC the co-ordinates of B are (0,0) AB = 2 ABC = /3 and the middle point BC has the co-ordinates (2,0) the centroid of the triangle is
1 3 A) 2 , 2
5 1
B) 3 , 3
4 3 1 C) 3 , 3
D) none
115. The four sides of a quadrilateral are given by the equation xy (x-2) (y-3) = 0. The equation of the line parallel to x – 4y = 0 that divides the quadrilateral in two equal areas is A) x – 4y + 5 = 0
B) x – 4y –5 = 0
C) 4y = x + 1
D) 4y + 1 = 0
116. The range of values of the ordinate of a point moving on the line x = 1 always remaining in the interior of the triangle formed by the lines y = x and the x –axis, x +y =4 is A) (0,1)
B) [0,1]
C) [0,4]
D) (0,2)
117. If a pair of lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 is such that each pair bisects the angle between the other pair then 266
CO-ORDINATE GEOMENTRY A) pq = -1
1
1
C) p q 0
B) pq =1
1
1
D) p q 0
118. The four straight lines given by the equations 12x 2 + 7xy – 12y2 = 0 and 12x2 + 7xy – 12y2 – x + 7y –1 = 0 lie along the sides of a A) square 119
B) parallelogram
C) rectangle
D) rhombus
If P (1+ / 2 , 2 / 2 ) be any point on a line then the range of values of t for which the point P lies between the parallel lines x+2y=1 and 2x=4y=15 is A)
4 2 5 2 3 6
B) 0
5/ 2 6
C)
4 2 0 3
D) None of these
120. The point (4,1)undergoes the following two successive transformations : A) reflection about the line y = x B) rotation through a distance 2 unit along the positive x–axis. Then the final co–ordinates of the point are A) (4,2)
B) (3,4)
C) (1,4)
121. In the ABC, the coordinates of B are (0,0), AB=2, ABC=
D) (7/2, 7/2)
and the middle point of BC has 3
the coordinates (2,0). The centroid of the triangle is 1
3
5
A) 2 , 2
B)
,
3
4 3 1 , 3 3
1 3
C)
D) None of these.
122. There are two parallel lines, one of which has the equation 3x+4y=2. If the lines cut an intercept of length 5 on the line x+y=1 then the equation of the other line is A) 3x 4 y
6 2 2
B) 3x 4 y
62 2
C)3x+4y=7
D) None of these
123. Let P=(1,1) and Q=(3,2). The point R on the x–axis such that PR+RQ is the minimum is 5 3
A) ,0
1 3
B) , o
C) (3,0)
D)None of these
124. If () be an end of a diagonal of a square and the other diagonal has the equation x–y= then another vertex of the square can be A) (a–b,a)
B)(a,0)
C) (0,a)
D) None of these
125. The point (–4,1) undergoes the following three transformations successively (I) Reflection about the line y=x
267
IIT- MATHS (II) Transformation through a distance 2 units along the positive direction of x–axis (III) Rotation through an angle p/4 about the origin in the anticlockwise direction. The final position of the point is given by the co–ordinates 1
A)
2
,
7 2
B) 2,7 2
C)
1
,
2
7 2
D)
2 ,7 2
126. P is a point on either of two lines y– 3 x 2 at a distance of 5 units from their point of intersection. The co–ordinates of the foot of perpendicular from P on the bisector of the angle between them are
1 1 4 5 3 or 0, 4 5 3 ,( depending on which line the point P is taken) 2 2
1 45 3 2
A) 0, B) 0,
C) 0,
1 45 3 2
5 5 3 D) 2 , 2
127. The equation of a line through the point(1,2) whose distance from the point(3,1) has the greatest possible value is A) y=x
B) y=2x
C) y=–2x
D) y=–x
128. The point P(2,1) is shifted by 3 2 parallel to the line x+y=1, in the direction of increasing ordinate, to reach Q. The image of Q by the line x+y=1 is. A)(5,–2)
B)(–1,–2)
C)(5,4)
D)(–1,4)
268
CO-ORDINATE GEOMENTRY
SECTION -B MULTIPLE ANSWER TYPE QUESTIONS 1.
The points (2, 3) (0, 2) (4, 5) and (0,t) are concyclic of the value of t is A) 1
2.
3.
B) 1
C) 17
The point of intersection of the lines
D) 3
x y x y =1 and 1 lies on a b b a
A) x-y = 0
B) (x+y) (a+b) = 2ab
C) (lx + my) (a + b) = (1 + m) ab
D) (lx–my) (a+b) = (1-m) = ab
The equations (b-c)x + (c-a) y + a-b= 0 (b3-c3) x + (c3-a3)y + a3-b3 = 0 will represent the same line if A) b =c
4.
6.
269
C) x2-y2 = 2(ax+by)
B) bx = ay
B) = -3
C) = 4
B) x + y - 3 = 0
B)
D) P can be (a, b)
D) = -4
C) x - 3y - 5 = 0
D) x - 3y + 5 = 0
3 a / 2, a / 2
C) (0, -a)
D) 3 a / 2, a / 2
If the lines ax + by + c = 0, bx + cy =a = 0 and cx + ay + b = 0 are concurrent (a+b+c 0) then A) a3 + b3 + c3 - 3abc = 0 C) a = b = c
11.
D) (-1/4, 11/4)
If one vertex of an equilateral triangle of side a lies at the origin and the other lies on the line x 3y =0, the co-ordinates of the third vertex are A) (0, a)
10.
C) (7/2, 13/2)
Equation of a straight line passing through the point of intersection of x-y+1 = 0 and 3x+y-5 = 0 are perpendicular to one of them is A) x + y + 3 = 0
9.
B) (3/4, -3/2)
If the lines x - 2y - 6 = 0, 3x + y -4 = 0 and x + 4y + 2 = 0 are concurrent, then A) = 2
8.
D) a + b + c = 0
The points (k, 2-2k), (-k+1, 2k) and (-4-k, 6-2k) are collinear for 1 A) any value of k B) k=1 C) k = D) no value of k 2 If the point P(x, y) be equidistant from the points A(a+b, a-b) and B(a-b, a+b) then A) ax = by
7.
C) a = b
The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x+3. The co-ordinates of the third vertex can be A) (-3/2, 3/2)
5.
B) c = a
B) a = b D) a2 + b2 + c2 - bc - ca - ab = 0
If the co-ordinates of the vertices of a triangle are rational numbers then which of the following points of the triangle will always have rational co-ordinates
A) centroid 12.
B) incentre
Let S1, S2 .... be squares such that four each n ³ 1, length of a side of Sn equals the length of a diagonal of Sn+1. If the length of a side of S1 is 10cmm, then for which of the following values of n is the area of Sn less than 1 sq. cm ? A) 7
13.
B) 8
C) 9
1 unit2 then 8
A) b, a, c, are in G.P C) b,
15.
A) bisector of the angle including origin
B) bisector of acetic angle
C) bisector of obtuse angle
D) none of these
Two roads are represented by the equations y - x = 6 and x + y = 8. An inspection bunglow has to be so constructed that it is at a distance of 100 from each of the roads. Possible location of the bunglow is given by B) (1-100 2 , 7)
C) (1, 7 + 100 2 )
D) (1, 7 - 100 2 )
Angles made with the x-axis by two lines drawn through the point (1, 2) cutting the line x + y =
A)
5 and 12 12
6 /3 from the point (1, 2) are B)
7 11 and 12 12
C)
3 and 8 8
D) none of these
If (a, b) be an end of a diagonal of a square and the other diagonal has the equation x-y = a then another vertex of the square can be A) (a–b, a)
18.
D) b, -2a, c are in G.P.
Consider the straight lines x + 2y + 4 = 0 and 4x + 2y - 1 = 0. The line for 6x + 6y + 7 = 0 is
4 at a distance
17.
B) b, 2a, c are in G.P.
a , c are in A.P.z 2
A) (100 2 + 1, 7) 16.
D) 10
If bx + cy = a, where a, b, c are the same sign, be a line such that the area enclosed by the lime and the axes of reference is
14.
C) circumcentre
IIT- MATHS D) orthocentre
B) (a, 0)
C) (0, –a)
D) (a+b, b)
The points (p+1, 1), (2p+1, 3) and (2p+2, 2p) are collinear if A) p = -1
B) p = 1/2
C) p = 2
D) y = -
1 2
270
CO-ORDINATE GEOMENTRY
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
B
B
A
A
C
C
B
D
B
B
C
16.
17. 18. 19.
20.
21. 22. 23
24. 25.
26.
12. 13.
B
C
27. 28.
14. 15.
C
A
29. 30.
A
C
C
A
B
A
D
C
B
B
A
C
A
B
B
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
A
A
B
A
B
A,B B,D C
A
A,B
B
B
A
A
C
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
A
B
D
B
A
C
C
B
D
A
A
A
B
A
61
62
63
64
65
66
67
68
69
70
71
72 73
74
75
B
C
B
C
A
C
D
D
B
A
C
A
271
C
A
IIT- MATHS
76
77
78
79
80
81
82
83
84
85
86
87 88
89
90
B
A
A
B
B
A
C
B
D
B
C
B
B
B,D
91
92
93
94
95
96
97
98
99 100 101 102 103 104 105
A
B
B
A
C
B
A
C
B
106 107 108 109
B
121
A
C
A
A
122 123 124
B
B
B
A
D
110 111 112 113 114 115 116
C
B
D
B
A
B
A
B
A
B
B
D
117 118 119 120
B
B
B
D
125 126 127 128
D
272
CO-ORDINATE GEOMENTRY
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS
1.
2.
4.
5.
6.
10.
11.
12. 13.
14. 15.
AC
AB AB AC CD CD
BC
BD AD BD AB AB CD CD
AC D
BC BD D
AB AB CD
16.
17. 18.
AB
BD CD
273
3.
7.
8.
9.
IIT- MATHS
THREE DIMENSIONAL GEOMETRY Distance bet ween any two point s P x1 , y1 , z1 and Q x 2 , y 2 , z 2 in space is given by
2
2
PQ x 2 x1 y 2 y1 z 2 z1
1 2 2
___
If P x1 , y1 , z1 , Q x 2 , y 2 , z 2 are two points, the points R which divides PQ in ratio l : m is given mx1 lx 2 my1 ly 2 mz1 lz 2 , , by R lm lm lm
___
If the ratio l : m in positive, R lies between P and Q on PQ and if the ratio of negative, R lies on ___ ___ extension of PQ when l : m is positive, we say R divides PQ internally and when the ratio is negative, ___ we say R divides PQ externally.. x1 x 2 y1 y 2 z1 z 2 ___ , , (i) Mid point of PQ = 2 2 2 (ii) If P x1, y1 , z1 Q x 2 , y 2 , z 2 R x3 , y 3 , z 3 are vertices of a triangle, the centroid (or the center of gravity) of DPQR is G. Then (iii)If P x1, y1 , z1 Q x2 , y2 , z2 R
x3 , y3 , z3 and S x4 , y4 , z4
are the vertices of a
xi yi zi tetrahedron, its center of gravity (or) centroid is given by 4 , 4 , 4
PARALLEL SHIFTING OF AXES If the origin is changed to x1 , y1 , z1 through parallel shifting of axes [i.e. the new axes are parallel to original axes and passing through the point x1 , y1 , z1 ] then x X+ x1 , y Y+ y1 , z = Z+ z1 where P x, y, z changes to (X, Y, Z ) in new axes. If a line makes angles , , with X, Y, Z axes respectively, the triple cos , cos , cos is called a triple of the direction cosines of the given line. (i)
If
cos , cos , cos is
a t riple of direct ion cosines of a line, we get cos cos cos 1. Note that direction cosines (D.C’s) of a line are components of a unit vector parallel to the given to the given line and two such triplets of D.C’s exist for any line namely cos , cos , cos , cos180 , cos180 , cos180 2
(ii)
2
2
If ‘O’ is the origin and P = x1 , y1 , z1 then D.C’s of OP are
x1 y1 z1 , , 2 2 2 2 x2 y2 z2 x1 y1 z1 x1 y12 z12 1 1 1
274
CO-ORDINATE GEOMENTRY (iii) Any triple of numbers proportional to the D.C’s of a line is called a triple of direction ratios (D.R’s) of the line. i.e. If l , m, n is a triple of D.C’s of a line, then lk , mk , nk , k R -{0} is a triple of D.R’s of the given line. If a, b, c is a triple of D.R’s of a line, then
(iv)
a b c , , 2 2 2 a2 b2 c2 a2 b2 c2 a b c
are triple of D.C’s of the given line.
If P x1 , y1 , z1 and Q x 2 , y 2 , z 2 are two points, a triple of D.R’s of PQ is given by x2 x1 , y 2 y1 , z2 z1
(v)
If two rays have D.C’s l1 , m1 , n1 and l 2 , m2 , n2 respectively, the angle between them is given by
l m
cos l1l 2 m1m2 n1 n2 (or) sin
1
l 2 m1
2
2
The lines are perpendicular if and only if l1l 2 m1m2 n1n 2 0 l1 m1 n1 The lines are parallel if and only if l m n 2 2 2 If two rays have D.R’s a1 , b1 , c1 a 2 , b2 , c 2 respectively, the angle between them is given by
(vi)
cos
a1a 2 b1b2 c1c 2 a12 b12 c12 a 22 b22 c 22
(or)
sin
a b
1 2
2
a 2 b1
a12 b12 c12 a 22 b22 c 22
The lines are perpendicular a1a 2 b1b2 c1c 2 0 a1 b1 c1 The lines are parallel a b c 2 2 2
PROJECTION Projection of a point P x, y, z in XOY plane is x, y,0 , in YOZ plane is 0, y, z and in XOZ plane is x,0, z
Projection of the line segment joining P x1 , y1 , z1 and Q x 2 , y 2 , z 2 (i)
on X - axis = x 2 x1
(ii) on Y - axis = y 2 y1 (iii) on Z - axis = z 2 z1 (iv) on any line with the D.C’s l , m, n is x 2 x1 l y 2 y1 m z 2 z1 n
PLANE The line joining two points on a plane totally lie on the same plane. (i)
275
The general equation of a plane is ax by cz d 0
(ii)
IIT- MATHS If a plane cuts the X, Y, Z axes at a,0,0 0, b,0 , 0,0, c respectively, its equation is given by x y z 1 (Intercept form) a, b, c are called the X, Y, Z - intercepts of the plane respeca b c tively.
(iii)
If the perpendicular from (0, 0, 0) to a plane has D.C’s l , m, n and is of length ‘p’ then the equation of the plane is lx my nz p (Normal form). Hence for a given plane |d|
ax by cz d 0 , the perpendicular distance from (0, 0,0) is (iv)
.
From the above we observe that the D.R’s of any normal to the plane ax by cz d 0 are proport ional to
a, b, c
and the D.C’s of any line normal to this plane are
a b c , , 2 2 2 a2 b2 c2 a2 b2 c2 a b c
(v)
a2 b2 c2
.
The equation of the plane through x1, y1 , z1 , whose normal has D.R’ss a, b, c is given by a x x1 b y y1 c z z1 0 . Also the equation of the plane parallel to Ax By Cz D 0 through x1 , y1 , z1 is A x x1 B y y1 C z z1 0
(vi)
The equation of the plane passing through three non collinear points A x1 , y1 , z1 , B x 2 , y 2 , z 2
x x1 x x1 and C x3, y 3, z 3 is 2 x3 x1
y y1 y 2 y1 y3 y1
z z1 z 2 z1 0 z3 z1
STRAIGHT LINE If P x, y, z is any point in the space, (i)
foot of the perpendicular from P to X -axis is x,0,0 and the perpendicular distance of P to X-axis is
(ii)
foot of the perpendicular from P to Y -axis is 0, y ,0 and the perpendicular distance of P to Y-axis is
(iii)
x2 z 2
foot of the perpendicular from P to Z -axis is 0,0, z and the perpendicular distance of P to Z-axis is
(iv)
y2 z2
x2 y2
If R x, y, z is any point on the line passing through P x1 , y1 , z1 and Q x 2 , y 2 , z 2 we get that
x x1 , y y1 , z z1 and x2 x1 , y 2 y1 , z 2 z1 are two triples of D.R’s of the line PQ and 276
CO-ORDINATE GEOMENTRY they must be proportional (i.e.)
x x1 y y1 z z1 t (say) x 2 x1 y 2 y1 z 2 z1
Hence we get x x1 t x 2 x1 , y y1 t y 2 y1 , z z1 t z 2 z1 . Therefore any point on the line can be taken in the form for some t R.
277
IIT- MATHS
SOLVED PROBLEM ILLUSTRATION : 01 The perpendicular distance of (3, 4, 5) from the Z-axis is (a)
26
(b)5
(c) 17
(d) 10
Ans : (B) Solution: The perpendicular distance to Z -axis =
x 2 y 2 9 16 5
ILLUSTRATION : 02 If D.C’s of two liens satisfy the relations 3l m 5n 0 and 6mn 2nl 5lm 0 , the angle between them is 1 1 (a) Cos 6
1 1 (b) Cos 3
(c)
1 3 (d) Cos 4
2
Ans : (A) Solution : Eliminating m from the given relations we get l 2 3 ln 2n 2 0 i.e. l 2n l n 0 ;
Hence using 1st leation
i.e.
l n l n or 2 1 1 1
l m n l m n or hold. 2 1 1 1 2 1
The angle between these lines is given by
cos
2 2 1 4 11 1 4 1
1 6
ILLUSTRATION : 03 The foot of the perpendicular from (1,2,3) to the line joining the points (6,7,7) and (9,9,5) is (a) (5, 3, 9)
(b) (3, 5, 9)
(c) (3, 9, 5)
(d) (3, 9, 9)
Ans : (B) Solution : Any point on the line joining the given points can be taken as 6 3t, 7 2t, 7 - 2t If it is required foot of the perpendicular of (1, 2, 3) we get 3(5 3t) 2(5 2t) - 2(4 - 2t) 0 i.e, t = -1
278
CO-ORDINATE GEOMENTRY ILLUSTRATION : 04 If A (1, 2, 3), B(6, 7, 8) C(1,2, 5) and D (3, 0, 4) are given points, then the projection of
AB on CD is
(a) 1/3
(b) 4/3
(c) 25/3
(d) 5/3
Ans : (D) Solution :
D.R’s of CD are (2, -2, -1) or (-2, 2, 1)
2 2 1 D.C’s of CD are 3 , 3 , 3 2 2 1 2 2 1 5 ___ Projection of AB on CD = 6 1 7 2 8 3 = 5 3 3 3 3 3 ILLUSTRATION : 05 The equation of the plane passing through the three points are (-2, -2, 2), (1, 1, 1) and (1, -1, 2) is (a) x 3 y 6 z 8 0
(b) x 3 y 6 z 14 0
(c) x 3 y 6 z 4 0
(d) x 3 y 6 z 20 0
Ans : (A) Solution : The required plane equation is
x 1 y 1 z 1 x 1 y 1 z 1 3 3 1 3 3 1 = 0; i.e. x 1 3 2 y 10 3 z 16 0 0 2 1 0 2 1 i.e. x 3 y 6 z 8 0 ILLUSTRATION : 06 If (2, 3, -1) is the foot of the perpendicular from (4, 2, 1) to a plane, then equation of the plane is (a) 2 x y 2 z 3 0
(b) 2 x y 2 z 9 0
(c) 2 x y 2 z 5 0
(d) 2 x y 2 z 1 0
Ans : (D) Solution : The line joining the given points is normal to the plane. 279
IIT- MATHS D.R’s of normal are (2, -1, 2) and (2, 3, -1) lies in the plane. The equation of the plane is 2 x y 2 z 4 3 2 ILLUSTRATION : 07 The equation of the plane parallel to x y 2 z 3 0 through (1, 2, 4) is (a) x y 2 z 11 0 (b) x y 2 z 6 0 (c) x y 2 z 11 0 (d) x y 2 z 8 0 Ans : (C) Solution : Plane parallel to the given plane can be taken in the form x y 2 z k 0 . This pass through (1,2, 4).
k 1 2 8 11
ILLUSTRATION : 08 The point which is equidistant from A(3, 4, -1) and B(1, -2, 5) on Y-axis is (a) (0,1, 0)
1 (b) 0, ,0 3
1 5 (c) 0, ,0 (d) 0, ,0 3 3
Ans : (C) Solution : ___
The plane that perpendicularly bisects AB is 2 x 2 6 y 1 6 z 2 0 (i.e.) x 3 y 3z 1 0 1 This cuts Y-axis at 0, ,0 3
ILLUSTRATION : 09 The plane 2 x 2 y 3z 14 0 and the line joining (1, 2, 4) and (3, 3, 0) intersect at (a) (5, 2, 0)
(b) (5, 4, -4)
(c) (-3, -1, -6)
(d) (10, -15, 12)
Ans : (B) Solution Any point on the given line is 1 2t , 2 t , 4 4t This lies in the given plane; i.e. 21 2t 22 t 34 4t 14 0 (i.e.,) 4t 2t 12t 2 4 12 14 0 (i.e.) 14t 28 0 t = 2 Intersection point is (5, 4, -4)
280
CO-ORDINATE GEOMENTRY ILLUSTRATION : 10 The equation of the plane through (2, -3, 1) which is perpendicular to line joining the points (3,4, -1) and (2, -1, 5) is (a) x 5 y 6 z 19 0 Ans : (A)
(b) x 5 y 6 z 7 0 (c) x 5 y 6 z 23
Solution: DR’s of line joining (3, 4, -1) (2, -1, 5) are 1, 5, -6 The required plane is x 5 y 6 z 2 15 6 19
281
(d) x 5 y 6 z 11 0
IIT- MATHS
SECTION - A SINGLE ANSWER TYPE QUESTIONS 1.
The projection of the line PQ joining the points P(3,4,5) Q(4,6,3) on x-axis is a) 1
2.
a b c 2 2 2
c) 21
d) 7/9
b) 5
c) 5 2
d) 2
5
b) 7 : 5
c) 9 : 11
d) 11: 9
b) 4x+y-3z-26=0
c) 2x-4y+3z+23=0
d) 3x+5y-2z+12=0
b) (-1,0,-7)
c) (1,0,-7)
d) (7,0,1)
b) -4
c) ± 4
d) 0
a b c 2 2 2
b) , ,
a b c , 2 2 2
c) ,
a b c , , 2 2 2
d)
The foot of the perpendicular drawn from the point A (1,2,1) to the line joining B(1,4,6) and C(5,4,4) is a) (-3,-4,5)
12.
b) 7/6
The co-ordinates of a point equidistant from the four points O(0,0,0) A(a,0,0) B(0,b,0) and C(0,0,c) is a) , ,
11.
9 3 3 , 2 2 2
d) ,
The points A(1,-1,1) B(2,a,5) and C(5,-13,11) are collinear then a = a) 4
10.
9 7 3 , , 2 2 4
c)
The point of intersection of the line through (-2,3,4) (1,2,3) with the xoz plane a) (1,0,7)
9.
3 7 7 2 2 2
b) , ,
The equation of the plane passing through (2,3,-5) and perpendicular to the planes x+2y+2z-8=0 and 3x+3y+2z+5=0 is a) 2x-3y+2z+15=0
8.
d) (0, 0, 6)
The ratio in which the plane 2x–3y+5z–2=0 divides the line segment joining (1,2,3) (2,1,–2) is a) 3 : 5
7.
c) (1/2, 1/3,1/4)
If the projections of a line segment on the axes are 3,4,5 then the length of the line segment is a) 12
6.
b) (11,-16,2)
The projection of the join of the two points (2,5,6) (3,2,7) on the line whose D.r’s are (6,-3,6) is a) 7/3
5.
d) –1
If the orthocentre and circumcentre of a triangle are (–3, 5, 2), (6, 2, 5) then its centroid is a) (3, 3, 4)
4.
c) –2
The harmonic conjugate of A (2, 3, 4) with respect to B (3, –2, 2), C (6, –17, –4) is a) (18/5, –5,4/5)
3.
b) 3
b) (3,-4,5)
c) (-3,4,-5)
d) (3,4,5)
The angle between the points passing through the points (8,2,0) (4,6,-7) and (-3,1,2) (-9,-2,4) is
282
CO-ORDINATE GEOMENTRY 2
13.
14.
1 b) Cos
a)
l1 l 2 m1 m 2 n 1 n 2 , , 2 2 2
b) l1 + l2, m1 + m2 , n1 + n2
c)
l1 l 2 m 1 m 2 n 1 n 2 , , 3 3 3
d)
b)
1 2
c)
66
b) x2+y2–6y+ 25 = 0 d) x2 + y2 – 6y – 8z + 25 = 0
b) (1,2,2) (2,1,2) d) (1,2,-2), (2,-1,-2)
b) 45o
c) 60o
b) 2x-3y+3z+5= 0
283
d) –2x-3y-3z+5 = 0
b) 2x + 4y – 3z – 39 = 0 d) 2x + 4y – 3z = 0
The equation of the plane lying mid way between 2x + 3y – 6z + 1 = 0 and 2x+3y-6z+7 = 0 is b) 2x-3y+6z – 4 =0 d) 2x+3y+6z-8 = 0
A plane passing through the fixed point (a,b,c) cuts the coordinate axes at A,B,C. Then the locus of the centroid of the triangle is a)
22.
c) 12x+6y-5z = 1
If (2, 4, –3) is foot of perpendicular drawn from origin to plane, then the equation if the plane is
a) 2x+3y-6z+4 = 0 c) 2x+3y-6z+8 = 0 21.
d) 90o
Equation of the plane passing through the points (1,1,0) (1,2,1) and (-2,2,-1) is
a) 2x + 4y – 3z – 29 = 0 c) 2x + 4y + 3z + 29 = 0 20.
66
If the d.c’s l,m,n of two lines are connected by the relations l+m+n = 0, l2 – m2 + n2 = 0 then the angle between the lines is
a) 2x+3y-3z = 5 19.
d)
If the d.c’s l,m,n of two lines are connected by the relations 2l+2m-n = 0 and mn+nl+mn =0 then the d.r’s of the two lines are
a) 30o 18.
17 2
The equation of the locus of the point which moves in such a way that the sum of its distances from (2,3,4) and (-2,3,4) is 4 is
a) (1,-2,-2) (2,-1,2) c) (1,-2,2) (2,1,-2) 17.
l1 l 2 m 1 m 2 n 1 n 2 , , 2 2 2
The distance between orthocentre and circumcentre of the triangle formed by (1,2,3)(3,-1,5)(4, 0, –3) is
a) y2+z2–6x–8z+25=0 c) x2 + y2 + z2 – 25 = 0 16.
1 d) Cos 63
1 c) Cos
If (l1, m1, n1), (l2, m2, n2) are d.c’s of two lines inclined at an angle 1200, then the d.c’s of the line bisecting the angle between them are
a) 0 15.
4
2 63
1 63
1 a) Cos 63
x y z =3 a b c
a
b
c
b) x y z 3
c) ax+by+cz = 3
d) x2 + y2 + z2 = a2+b2+c2
A variable plane is a at a constant distance 3p from the origin and meets the axes in A, B and C. The locus of the centroid of the triangle ABC is
IIT- MATHS 1
1
1
1
1
1
1
3
1
1
1
9
1
1
1
16
a) x 2 y 2 z 2 p 2 b) x 2 y 2 z 2 p 2 c) x 2 y 2 z 2 p 2 d) x 2 y 2 z 2 p 2 23.
The end points of a diagonal of a rectangular parallelopiped with faces parallel to the coordinate planes are (2,3,5) and (5,7,10). The lengths of its edges are a) 5,4,3
24.
30.
b) (1,1,5)
b) (-1/2, 2, 0)
c) (–3, 5, 2)
b) 450
c) 600
x 1 y 1 z 10 is 2 3 8k
a) (3, -4, -2)
c)(1,-1,-10)
d) (1,-1,-5)
d) (1/2, -2, 0)
d) (3, 5, 2)
b)(5, -8,-4)
d) None of these
d)(2,-3,8)
If the foot of the perpendicular from the origin to a plane is (a,b,c) the equation of the plane is x y z 3 a b c
b) ax by cz 3 d) ax by cz a b c
The angle between the lines whose direction cosines are given by the equations l2+m2–n2 =0, l+m+n = 0 is a) /6
b) /4
c) /3
d)/2
The volume of the tetrahedron included between the plane 3x+4y-5z-60=0 and the coordinate planes is a) 60
33.
c) (-1/2, -2, 0)
b) (3, 5, –2)
c) ax by cz a 2 b 2 c 2
32.
c) (1,1,-5)
The reflection of the point P(1,0,0) in the line
a)
31.
d) None
A line makes an angle of 600 with each of x and y-axis, the angle which it makes with the z -axis is a) 300
29.
c) 13
The image of (1, 3, 4) in the plane 2x – y +z + 3 = 0 is a) (3, –5, 2)
28.
b) 2 13
The circumcentre of the triangle formed by the points A(1,1,0) B(1,2,1) and C(0,-5,1) is a) (1/2, 2, 0)
27.
d) 3,4,5
A point lying on the line joining the points (-3,5,4) and (1,-1,5) has its x-coordinate as 1 then the point is a) (1,-1,5)
26.
c) 3,5,4
The extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to reference planes are (-2,4,6) and (3,16,19). The length of the base diagonal is a) 13
25.
b) 4,5,3
b) 600
c) 720
d) none of these
The plane passing through the point (-2,-2,2) and containing the line joining the points (1,1,1) and (1,-1,2) makes intercepts on the coordinates axes the sum of the whose lengths is a) 3
b) 4
c) 6
d) 12 284
CO-ORDINATE GEOMENTRY 34. A line segment has length 63 and direction ratios are (3, -2,6). If the line makes an obtuse angle with x-axis, the components of the line vector are a) 27, -18,54 35.
The lines
b) x 2 y 2 z 2 2
c) x y z 1
d) x y z 2
b) 3:2
c) 4:5
d) -7 :8
b) r 2 x ry z 3r 2
c) x ry r 2 z 3
d) r 2 x ry z 3
b) 19/2
c)-22/3
d) 26/3
Equation of the plane passing through the origin and containing the lines whose direction ratios are 1, -2, 2 and 2, 3, -1 is a) x 2 y 2z 0
41.
d) k = 3 or -3
Algebraic sum of the intercepts made by the plane x+3y-4z+6 =0 on the axes is a) -13/2
40.
c) k = 0 or -3
A plane meets the coordinates axes in A,B,C such that the centriod of the triangle is the point (1, r, r2), the equation of the plane is a) x ry r 2 z 3r 2
39.
b)k =1 or -1
The ratio in which the yz plane divides the line joining the points (-2,4,7) and (3,-5,8) is a) 2 :3
38.
d)27,-18,-54
A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = 1 , y = 1 , z= 1 is 10 units. The locus of the point is a) x 2 y 2 z 2 1
37.
c) -27,18,-54
x 2 y 3 z 4 k 1 y 4 z 5 and are coplanar if 1 1 k k 2 1
a) k = 0 or -1 36.
b) -27,18,54
b) 2x 3y z 0
c) x 5 y 3z 0
d) 4x 5y 7 z 0
If a line makes angles , , , with four diagonals of a cube, then cos 2 cos 2 cos 2 cos 2 is equal to a) 1/3
42.
285
b)3
c)6
d)None of these
b) xyz 8k 3
c) x y z 6k
d) x 3 y 3 z 3 64k 3
A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point (a, b, c). Then the equation of the plane is a)
45.
d) 8/3
A variable plane makes with coordinate planes a tetrahedron of constant volume 64k3. The locus of the centroid of the tetrahedron is a) xyz 6k 3
44.
c) 4/3
The area of the triangle with vertices A(3,4,-1) B(2,2,1) and C(3,4,-3) is a) 4 5
43.
b) 2/3
x y z 3 a b c
b)
a b c 3 x y z
c) ax + by + cz = 3
d) none of these
The equation of the plane through the points (1, 0, -1) and (3, 2, 2) and parallel to the line
IIT- MATHS
x 1 y 1 z 1 is 1 2 3 a) 4x+y+2z = 6 46.
b) 3/2
65
c) 65 / 3
c) 3
d) 1
The centre of the circle given by r ( i 2 j 2k ) = 15 and r ( j 2k ) 4 is b) (-1, 3, 4)
c) (1, -3, 4)
d) (1, 3, -4)
The d.c’s of a line which makes equal angles with the axes is 1 1 1 , b) , 3 3 3
a) 1, 1, 1 1
1
1
, , c) 2 2 2
50.
d) none of these
b) 4
a) (1, 3, 4) 49.
d) none of these
If a, b, g are angles which a line makes with the axis then the value of sin2a + sin2b + sin2g is equal to a) 2
48.
c) 4x-y-2z = 6
The area of the triangle whose vertices are (0, 0, 0), (3, 4, 7) and (5, 2, 6) is a) 3/ 74
47.
b) 4x-y+2z = 6
d) None of these
When a right handed rectangular cartesian system OXYZ is rotated about the Z-axis through an angle p/4 in the counter - clockwise direction it is found that a vector a has the components 2 3 , 3 2 and 4. The components of a in the OXYZ coordinate system are a) 5, -1, 4
51.
b) 5, -1, 4 2
1
1
1
1
d) none of these
b) 1
c) -1
d) 0
The coordinates of the foot of the perpendicular from the point A (1, 8, 4) to the line joining B(0, -1, 3) and C(2, -3, -1) is a) (-5/3, 2/3, 19/3)
b) (5/3, 2/3, 19/3)
c) (5/3, -2/3, 19/3)
d) none
The locus of a point which moves so that the difference of the squares of its distance from two given points is constant is a a) straight line
55.
1
The value of x for which the points A (1,0,3), B(-1,3,4) C(1,2,1) and D(x,2,5) are coplanar is a) 2
54.
1
b) x y z = 1
c) 1 x 1 y 1 z =1
53.
d) -1, 5, 4
The coplanar points A, B, C, D are (2-x, 2, 2), (2, 2-y, 2), (2, 2, 2-z) and (1, 1, 1) respectively, then a) x + y + z = 1
52.
c) -1, -5, 4 2
b) plane
c) sphere
P (1,1,1) and Q () are two points in the space such that PQ =
d) none
27 , the value of can be 286
CO-ORDINATE GEOMENTRY a) -4 b) 2 56.
The lines
b) 14 / 3
b) -1
The lines
d) -2
c) -69
d) 28
X 1 Y 1 Z 2 X 1 Y Z 1 , are 1 2 1 2 1 4 b) intersecting lines
c) perpendicular lines
d) none of these
The projections of a line on the axes are 9, 12 and 8. The length of the line is a) 7
b) 17
c) 21
d) 25
A variable plane makes with the coordinate planes, a tetrahedron of constant volume 64K3. Then the locus of the centroid of tetrahedron is the surface. a) xyz = 6k2
64.
c) 2
b) 73
a) parallel lines
63.
d) 5/ 3
The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x + 3y + 10z = 25. The equation of the plane in its new position is x-4y+6z=k, where k is a) 106
62.
c) 16/ 3
AP BQ CR DS equals. PB QC RD SA
a) 1
61.
d) none
P, Q, R, S are four coplanar points on the sides AB, BC, CD, DA of a skew quadrilateral. The product
60.
b) intersect at (4, 0, -1) c) intersect at (1, 1, -1)
The distance of the point A (-2, 3, 1) from the line PQ through P (-3, 5, 2) which make equal angles with the axes is a) 2/ 3
59.
b) x2 + y2 + z2 + x + y + z = 0 d) x2 + y2 + z2 - x - y - z - 2 = 0
x 1 y 1 z 1 x 4 y 0 z 1 and 3 1 0 2 0 3
a) do not intersect 58.
d) none of these
The equation of a sphere which passes through (1, 0, 0) (0, 1, 0) and (0, 0, 1) and whose centre lies on the curve 4 x y = 1 is a) x2 + y2 + z2 - x - y - z = 0 c) x2 + y2 + z2 + x + y + z = 0
57.
c) -2
b) xy + yz + zx = 6k2
The plane cont aining t he t wo lines
c) x2 + y2 + z2 = 8k2
d) none
X 3 Y 2 Z 1 X 2 Y 3 Z 1 and is 1 4 5 1 4 5
11x+my+nz = 28 where a) m = -1, n = 3 65.
c) m = -1, n = -3
d) m = 1, n = 3
A variable plane passes through the fixed point (a,b, c) and meet the axes at A,B,C. The locus of the point of intersection of the planes through A,B,C and parallel to the co-ordinate plane is a)
287
b) m = 1, n = -3
a b c 2 x y z
b)
a b c 1 x y z
c)
a b c 1 x y z
a
b
c
d) x y z 2
66.
IIT- MATHS A line passes through the points (6, -7, -1) and (2, -3, 1). The direction cosines of the line so directed that the angle made by it with positive direction of X-axis is acute, are a)
67.
2 2 1 , , 3 3 3
The point in which the line a) (7, -8, 26)
68.
2 2 1 , , 3 3 3
c)
2 2 1 , , 3 3 3
d)
2 2 1 , , 3 3 3
X 2 Y 1 Z 2 meets the plane x-2y+z = 20 is 3 4 12
b) (8, 7, 26)
c) (7, 8, 26)
d) none
If P(3, 2, -4) , Q (5, 4, -6) and R(9, 8, -10) are collinear then R divides PQ in the ratio a) 3 : 2 internally
69.
b)
b) 3 : 2 externally
c) 2 : 1 internally
The equation of the plane perpendicular to the line
d) 2 : 1 externally
X 1 Y 2 Z 1 and passing through 1 1 2
the point (2, 3, 1) is a) r.i j 2k 1 70.
1 2 2 , , 3 3 3
b)
d)
1 2 2 , , 3 3 3
c) x+y+z = 1
d) none
b) r2x + ry + z = 3r2
c) x + ry + r2z = 3
d) r2x + ry + z = 3
c) 7 : 8
d) 1 : 1
A line segment has length 63 and direction ratios are (3, -2, 6). If the line makes an obtuse angle with X-axis, the components of the line vector are
The lines a) k = -1
76.
1 2 2 , , 3 3 3
X 1 Y 3 Z 2 and the point (0, 7, -7) is 3 2 1
b) x+y+z = 2
b) 4 : 5
a) 27, -18, 54 75.
c)
The ratio in which the plane 2x - 1 = 0 divide the line joinig (-2, 4, 7) and (3, -5, 8) is a) 2 : 3
74.
d) none
A plane meets the coordinate axes in A, B, C such that the centroid of the triangle is the point (1, r, r2) the equation of the plane is a) x + ry + r2z = 3r2
73.
1 2 2 , , 3 3 3
The equation of plane containing the line a) x+y+z = 0
72.
c) r.i j 2k 7
A mirror and a source of light are situated at the origin O and at a point on OX respectively. A ray of light from the source strikes the mirror and is reflected. If the DR’s of the normal to the plane are 1, -1, 1 then DC’s of the reflected ray are a)
71.
b) r.i j 2 k 1
b) -27, 18, -54
c) -27, 18, -54
d) 27, -18, -54
X 2 Y3 Z4 X 1 Y 4 Z 5 and are coplanar if 1 1 k k 2 1 b) k = +3
c) k = 1
d) k = 0
If P1, P2, P3 denote the distances of the plane 2x - 3y + 4z + 2 = 0 from the planes 2x-3y+4z+6 = 0, 4x-6y+8z+3 = 0 and 2x - 3y+4z-6 = 0 respectively then 288
CO-ORDINATE GEOMENTRY a) P1 + 8P2 - P3 = 0 b) P3 = 16P2 77.
b) y-axis
b) 4 planes
r r r
b) x , y , z
The line
85.
86.
289
x y z , , r r r
d) none of these
b) sin2 + sin2 + sin2 = 1 d) cos2+cos2+sin2=1
b) (-3, 5, 2)
c) (3, -5, 2)
d) (3, 5, -2)
X 2 Y 1 Z 1 intersects the curve xy = c2, z = 0 if c = 3 2 1 b) ± 1/3
c) ±
5
d) none
The number of spheres of radius r touching the coordinate axis is a) 4
84.
d) 6 planes
The image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0 is
a) ± 1 83.
c) 5 planes
If be the angles which a line makes with the coordinate axes, then
a) (3, 5, 2) 82.
d) yz- plane
c)
a) sin2 + cos2 + sin2 = 1 c) cos2+cos2+cos2 = 1 81.
c) z-axis
If (x, y, z) be the coordinates of a point P and OP = r then the direction cosines of OP are a) rx, ry, rz
80.
d) (A) , (B) , (C)
Tetrahedron is bounded by a) 3 planes
79.
2a
Graph of the equation y2 + z2 = 0 in three dimensional space is a) x-axis
78.
c) P1 + 2P2 + 3P3 =
b) 6
c) 8
d) none
If 1 , m1, n1 and 2 , m2, n2 are DC’s of the two lines inclined to each other at an angle q, then the DC’s of the internal bisector of the angle between these lines are a)
1 2 m1 m 2 n 1 n 2 , , 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2
b)
1 2 m1 m 2 n 1 n 2 , , 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2
c)
1 2 m1 m 2 n 1 n 2 , , 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2
d)
1 2 m1 m 2 n 1 n 2 , , 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2
If 1 , m1, n1 and 2 , m2, n2 are DC’s of the two lines inclined to each other at an angle q then the DC’s of the external bisector of the angle between the lines are a)
1 2 m1 m 2 n 1 n 2 , , 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2
b)
1 2 m1 m 2 n 1 n 2 , , 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2
c)
1 2 m1 m 2 n 1 n 2 , , 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2
d)
1 2 m1 m 2 n 1 n 2 , , 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2
1 3 3 , , then other end is If one end of a diameter of sphere x2 + y2+z2 - 2x - 2y - 2z + 2 = 0 is 1 2 2 2
IIT- MATHS
a) 1
1 1 , , 2 2 2
1
87.
88.
b) 1
1 3 3 , , 2 2 2
a)
x 1 y 2 z 3 1 1 1
c)
x 1 y 2 z 3 2 3 1
b)
x 1 y 2 z 3 1 1 1
a b c
2 abc
b) | a x b b x c c x a |
|axbbxccxa |
d) none
A (3, 2, 0), B(5, 3, 2) and C(-9, 6, -3) are the vertices of a triangle ABC. If the centroid of ABC meets BC at D, then coordinates of D are 19 57 17 , , 8 16 16
19 57 17 , , 8 16 16
b)
19 57 17 , , 8 16 16
c)
b) 2x - y = 5
c) 2x + z = 5
The points A (0, 0, 0), B(2, 0, 0), C(1, a) rhombus
b) square
c) a regular tetrahedron
d) none
If P is the length of perpendicular from the origin on to the plane where intercepts on the axes are a, b, c then d) none
A line makes angle a, b, g, d with the four diagonals of a cube, then the value of cos2+cos2+ cos2 + cos2 is equal to a) 3/4
94.
d) 2x - z = 5
1 2 2 , 1, are the vertices of a , 0) and D 3 3 3
a) a 1 b 1 c 1 p 1 b) a 2 b 2 c 2 p 2 c) a + b + c = p 93.
d) none
The Cartesian equation of the plane r ( 1 )i ( 2 ) ( 3 2 2 )k is a) 2x + y = 5
92.
d) none
The length of the perpendicular from the origin to the plane passing through three non-collinear points a , b, c is
a)
91.
,
d) none
c) a b c
90.
1 1 , 2 2 2
1
Equation of a line passing through (-1, 2, -3) and perpendicular to the plane 2x+3y+z + 5 = 0 is
a)
89.
c) 1
b) 1/4
The coordinates of a point on the line
c) 4/3
d) 2/3
x 1 y 1 =z at a distance 4 14 from the point 2 3
(1, -1, 0) are a) 9,13,4 c) (-7, 11, -4) 95.
b) - 8 14 +1, 12 14 -1, -4 d) 8+1, -12-1, 4
The position vectors of points A and B are and respectively. The equation of a plane is = 0. The
290
CO-ORDINATE GEOMENTRY points A and B a) lie on the plane c) are on opposite sides of the plane 96.
291
b) are on the same sides of the plane d) none of the above
The extremities of a diameter of a sphere lie on positive y and positive z-axis at distance 2 and 4 from the origin respectively, then a) sphere passes through origin
b) centre of the sphere is (0, 1, 2)
c) radius of the sphere is
d) all the above
5
IIT- MATHS
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
The line whose vector equations are r = 2i – 3j + 7k + (2i + pj + 5k) and
r = i + 2j + 3k + (3i – pj + pk)
are perpendicular for all values of l and m if p = a) –1 2.
b) 2
b) 2
The lines
d) – 5
b) (2, 3, -1)
c) (3, 1, 1)
d0 (4, 0, 7)
x 2 y3 z 4 x 1 y 4 z 5 and are coplanar if 1 1 k k 2 1
a) k = 0 5.
c) 5
The plane passing through the origin and containing the line whose direction cosines are proportional to 1, -2, 2 and 2, 3, -1 passes through the point a) (1, -2, 2)
4.
d) 6
A plane meets the coordinate axes in A, B, C such that the centroid of the triangle ABC is the point (1, r, r2). The plane passes through the point (4, -8, 15) if r is equal to a) –3
3.
c) 5
b) k = -1
c) k = 2
d) k = -3
An equation of the line passing through 3i – 5j + 7k and perpendicular to the plane 3x–4y+5z= 8 is a)
x 3 y5 z 7 3 4 5
b)
c) r = 3i – 5j + 7k + l (3i – 4j + 5k)
x 3 y 4 z 5 3 5 7
d) r = 3i – 4j – 5k + m (3i + 5j + 7k)
( are parameter) 6.
The coordinates of a point on the line
x 1 y 1 = z at a distance 4 14 from the point (1,-1, 2 3
0) are
7.
a) (9, -13, 4)
b) 8 14 1, 12 14 1, 4 14
c) (-7, 11, -4)
d) 8 14 1, 12 14 1, 4 14
If a plane is at a distance 3/2 from the origin O, and meets the axes in A, B and C, the coordinates of the centroid can be 1 a) 1, , 1 2
8.
1 b) , 1, 1 2
1 c) 1, 1,
2
d) (1, 1, 1)
If a plane passes through a fixed point (2, 3, 4) and meets the axes of reference in A, B and C, the point of intersection of planes through A, B, C parallel to the coordinate planes can be
292
CO-ORDINATE GEOMENTRY a) (6, 9, 12) 9.
Equation of a plane through the line a) 4y – 3z + 1 = 0
10.
c) (1, 1, -1)
d) (2, 3, -4)
x 1 y 2 z 3 and parallel to a coordinate axis is 2 3 4
b) 2x – z + 1 = 0
c) 3x – 2y + 1 = 0
d) 2x + 3y + 1 = 0
The line joining the points (2, -3, 1) and (3, -4, -5) cuts a coordinate plane at the point. a) (0, -1, 13)
11.
b) (4, 12, 16)
b) (0, 0, 1)
c) (-1, 0, 19)
d) (8, -9, 0)
If l, m, n are the direction cosines of the line of shortest distance between the lines
x 3 y 15 z 9 x 1 y 1 z 9 and then 2 7 5 2 1 3 a) 3l – 25m + 9n = 0 c) l = m = n = 1/ 3 12.
The foot of the perpendicular from the origin to the join of A(-9, 4, 5) and B (11, 0, -1) lies on the plane a) 2x + y + z = 6
13.
b) x – y + z = 1
c) x + y + z = 1
d) x – y – z = 1
If a plane p passes through the point (1, 2, 3), direction cosines of the normal to p are l, m, n; and it contains the line joining the origin to the point (1, 1, 1), then a) l + 2m + 3n = 0
293
b) 2l – 7m + 5n = 0 d) 2l + m – 3n = 0
b) l + m + n = 0
c) l + m – n = 0
d) l – m + 2n = 0
IIT- MATHS
COMPREHENSION TYPE QUESTIONS PASSAGE 1: A (-2, 2, 3) and B (13, -3, 13) L is a line through A 1.
A point P moves in the space such that 3PA = 2PB, then the locus of P is a) x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0 c) x2 + y2 + z2 + 28x – 12y – 10 z + 247 = 0
2.
Coordinates of the point P which divides the join of A and B in the ratio 2 : 3 internally are a) (33/5, -2/5, 9)
3.
4.
b) x2 + y2 + z2 – 28x + 12y + 10z – 247 = 0 d) x2 + y2 + z2 - 28x + 12y – 10 z + 247 = 0
b) (4, 0, 7)
c) (32/5, -12/5, 17/5)
d) (20, 0, 35)
Equation of a line L, perpendicular to the line AB is a)
x 2 y 2 z3 15 5 10
b)
x 2 y 2 z3 3 13 2
c)
x 2 y 2 z3 3 13 2
d)
x 2 y 2 z3 15 5 10
Direction ratios of the normal to the plane passing through the origin and the points A and B are a) 15, -5, 10
b) 11, -1, 16
c) 3, 13, 2
d) 7, 13, -4
PASSAGE 2: a = 6i + 7j + 7k, b = 3i + 2j – 2k, P(1, 2, 3). 5.
The position vector of L, the foot of the perpendicular from P on the line r = a + lb is a) 6i + 7j + 7k
6.
c) 3i + 5j + 9k
d) 9i + 9j + 5k
The image of the point P in the line r = a + b is a) (11, 12, 11)
7.
b) 3i + 2j – 2k
b) (5, 2, -7)
c) (5, 8, 15)
d) (17, 16, 7)
If A is the point with position vector a the Area of the PLA in sq. units is equal to a) 3 6
b) 7 17 / 2
c) 17
d) 7/2
PASSAGE 3: P(2,3, -4), b = 2i – j + 2k 8.
Vector equation of a plane passing through the point P perpendicular to the vector b is a) r.(2i – j + 2k) = -7 b) r. (2i – j + 2k) = 7 c) r.(2i + 3j – 4k) = -7 d) r . (2i + 3j – 4k) = 7
9.
Cartesian equation of a plane passing through the point with position vector b and perpendicular to the vector OP , O being the origin is a) 2x – y + 2z + 7 = 0 b) 2x – y + 2z – 7 = 0 c) 2x + 3y – 4z + 7 = 0 d) 2x + 3y – 4z – 7 = 0
10.
Sum of the lengths of the intercepts made by the plane on the coordinate axes is 294
CO-ORDINATE GEOMENTRY a) 14 b) 91/12
c) 9/7
d) 5/7
PASSAGE 4: L:
x 1 y 1 z 1 2 3 4
p1: x + 2y + 3z = 14, p2: 2x - y + 3z = 27
If the line L meets the plane p1 in the point P,and the coordinates of P are (), then is equal to
11.
a) 3 12.
b) 14
c) 28
d) 29
The line through P perpendicular to the plane p1 passes through the point a) (1, 1, 1)
b) (0, 1, 0)
c) (0, 0, 0)
d) (0, 0, 1)
If the line through P perpendicular to 1 meets the plane 2 in the point Q, then the coordinates of the mid-point of PQ are
13.
a) (1, 2,3)
b) (3, 6, 9)
c) (2, 3, 4)
d) (2, 4, 6)
MATCHING TYPE QUESTIONS 1. i)
x 2 y 7 z 5 3 4 2
a) Perpendicular to the plane 3x + 4y + 2z = 1
ii)
x 1 y 2 z 7 3 4 2
b) Passes through (2, 7, -5)
iii)
x 5 y 2 z 2 c) direction cosines are 2/ 30 5/ 30 1/ 30 1 3 4
iv)
x 1 y 1 z 1 2 5 1
2. ax + by + cz + d = 0,
d) lies in the plane 7x – y – z = 35
x y z m n
i) lines is perpendicular the plane
a) if al + bm + cn = 0
ii) line is parallel to the plane ifb) if al + bm + cn = 0 and aa + bb + cg + d = 0 iii) line lies in the plane c) if a/l = b/m = c/n
3. L :
295
x 2 y3 z4 is a line them 3 4 5
i) Point on the line at a distance 10 2 from (2, 3, 4)
a) (-1, -1, -1)
ii) Point on the line common to the plane x + y + z + 3 = 0
b) (2, 3, 4)
iii) Point on the line at a distance
c) (8, 11, 14)
29 from the origin
IIT- MATHS
MATCHING TYPE QUESTIONS 1. i)
x 2 y 7 z 5 3 4 2
a) Perpendicular to the plane 3x + 4y + 2z = 1
ii)
x 1 y 2 z 7 3 4 2
b) Passes through (2, 7, -5)
iii)
x 5 y 2 z 2 c) direction cosines are 2/ 30 5/ 30 1/ 30 1 3 4
iv)
x 1 y 1 z 1 2 5 1
2. ax + by + cz + d = 0,
d) lies in the plane 7x – y – z = 35
x y z m n
i) lines is perpendicular the plane
a) if al + bm + cn = 0
ii) line is parallel to the plane ifb) if al + bm + cn = 0 and aa + bb + cg + d = 0 iii) line lies in the plane c) if a/l = b/m = c/n
3. L :
x 2 y3 z4 is a line them 3 4 5
i) Point on the line at a distance 10 2 from (2, 3, 4)
a) (-1, -1, -1)
ii) Point on the line common to the plane x + y + z + 3 = 0
b) (2, 3, 4)
iii) Point on the line at a distance
c) (8, 11, 14)
29 from the origin
296
CO-ORDINATE GEOMENTRY
NUMERICAL SUBJECTIVE TYPE QUESTIONS 1.
A plane meets the coordinate axes in A, B, C such that the centroid of the DABC is the point (12,15,16). The sum of the squares of the intercepts made by the plane on the coordinate axes is
2.
If q is the angle between the line
x 1 y 1 z 2 and the plane 2x+y–3z+4=0, then 64 cosec2 q 3 2 4
is equal to 3.
If d is the distance between the point (-1, -5, -10) and the point of intersection of the line x 2 y 1 z 2 with the plane x – y + z = 5, then 3d3 is equal to. 3 4 12
4.
If Q is the foot of the perpendicular fro the point P(4,-5,3) on the line
x 5 y 2 z6 then 100 3 4 5
(PQ)2 is equal to 5.
A plane passes through (1, 2, 2) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. Square of the distance of the plane from the point (52, 53, 57) is.
6.
P, Q, R, S are the points (1, 2, -2), (8, 10, 11), (1, 2, 3) and (3, 5, 7) respectively. If s denotes the projection of PQ on RS then 29s2 + 29 is equal to.
7.
The lines
x 4 y 17 z 11 x 15 y 9 z 8 and intersect at the point P, then square of the 15 9 8 4 17 11
distance of P from the origin is. 8.
If the position vector of the point of intersection of the line r=(i+2j+3k)+l (2i+j+2k) and the planer r.(2i – 6j + 3k) + 5 = 0 is ai + bj + ck, then (50a + 70b + 75c)2 is equal to.
9.
If d is the shortest distance between the lines r = 3i + 5j + 7k + l (i + 2j + k) and r = -i – j – k m (7i – 6j + k) then 125 d2 is equal to.
10.
If the foot of the perpendicular from the origin on a plane is (11, 11, 11), then the sum of the square s of the intercepts made by the plane on the coordinate axes is equal to.
11.
If the point of intersection of the line r = (i + 2j + 3k) + l (2i + j + 2k) and the plane r.(2i – 6j + 3k) + 5 = 0 lies on the plane r. (i + 75j + 60k) - a = 0, then 19 a + 17 is equal to.
12.
If the line
x 1 y 1 z 1 intersect the curve 6x2+5y2=1, z=0; then 10n2–20n+k=0,where the 15 16 n
value of k is.
297
IIT- MATHS
SECTION - A SINGLE ANSWER TYPE QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A
A
A
A
C
C
B
B
D
B
D
16.
17. 18. 19.
20.
21. 22. 23
24. 25.
26.
12. 13.
C
B
27. 28.
14. 15.
B
A
29. 30.
A
C
A
A
C
B
A
D
A
B
B
C
B
B
C
31
32
33
34
35
36
37
38
39
40
41
42 43
44
45
C
B
D
B
C
B
D
B
A
D
C
D
A
A
C
46
47
48
49
50
51
52
53
54
55
56
57 58
59
60
B
A
A
B
D
B
C
A
B
C
A
B
A
A
61
62
63
64
65
66
67
68
69
70
71
72 73
74
75
C
B
A
C
C
A
B
B
B
D
A
B
C
D
B
D
298
CO-ORDINATE GEOMENTRY
76
77
78
79
80
81
82
83
84
85
86
87 88
89
90
D
A
B
D
C
B
C
C
B
C
A
C
A
C
91
92
93
94
95
96
C
B
C
C
C
D
A
SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1.
2.
3.
4.
A,D B,C A,B A,D C
5.
6.
7.
8.
9.
10.
11.
12. 13.
A,C A,C A,B A,B A,B A,C B,C A,B A,B C C,D C D C
COMPREHENSON TYPE QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A
B
C
D
C
C
B
A
C
B
B
MATCHING TYPE QUESTIONS 1.
(i) (b), (ii) (a), (iii) (d), (iv) (c)
2.
(i) (c), (ii) (a), (iii) (b)
3. 299
(i) (c), (ii) (a), (iii) (b)
12. 13.
C
D
IIT- MATHS
NUMERICAL SUBJECTIVE TYPE QUESTIONS
1.
5625
5.
5202
9.
2116
2.
1624
6.
8129
10.
3267
3.
6591
7.
1398
11.
2563
4.
1828
8.
7225
12.
2630
300
CO-ORDINATE GEOMENTRY
THREE DIMENSIONAL GEOMENTRY We know that the position of a point in a plane can be determined if the co–ordinates (x, y) of the point with refersence to two mutually perpendicular lines called x and y axes are known. But all points of space donot lie in a plane and so in order to locate a point in space two coordinate axes are insufficient. In order to locate a point in space we need three–co–ordinate axes. The position of a point in space can be determined with reference to three mutually perpendicular lines called x, y and z axes. In case of two dimensional geometry two mutually perpendicular lines are taken and they divide the plane (xy–plane) in four parts. The four parts are called the quadrants. The sign of co–ordinates of the points in the four parts are (+ , +), (–, +), (–, –) and (+, –) In case of three dimensional geometry we take three mutually perpendicular lines which divide the space in eight parts called octants. The sign of co–ordinates of the points in the 8 parts in which the space is divided are (+, +, +) (–, +, +) (+, –, +), (+, +, –) (–, –, +), (–, +, –), (+, –, –), (–, –, –).
Y
O X Z
CO-ORDINATES OF A POINT ON AXES 1.
Co–ordinates of a point P on x–axis will be of the form (a, 0, 0) where a is the distance of foot of perpendicular from P to x–axis from the origin with suitable sign. a is positive or negative according as it lies on positive or negative direction of x–axis.
2.
Co–ordinates of a point on y–axis is of the form (0, b, 0).
3.
Co–ordinates of a point on z–axis is of the form (0, 0, g).
4.
Projection of a Line segment on a Plane: Let AB be a line segment, and L and M be the feet of normals from A and B to a given plane. Then the line segment LM is called the projection of the line segment AB on the plane.
DISTANCE FORMULA Let P º (x1, y1, z1) and Q º (x2, y2, z2) Then PQ = 301
(x1 x 2 )2 (y1 y 2 ) 2 (z1 z 2 ) 2
IIT- MATHS We draw PL xy plane. Then L º (x1, y1, 0) and M º (x2, y2, 0) PL = z1, QM = z2 From DPHQ Let PQ =
PH 2 QH 2 =
2 2 2 LM 2 QH 2 = {(x1 x 2 ) (y1 y 2 ) } (z 2 z1 )
{ LM2 = (x1 – x2)2 + (y1 – y2)2] =
(x1 x 2 )2 (y1 y 2 ) 2 (z1 z 2 ) 2 .
SECTION FORMULAE I.
Section Formula for Internal Division Let P º (x1, y1, z1) and Q º (x2, y2, z2) Let R divide the line segment PQ internally in the ratio m : n
mx 2 nx1 my 2 ny1 nz 2 nz1 , , Then R º mn mn mn
II.
Section Formula for External Division Let P º (x1, y1, z1), Q º (x2, y2, z2) Let R divide the line segment PQ externally in the ratio m : n, then
PR m RQ n
Let R º (x, y, z) From P, Q, R draw PL, QM and RN perpendiculars to xy–plane Again draw PH RN and QK RN RH PR From similar DPHR and DQKR RK QR
z z1 m PR m Þ nz – nz1 = mz – mz2 z z 2 n QR n
(m – n) z = nz2 – nz1 z=
nz 2 nz1 mn
Similarly by drawing perpendiculars from P, Q, R to yz–plane we can show that x=
mx 2 nx1 mn
Again by drawing perpendiculars from P, Q, R to zx plane we can show that
302
CO-ORDINATE GEOMENTRY my 2 ny1 y= mn mx 2 nx1 ny 2 ny1 mz 2 nz1 , , Thus R º mn mn mn
ANGLE BETWEEN TWO LINES Let AB and CD be two given lines having direction cosines l1 , m1, n1 and l2, m2, n2 respectively and q be the angle between them. Let O be the origin. Through O we draw OP parallel to AB and OQ parallel to CD. Let P º (x1, y1, z1), Q º (x2, y2, z2) and OP = r1, OQ = r2 Since direction cosines of AB are l1, m1, n1 and OP || AB Therefore d.cs of OP are l1, m1, n1. Similarly d.cs of OQ will be l2, m2, n2
x1 y1 z1 x2 y2 z2 Now l1 = r , m1 r , n1 r and l2 = r , m 2 r , n 2 r 1 1 1 2 2 2 From PQO, r12 r22 [(x1 x 2 )2 (y1 y 2 )2 (z1 z 2 )2 ] OP 2 OQ 2 PQ 2 cosq = = 2.r1 r2 2.OP.OQ r12 r22 [(x12 y12 z12 ) (x 22 y 22 z 22 ) 2x1x 2 2y1y 2 2z1z 2 ] = 2r1r2
x1 x 2 y1 y 2 z1 z 2 r12 r22 (r12 r22 2x1x 2 2y1y 2 2z 1z 2 ) = = r .r r .r r .r 2r1r2 1 2 1 2 1 2 cos = l1l2 + m1m2 + n1n2) … (1) or
= cos–1 (l1l2 + m1m2 + n1n2)
ANGLE BETWEEN TWO LINES IN TERMS OF DIRECTION RATIOS Let a1, b1, c1 and a2, b2, c2 be the direction ratios of two lines AB and D respectively. Direction cosines of the two lines are given by l1
a1 a12 b12 c12
, m1
a2
and l2 =
2 2
2 2
a b c
2 2
b1 a12 b12 c12
, m2
c1
, n1
a12 b12 c12
b2 2 2
2 2
a b c
2 2
, n2
If be the angle between the two lines, then cos = l1 l2 + m1m2 + n1n2
303
c2 2 2
a b 22 c22
IIT- MATHS a1a 2 b1b 2 c1c 2
=
2 1
a b12 c12 a 22 b 22 c 22
The two lines will be perpendicular iff or, a1a2 + b1b2 + c1c2 = 0 sin =
(l1m 2 l2 m1 )2 (m1m 2 m 2 n1 ) 2 (n1l2 n 2l2 ) 2
The two lines will be parallel iff = 0 or sin = 0 or,
(a1b2 – a2b1)2 + (b1c2 – b2c1)2 + (c1a2 – c2a1)2 = 0
or,
(a2b2 – a2b1)2 = 0, (b1c2 – b2c1)2 = 0, (c1a2 – c2a1)2 = 0
or,
a1 b1 c1 sin tanq = = a 2 b2 c2 cos
(a1b 2 a 2 b1 ) 2 (b1c 2 b 2c1 ) 2 (c1a 2 c 2a1 ) 2 a1a 2 b1b 2 c1c 2
PROJECTION OF THE LINE SEGMENT JOINING TWO POINTS ON A LINE Projection of the line segment joining (x1, y1, z1) and (x2, y2, z2) on the line having direction cosines l, m, n. = l(x2 – x1) + m(y2 – y1) + n (z2 – z1) The length of projection of the line segment joining (x1, y1, z1) and (x2, y2, z2) on the line having direction cosines l, m, n. = |l(x2 – x1) + m(y2 – y1) + n(z2 – z1)|
GENERAL EQUATION OF A PLANE General equation of the first degree in x, y, z always represents a plane. Let the general equation of second degree in x, y and z be ax + by + cz + d = 0
… (1)
where at least one of a, b, c is non–zero. Locus (1) will be a plane if the line joining any two points on the surface fully lies on the surface. Let A(x1, y1, z1) and B(x2, y2, z2) be any two points on locus (1), then ax1 + by1 + cz1 + d = 0 … (1) and ax2 + by2 + cz2 + d = 0
… (2)
m.(2) + n.(1) a(mx2 + nx1) + b(my2 + ny1) + c(mz2 + nz1) + (m + n) d = 0
mx 2 nx1 my 2 ny1 mz 2 nz1 or, a b c +d=0 mn mn mn
… (3)
mx 2 nx1 my 2 ny1 mz 2 nz1 Let P () where denotes , , it is clear from mn mn mn
(3) that P() lies on locus (1). But P is the point dividing the line segment joining A and B in the ratio
304
CO-ORDINATE GEOMENTRY m : n and m and n are arbitrary. Hence all points on the line joining A and B lies on the locus (1). Thus locus (1) is a plane. Note 1: Equation of the xy–plane Let the equation of the xy–plane be ax + by + cz + d = 0 … (1) Since O(0, 0, 0) lies in the xy–plane
d=0
Since (1, 0, 0) lies on x–axis, therefore it also lies on the xy–plane
a=0
Again (0, 1, 0) lies on y–axis, therefore it also lies on xy–plane
b=0
Hence from (1), equation of the plane becomes cz = 0 or, z = 0 Thus equation of xy–plane is z = 0
Equation of a plane in Intercept Form Equation of the plane which cuts intercepts a, b, c on x, y and z–axis respectively is x y z 1 a b c
Equation of a plane in Normal Form Equation of the plane upon which the length of perpendicular from origin is p and l, m, n be the direction cosines of the normal to it is lx + my + nz = p
Equation of a plane parallel to a given plane: General equation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0. Where k is a constant.
Angle between two planes: Angle between two planes is equal to the angle between their normals Angle q between two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by
305
IIT- MATHS a1a 2 b1b 2 c1c 2
cos =
2 1
a b12 c12 a 22 b 22 c 22
here a1, b1, c1 are the direction ratios of the normal to the first plane and a2, b2 c2 are the direction ratios of the normals to the second plane.
Distance of a Point From a plane Length of perpendicular from point (, , ) to the plane ax + by + cz + d = 0 is given by P=
a b c d a 2 b2 c2
Equation of the planes bisecting the angle between two planes Equation of the planes bisecting the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is a1x b1y c1z d1
=
2 1
2 1
2 1
a 2 x b2 y c2z d 2 a 22 b 22 c 22
a b c
Bisector of the angle between two planes containing the origin Let the equation of the two planes be a1x + b1y ++ c1z + d1 = 0 … (1) and a2x + b2y + c2z + d2 = 0
… (2)
where d1 and d2 are positive then equation of the bisector of the angle between planes (1) and (2) containing the origin is a1x b1y c1z d1 2 1
2 1
2 1
a 2 x b2 y c2z d 2
a b c
a 22 b 22 c 22
Bisector of the acute and obtuse angle between two planes Let the equation of the two planes be a1x + b1y + c1z + d1 = 0
… (1)
and a2x + b2y + c2z + d2 = 0
… (2)
where d1, d2 > 0 (i) If a1a2 + b1b2 + c1c2 > 0, then origin lies in the obtuse angle between two planes and the equation of bisector of the acute angle between two planes is
306
CO-ORDINATE GEOMENTRY a1x b1y c1z d1 a x b2 y c2z d 2 2 2 2 2 a1 b1 c1 a 22 b 22 c22 (ii) If a1a2 + b1b2 + c1c2 < 0, then origin lies in the acute angle between two planes and the equation of bisector of the acute angle between two planes is a1x b1y c1z d1 2 1
2 1
2 1
a b c
a 2 x b2 y c2z d 2 a 22 b 22 c 22
.
Otherwise, the bisector of the acute angle makes with either of the planes an angle which is less then 45° and the bisector of the obtuse angle makes with either of them an angle which is greater than 45°. This gives a test for determining which angle, acute or obtuse, each bisecting plane bisects.
307
IIT- MATHS
THE STRAIGHT LINES Introduction: We know that 1. one and only one line can be drawn through a given direction. 2. One and only one line can be drawn through two given points. 3. Intersection of two non parallel planes is a unique line. Thus a straight line in space will be determined uniquely if (i)
it passes through a fixed point and is parallel to a fixed line
(ii)
it passes through two fixed points.
(iii)
it is the intersection of two given non parallel planes.
Equation of the line passing through point (x1, y1, z1) and having direction rations (or direction cosines) l, m, n is x x1 y y1 z z1 = l m n
Equation of the line joining points (x1, y1, z1) and (x2 , y2, z2) is
x x1 y y1 z z1 x1 x 2 y1 y 2 z1 z 2 The Equation of a Line (General Form) Intersection of two planes is a straight line, therefore in general a line is obtained by the intersection of two planes. Equation of a line as the intersection of two planes Equation of the line, which is the intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is a1x + b1y + c1z + d1 = 0
… (1)
and a2x + b2y + c2z + d2 = 0
… (2)
Equation (1) and (2) taken together is the equation of line of intersection of planes (1) and (2) in asymmetrical form. Angle between a line and a plane. Angle between a line and plane is the complement of the angle between the line and the normal to the plane. Let the given line PQ be
308
CO-ORDINATE GEOMENTRY x x1 y yi z z1 … (1) l m n and the given plane be N
Q 90°
ax + by + cz + d = 0
… (2)
P
L
Let q be the angle between the line and the plane, then the angle between the line and normal PN to the plane is 90° –. Direction ratios of PN are a, b, c and direction ratios of line PQ are l, m, n al bm cn Now cos(90° – )
a 2 b 2 c2 l 2 m 2 n 2 al bm cn
or sin =
a 2 b2 c2 l 2 m2 n 2
General Equation of the Plane containing a line Equation of any plane containing line x x1 y y1 z z1 is l m n
A(x – x1) + B(y – y1) + C(z – z1) = 0 Where Al + Bm + Cn = 0 Equation of the Plane containing two given lines Equation of the plane containing lines
x x1 y y1 z z1 x x 2 y y2 z z 2 and l1 m1 n1 l2 m2 n 2 is
or,
x x2
y y2
z z2
l1 l2
m1 m2
n1 n2
0
To Find the shortest distance between two lines
309
x x1
y y1
z z1
l1 l2
m1 m2
n2 n2
=0
IIT- MATHS
x x1 y y1 z z1 x x 2 y y2 z z 2 and l1 m1 n1 l2 m2 n2 Given lines are
x x1 y y1 z z1 … (1) l1 m1 n1 x x 2 y y2 z z 2 And CD : l m n 2 … (2) 2 2 AB :
A(x1, y1, z1)
B
90° 90°
C
M
D
(x2, y2, z2)
Let l, m, n be the direction ratios of the shortest distance LM. Since LM AB and LM CD ll1 + mm1 + nn1 = 0
… (3)
andll2 + mm2 + nn2 = 0
… (4)
From (3) & (4),
l m n m1n 2 m 2 n1 n1l2 n 2l1 l1m 2 l2 m1 =
1 (m1n 2 m 2 n1 )2
From the above equation l, m & n can be found out and the shortest distance between the lines is projection of line joining the point (x1, y1, z1) & (x2, y2, z2) on the line with the direction cosines l, m & n.
Image of a Point in a Plane To find the image of the point P(a, b, g) in the plane ax + by + cz + d = 0 Given plane is ax + by + cz + d = 0 … (1) P
Q P (a, b, g) 310
CO-ORDINATE GEOMENTRY Let Q(x1 ,y1 , z1) be the image of point P in the plane (1) Let PQ meet plane (1) at L. Direction ratios of normal to plane (1) are a, b, c Since PQ plane (1), therefore direction ratios of PQ are a, b, c equation of lien PQ is
x y z = r (say) a b c
Co–ordinates of any point on line PQ may be taken as (ar + a, br + b, cr + g) Let Q (ar + a, br + b, cr + g) Since L is the middle point of PQ
ar br cr L , , 2 2 2 Since L lies on plane (1)
ar br cr a b c d = 0 2 2 2 (a2 + b2 + c2)
r = –(aa + bb + cg + d) 2
Thus Q º (a + ar, b + br, g + cr) Where r = –
311
2(a b c d) a 2 b2 c2
r=–
2(a b c d) a 2 b2 c2