PAPER 1 1
The functions f and g are defined by 1 f : x → —, x ∈ R \ {0}; x g : x → 2 x – 1, x ∈ R.
2 3
4 5
6
7
Find f ° g and its domain. 3 ( x 2)2 5 2 x – 2) Show that ——— dx = — + 4 ln — 2 2 3 3 x
[4 marks]
∫
[4 marks]
Using definitions, show that, for any sets A, B and C , ). A ∩ ( B B ∪ C ) ⊂ ( A A ∩ B) ∪ ( A A ∩ C ). 1 If z z is a complex number such that z |z| = 1, find the real part of ——–. 1 –z 1 The polynomial p( x x) = 2 x 3 + 4 x 2 + — x – k has factor ( x x + 1). 2 (a) Find the value of k . (b) Factorise p( x x) completely. sin x – cos x d 2 y dy If y = 2 y —–. y = —————–, show that —— 2 sin x + cos x dx dx
[5 marks] [6 marks]
[2 marks] [4 marks] [6 marks]
1 0 0 Matrix A is given by A = 1 –1 0 . 1 –2 1 2 (a) Show that A = I, where I is the 3 × 3 identity matrix, and deduce A–1. 1 4 3 (b) Find matrix B which satisfies BA = 0 2 1 . –1 0 2
[4 marks] [4 marks]
8
The lines y = 2 x and y = x intersect the curve y2 + 7 xy = 18 at points A and B respectively, where A and B lie in the first quadrant. (a) Find the coordinates coordinates of A [4 marks] A and B. (b) Calculate the perpendicular distance of A to OB, where O is the origin. [2 marks] (c) Find the the area of the OAB triangle. [3 marks]
9
Find the solution set of the inequality 4 3 —— > 3 – —. x – 1 x
10
[10 marks]
x
Show that the gradient of the curve y = ——– is always decreasing. [3 marks] x 2 – 1 Determine the coordinates of the point of inflexion inflexion of the curve, and state the intervals intervals for which the curve is concave upward. [5 marks] Sketch the curve. [3 marks] Actual 2008 STPM Mathematics Examination Paper
1
11
Sketch, on the same coordinate axes, the curves y = 6 – e x and y = 5e– x, and find the coordinates of the points of intersection. [7 marks] Calculate the area of the region bounded by the curves. [4 marks] Calculate the volume of the solid formed when the region is rotated through 2 π radians about the x-axis. [5 marks]
12
At the beginning of this year, Mr. Liu and Miss Dora deposited RM10 000 and RM2000 respectively in a bank. They receive an interest of 4% per annum. Mr Liu does not make any additional deposit nor withdrawal, whereas, Miss Dora continues to deposit RM2000 at the beginning of each of the subsequent years without any withdrawal. (a) Calculate the total total savings of Mr. Liu Liu at the end of nth year. [3 marks] th (b) Calculate the total total savings of Miss Dora Dora at the end of n year. [7 marks] (c) Determine in which year the the total savings of Miss Dora exceeds exceeds the total savings of Mr. Liu. [5 marks]
PAPER 2 1
Show that the substitution u = x 2 + y transforms the differential equation dy
(1 – x)—– + 2 y + 2 x = 0 dx
into the differential equation du dx
(1 – x) —– = –2u. 2
3
[3 marks]
In triangle ABC , the point X divides BC internally in the ratio m : n, where m + n = 1. Express AX 2 in terms of AB [5 marks] AB, BC , CA, m and n. θ 2t 1 – t 2 θ If t = tan —, show that sin θ = ——– and cos = ——–. 2 1 + t 2 1 + t 2
[4 marks]
Hence, find the values of θ between 0° and 360° that satisfy the equation 10 sin θ – 5 cos θ = 2. 4
[3 marks]
The diagram below shows the circumscribe circumscribedd circle of the triangle ABC . B
R
Q
A
C P T
The tangent to the circle at A meets the line BC extended to T . The angle bisector of the angle ATB cuts AC at P, AB at Q and the circle at R. Show that (a) triangles APT and BQT are similar, [4 marks] (b) PT • BT = QT • AT , [2 marks] (c) AP = AQ. [4 marks] 5
2
The position vectors of the points A, B and C , with respect to the origin O, are a, b and c respectively. The points L, M , P and Q are the midpoints of OA, BC , OB, and AC respectively. Actual 2008 STPM Mathematics Examination Paper
1 1 (a) Show that the position position vector of of any point on the line LM is —a + —λ (b + c – a) for 2 2 some scalar λ , and express the position vector of any point on the line PQ in terms of [6 marks] a, b and c. (b) Find the position vector vector of the point of intersection intersection of the line LM and the line PQ. [4 marks] 6
A 50 litre tank is initially filled with 10 litres of brine solution containing 20 kg of salt. Starting from time t = 0, distilled water is poured into the tank at a constant rate of 4 litres per minute. At the same time, the mixture leaves the tank at a constant rate of k litres per minute, where k > 0. The time taken for overflow to occur is 20 minutes. (a) Let Q be the amount of salt in the tank at time t minutes. Show that the rate of change of Q is given by dQ
Q k
—– = – ——————. 10 + (4 – k )t dt Hence, express Q in terms of t . [7 marks] (b) Show that k = 4, and calculate the amount of salt in the tank at the instant overflow occurs. [6 marks] (c) Sketch the graph of Q against t for 0 ≤ t ≤ 20. [2 marks] 7
8
9
10
There are 12 towels, two of which are red. If five towels are chosen at random, find the probability that at least one is red. [4 marks] 1 The random variable X has a binomial distribution with parameters n = 500 and p = —. 2 Using a suitable approximate distribution, find P(| X )| ≤ 25). [6 marks] X – E( X X )| In a basket of mangoes and papayas, 70% of mangoes and 60% of papayas are ripe. If 40% of the fruits in the basket are mangoes, (a) find the percentage percentage of the fruits fruits which are ripe, ripe, [3 marks] (b) find the percentage of the the ripe fruits which are mangoes. mangoes. [4 marks] A sample of 100 fuses, nominally rated at 13 amperes, are tested by passing increasing electric current through them. The current at which they blow are recorded and the following cumulative frequency table is obtained. Current (amperes)
Cumalative frequency
< 10
0
< 11
8
< 12
30
< 13
63
< 14
88
< 15
97
< 16
99
< 17
100
Calculate the estimates of the mean, median and mode. Comment on the distribution. [8 marks] Actual 2008 STPM Mathematics Examination Paper
3
11
The continuous random variable X has probability density function ⎫0, x < 0 ⎪ 5 ⎪— – x, 0 ≤ x < 1, f ( x x) = ⎬ 4 ⎪ 1 x ≥ 1. ⎪——, 2 4 x ⎭
(a) Find the cululative cululative distribution distribution function function of X [7 marks] X . (b) Calculate the probability probability that at least one of two independent independent observed values of X is greater than three. [4 marks] 12
A car rental shop has four cars to be rented out on a daily basis at RM50.00 per car. The average daily demand for cars is four. (a) Find the probability probability that, on a particular day, (i) no cars are requested, [2 marks] (ii) at least four requests for cars are received. [2 marks] (b) Calculate the expected daily daily income received from from the rentals. [5 marks] (c) If the shop wishes to have one more car, car, the additional cost incurred incurred is RM20.00 per day. Determine whether the shop should buy another car for rental. [5 marks]
SUGGESTED ANSWERS
PAPER 1
1 1. f : x → —, x ∈R \ {0}
A \ B B means A – B or A ∩ B' .
x g : x → 2 x – 1, x ∈ R f ° g = fg( x x) = f (2 (2 x – 1)
1 1 = ———, x ≠ — 2 x – 1 2 1 The domain of f f ° g is x x : x ∈ R, x ≠ — . 2 3 ( – 2)2 x x 2. ——— dx 2
∫
2
3
x x 2 – 4 x + 4 ————–— dx x 2
4 = ∫ 1 – — + 4 x dx x x = x x – 4 ln x |x| + 4 —— –1 4 = x x – 4 ln x |x| – — x 4 4 = 3 – 4 ln 3 – — – 2 – 4 ln 2 – — 3 2 =
∫ 2 3
–2
2
–1
3 2
3 2
5 = — + 4 ln 2 – 4 ln 3 3 5 = — + 4(ln 2 – ln 3) 3 5 2 = — + 4 ln — [shown] 3 3
4
Actual 2008 STPM Mathematics Examination Paper
3. By using basic definitions of sets, let ), x ∈ A ∩ ( B B ∪ C ), then, x ∈ A and x ∈ ( B B ∪ C ) x ∈ A and ( x x ∈ B or x ∈ C ) ( x x ∈ A and x ∈ B) or ( x x ∈ A and x ∈ C ) ( x x ∈ A ∩ B) or ( x x ∈ A ∩ C ) x ∈ ( A A ∩ B) ∪ ( A A ∩ C ) Thus, all elements in A ∩ ( B B ∪ C ) are also found in ( A ). A ∩ B) ∪ ( A A ∩ C ). Hence, A ∩ ( B B ∪ C ) ⊂ ( A A ∩ B) ∪ ( A A ∩ C ) [shown] 4. Let z = x + yi. |z| = 1 z x 2 + y 2 = 1 … (1) x 2 + y 2 = 1 1 1 ——– = ————— 1 – z 1 – ( x x + yi) 1 = ————— 1 – x – yi 1 – x + yi = —————————— (1 – x – yi)(1 – x + yi) 1 – x + yi = —————— (1 – x)2 + y 2 1 – x + yi = ——————— 1 – 2 x + x 2 + y 2 1 – x + yi = ————— 1 – 2 x + 1 1 – x + yi = ————– 2 – 2 x
From (1)
1 – x + yi = ————— 2(1 – x) 1 – x y = ———— + ———— i 2(1 – x) 2(1 – x) 1 y = — + ————i 2 2(1 – x) 1 1 Hence, the real part of ——– is —. 1 – z 2
d 2 y dx dy dy + —– – 1 + —– (cos x – sin x) dx dx + y(–sin x – cos x) = 0 2 d y (sin x + cos x) —— dx 2 dy + 2—– – 1 (cos x – sin x) dx + y(–sin x – cos x) = 0 2 d y dy (sin x + cos x) —— = 2—– – 1 (sin x – cos x) + 2 dx dx y(sin x + cos x) sin x – cos x d 2 y dy —— = 2—– – 1 ——–—––— 2 sin x + cos x dx dx sin x + cos x + y —————— sin x + cos x 2 d y dy —— = 2—– – 1 y + y 2 dx dx 2 d y dy —— = 2 y —– [shown] 2 dx dx
(sin x + cos x) —— 2
1 5. (a) p( x x) = 2 x 3 + 4 x 2 + — x – k 2 Since ( x x + 1) is a factor of p p( x x), then p(–1) = 0 1 2(–1)3 + 4(–1)2 + —(–1) – k = 0 2 1 –2 + 4 – — – k = 0 2 3 — – k = 0 2 3 k = — 2 1 3 (b) Therefore, p( x x) = 2 x 3 + 4 x 2 + — x – — 2 2 3 2 x 2 + 2 x – — 2 1 3 x + 1 2 x 3 + 4 x 2 + — x – — 2 2 2 x 3 + 2 x 2
7. (a)
2
=
A
1 0 0 = 1 –1 0 1 –2 1 1 0 0 = 0 1 0 0 0 1 = I [shown] 2 A = I AA = I
1 0 0 1 –1 0 1 –2 1
–1
=I By comparison,
(b)
BA
=
–1
=
BI
=
B
=
BAA
dy
dx y(cos x – sin x) = cos x + sin x dy (sin x + cos x) —– – 1 + y(cos x – sin x) = 0 dx
d 2 y dy (sin x + cos x) ——2 + —– – 1 (cos x – sin x) dx dx dy + y(–sin x – cos x) + (cos x – sin x) —— = 0 dx
AA
(sin x + cos x)—– +
AA
3 Hence, p( x x) = ( x x + 1) 2 x 2 + 2 x – — 2 2 4 x + 4 x – 3 = ( x x + 1) —————— 2 1 = — ( x x + 1)(2 x + 3)(2 x – 1) 2 sin x – cos x y = —————— sin x + cos x y = sin x – cos x (sin x + cos x) y
3 3 – — x – — 2 2 3 3 – — x – — 2 2 0
6.
1 2 x 2 + — x 2 2 x 2 + 2 x
=
1 0 –1 1 0 –1 1 0 –1 1 0 –1 8 3 1
8. (a) y = 2 x y = x y 2 + 7 xy = 18
–1
A
=
A
1 0 = 1 –1 1 –2 4 3 2 1 0 2 4 3 2 1 A–1 0 2 4 3 2 1 A–1 0 2 4 3 1 0 2 1 1 –1 0 2 1 –2 –10 3 –4 1 –4 2 ............. (1) ............. (2) ............. (3)
0 0 1
0 0 1
Actual 2008 STPM Mathematics Examination Paper
5
Substituting (1) into (3), (2 x)2 + 7 x(2 x) = 18 4 x 2 + 14 x 2 = 18 18 x 2 = 18 x 2 = 1 x = ±1 x = –1 is not accepted because point A lies in the first quadrant. Thus, x = 1. When x = 1, y = 2(1) = 2 Hence, the coordinates of point A are (1, 2). Substituting (2) into (3), x 2 + 7 x 2 = 18 8 x 2 = 18 9 x 2 = — 4 3 x 2 = ± — 2 3 x = – — is not accepted because point B 2 lies in the first quadrant. 3 Thus, x = —. 2 3 3 When x = —, y = — 2 2 Hence, the coordinates of point B are 3 3 —, — . 2 2
3 —–0 2 y – 0 (b) The equation of OB is ——— = ———– 3 x – 0 —–0 2 y = x x – y = 0 The perpendicular distance from A(1, 2) to OB
|1 – 2| = ————— 12 + (–1)2 1 = —– 2 2 = —– units 2 1 (c) Area of ∆OAB = — 2
3 0 1 — 0 2 3 0 2 — 0 2
1 3 =——–3 2 2
3 = –— 4
3 = — units2 4 6
Actual 2008 STPM Mathematics Examination Paper
4 9. y = ——— x – 1
4 ⎫ ——–, x > 1 ⎪ x – 1 y = ⎬ 4 ⎪ – ——– , x < 1 ⎭ x – 1
As y → ∞, x – 1 → 0 x → 1 Thus, x = 1 is the asymptote. As x → ±∞ , y → 0. 4 The graph of y y = ——– is as shown. x – 1
y
4 y = ——— x–1
3 y = 3 – —– x 4
3
A
4 y = – ——— x – 1
O
1
3 y = 3 – —– x
3
x
3
y = 3 – — x As y → ∞, x → 0 Thus, x = 0 (the y-axis) is the asymptote. As x → ± ∞, y → 3.
3 The graph of y y = 3 – — is as shown. x
The x-coordinate of point A is obtained by solving the following equations simultaneously. 4 ... (1) y = ——– x – 1 3 ... (2) y = 3 – — x
4 3 ——— = 3 – — x – 1 x 4 3 x – 3 ——– = ——— x – 1 x (3 x – 3)( x x – 1) = 4 x 3 x 2 – 6 x + 3 – 4 x = 0 3 x 2 – 10 x + 3 = 0 (3 x – 1)( x x – 3) = 0 1 x = — or 3 3 1 x = — is not accepted 3 Thus, x = 3 The solution set for which 4 3 > 3 – — is given by the part of the ——– –1 4 graph where the curve = ——– –1 x
x
y
x
3 is above the curve y = 3 – —, that is, x
{ x : 0 < x < 1 or 1 < x < 3}.
x
10. y = ——— x 2 – 1 ( x dy x 2 – 1)(1) – x(2 x) —– = ———————— ( x dx x 2 – 1) 2 – x 2 – 1 = ———— ( x x 2 – 1)2 –( x x 2 + 1) = ———— x 2 – 1)2 ( x dy dx
Since —– < 0 for all real values of x x, then the gradient of the curve is always decreasing. [shown] –( x d 2 y x 2 – 1)2(2 x) + ( x x 2 + 1)(2)( x x 2 – 1)(2 x) —— = ———————————————–– ( x dx 2 x 2 – 1) 4 –2 x( x x 2 – 1)[ x x 2 – 1 – 2( x x 2 + 1)] = ————————————— ( x x 2 – 1)4 –2 x (– x 2 – 3) = —————— ( x x 2 – 1)3 2 x( x x 2 + 3) = ————— ( x x 2 – 1)3
+
+
+
+
–
–
–
+
–
–
+
+
–
+
+
+
– –1
+ 0
– 1
+
x2 + 3 > 0 ( – 1) 3 > 0 x
x > 0 ( + 1) 3 > 0 x
x
Hence, the intervals for which the curve concaves upwards are –1 < x < 0 or x > 1 and the intervals for which the curve concaves downwards are x < –1 or 0 < x < 1. x
The curve y = ——— is as shown below. x 2 – 1 y
–1 O
1
x
d 2 y dx
When —— = 0, 2 2 x( x x 2 + 3) ————– = 0 ( x x 2 – 1) 3 x = 0 0 When x = 0, y = ——— =0 2 0 –1 ( x x 2 – 1)3(6 x 2 + 6) – (2 x)( x d y x 2 + 3)(3)( x x 2 – 1)2(2 x) —— = ———————————— ( x dx 3 x 2 – 1)6 x 2 – 1)3( x x 2 + 1) – (12 x 2)( x x 2 + 3)( x x 2 – 1)2 6( x = ———————————————— ( x x 2 – 1)6 6( x x 2 – 1)2[( x x 2 – 1)( x x 2 + 1) – (2 x 2)( x x 2 + 3)] = ———————————————— ( x x 2 – 1)6 2 2 4 6( x x – 1) ( x x – 1 – 2 x 4 – 6 x 2) = ———————————— ( x x 2 – 1) 6 6(– x 4 – 1 – 6 x 2) = ——————— ( x x 2 – 1)4 6[–04 – 1 – 6(0)2] d 3 y When x = 0, —— = ———————— = –6 (02 – 1) 4 dx 3 (that is ≠ 0) 3
11. y = 6 – e x On the x-axis, y = 0 6 – e x = 0 e x = 6 x = ln 6 Thus, the curve y = 6 – e x intersects the x-axis at (ln 6, 0). On the y-axis, x = 0 y = 6 – e0 y = 5 Thus, the curve y = 6 – e x intersects the y-axis at (0, 5). As x → ∞, y → –∞ As x → – ∞, y → 6 y 6 5
d 2 y d 3 y dx dx when x = 0, then (0, 0) is the point of inflexion.
y = 5e– x
Since —— = 0 and —— ≠0 2 3
When the curve concaves upwards. d 2 y —— >0 dx 2
2 x( x x 2 + 3) ————— >0 ( x x 2 – 1) 3 2 x( x x 2 + 3) ——————— > 0 [( x x + 1)( x x – 1)]3 2 x( x x 2 + 3) ——————— >0 ( x x + 1)3( x x – 1)3
O
y = 6 – e x
(ln 5, 1) ln 6
x
y = 5e– x On the y-axis, x = 0 y = 5(e0) y = 5 Therefore, the curve y = 5e– x intersects the y-axis at (0, 5). As x → ∞, y → 0. As x → –∞, y → ∞ The curve y = 6 – e x and y = 5e– x are as
shown. Actual 2008 STPM Mathematics Examination Paper
7
y = 6 – e x y = 5e– x
... (1) ... (2) Substituting (1) into (2), 6 – e x = 5e– x 6e x – (e x)2 = 5 Letting e x = p, 6 p – p2 = 5 2 p – 6 p + 5 = 0 ( p p – 1)( p p – 5) = 0 p = 1 or 5 When p = 1, e x = 1 x = ln 1 x = 0 When x = 0, y = 6 – e0 = 5 When p = 5, e x = 5 x = ln 5 When x = ln 5, y = 6 – eln 5 = 6 – 5 = 1 Hence, the points of intersection are (0, 5) and (ln 5, 1). Area of the shaded region =
ln 5
∫
[(6 – e x) – 5e– x] dx
0
5 = 6 x – e x – —— e– x (–1)
= 6 x – e
x
5 + —– x e
ln 5 0
ln 5 0
5 5 = 6 ln 5 – eln 5 + —— – 0 – e0 + —– ln 5 0
e
e
5 = 6 ln 5 – 5 + — – (–1 + 5) 5 = 6 ln 5 – 5 + 1 + 1 – 5 = (6 ln 5 – 8) units2
∫
= π
[(6 – e x)2 – (5e– x)2] dx
0 ln 5
∫
= π
0
[36 – 12e x + e2 x – 25e–2 x] dx
1 25 = π 36 x – 12e x + — e2 x – —— e–2 x 2 (–2)
1 25 = π 36 x – 12e x + — e2 x + ——– 2 2e2 x
ln 5 0
ln 5 0
1 25 = π 36 ln 5 – 12eln 5 + — e2 ln 5 + ——–– 2 2e2 ln 5 1 25 – 0 – 12e0 + — e0 + —— 2 2e0
1 25 = π [36 [36 ln 5 – 12(5) + —(25) + ——— 2 2(25) 1 25 – –12 + — + —– 2 2 = π (36 (36 ln 5 – 48) = 12(3 ln 5 – 4)π units3
8
At the end of the nth year, the total savings, U n = 1.04n × 10 000 = RM10 000(1.04n) (b) For Miss Dora: At the end of the 1st year, the total savings, U 1 = 1.04 × 2000 At the beginning of the 2nd year, the total savings = (1.04 × 2000) + 2000 At the end of the 2nd year, the total savings, U 2 = 1.04[(1.04 × 2000) + 2000] = 1.042 × 2000 + 1.04 × 2000 At the beginning of the 3rd year, the total savings = (1.042 × 2000 + 1.04 × 2000) + 2000 At the end of the 3rd year, the total savings, U 3 = 1.04[1.042 × 2000 + 1.04 × 2000 + 2000] = 1.043 × 2000 + 1.042 × 2000 + 1.04 × 2000
At the end of the nth year, the total savings, U n = 1.04n × 2000 + ... + 1.042 × 2000 + 1.04 × 2000 = 2000(1.04n + ... + 1.042 + 1.04) 1.04 (1.04n – 1) = 2000 ——–——–—— 1.04 – 1 = 52 000(1.04n – 1) (c) When the total savings of Miss Dora exceeds the total savings of Mr. Liu, 52 000(1.04n – 1) > 10 000 (1.04n) 5.2(1.04n – 1) > 1.04n 5.2(1.04n) – 5.2 > 1.04n (5.2 – 1)(1.04n) > 5.2 4.2 (1.04n) > 5.2 5.2 1.04n > —— 4.2 5.2 n ln 1.04 > ln —— 4.2 5.2 ln —— 4.2 n > ————— ln 1.04 n > 5.45 Smallest integer value of n = 6 Hence, the total savings of Miss Dora exceeds the total savings of Mr. Liu at the end of the 6th year.
Volume of the solid generated ln 5
12. (a) For Mr. Liu: At the end of the 1st year, the total savings, U 1 = 1.04 × 10 000 At the end of the 2nd year, the total savings, U 2 = 1.04(1.04 × 10 000) = 1.042 × 10 000 At the end of the 3rd year, the total savings, U 3 = 1.04(1.042 × 10 000) = 1.043 × 10 000
Actual 2008 STPM Mathematics Examination Paper
Based on ∆ ABC , 2t sin θ = ——— [shown] and 1 + t 2 1 – t 2 cos θ = ——— [shown] 1 + t 2 10 sin θ – 5 cos θ = 2 2t 1 – t 2 10 ——–2 – 5 ——— =2 1 + t 1 + t 2 10(2t ) – 5(1 – t 2 ) = 2(1 + t 2 ) 20t – 5 + 5t 2 = 2 + 2t 2 3t 2 + 20t – 7 = 0 (3t – 1)(t + 7) = 0 1 t = — or t = –7 3 1 When t = —, 3 θ 1 tan — = — 2 3 θ — = 18.43° 2 θ = 36.9° [correct to one decimal place] When t = –7, θ tan — = –7 2 θ — = 98.13° 2 θ = 196.3° [correct to one decimal place]
Paper 2
1.
2
u = x + y du dx
dy dx
—– = 2 x + —– dy dx
du dx
—– = —– – 2 x
dy dx
(1 – x)—– + 2 y + 2 x = 0
du dx
(1 – x) —– – 2 x + 2(u – x2) + 2 x = 0 du dx
(1 – x)—– – 2 x(1 – x) + 2u – 2 x2 + 2 x = 0 du dx
(1 – x)—– – 2 x + 2 x2 + 2u – 2 x2 + 2 x = 0 du dx
(1 – x)—– + 2u = 0 du dx
(1 – x)—– = –2u [shown]
m m BX = — BC 1
A
2. BX = ——— BC m+n
BX = mBC m
B
X
n
C
In ∆ ABC , by using the cosine rule, 2
2
4.
θ 2 tan — 2 3. tan θ = —————— θ 1 – tan2 — 2
θ
1 – t2
= (1 – t 2 )2 + (2t )2 2
4
2
= 1 – 2t + t + 4t = 1 + 2t 2 + t 4 = (1 + t 2 )2 = 1 + t 2
2t
θ
C
P A
B 1 + t2
θ
Q
AB = AC 2 + BC 2
B
R
AB2 + BC 2 – CA2 – 2( AB)(mBC ) ———–———— 2( AB AB)( BC BC ) = AB2 + m2 BC 2 – m( AB AB2 + BC 2 – CA2) = AB2 + m2 BC 2 – mAB2 – mBC 2 + mCA2 = (1 – m) AB AB2 + (m2 – m) BC BC 2 + mCA2 = (1 – m) AB AB2 + m(m – 1) BC BC 2 + mCA2 2 2 = nAB + m(–n) BC BC + mCA2 = nAB2 – mnBC 2 + mCA2
A
2
AB + BC – CA cos ∠ ABC = ——————— 2( AB AB)( BC BC ) ∆ In ABX , by using the cosine rule, AX 2 = AB2 + BX 2 – 2( AB AB)( BX BX ) cos ∠ ABC 2 2 = AB + (mBC )
2t tan θ = ——— 1 – t 2
C T
(a) ∠ BTQ = ∠ ATP [ RQPT RQPT is the angle bisector of ∠ ATB] ∠PAT = ∠QBT [Alternate segment theorem] APT BQT APT (b) Since ∆ —— are similar, then BQT AT PT —– = —– BT QT PT • BT = QT • AT [shown] APT (c) Since ∆—— are similar, BQT then ∠ APT = ∠ BQT . Let ∠ APT = ∠ BQT = θ . ∠ APQ = 180° – θ [Angles on a straight
Hence, ∆ —— are similar [shown].
line] ∠ AQP = 180° – θ [Angles on a straight line] Since ∠ APQ = ∠ AQP, then ∆ APQ is an isosceles triangle where AP = AQ [shown]. Actual 2008 STPM Mathematics Examination Paper
9
5.
y B
M X
P
C c
T
Y
Q
b
A a
L x
O
(a) Let X be a point on the line LM . → → → OX = OL + LX → → = OL + λ LM LM , where LX = λ LM LM → → → → = OL + λ ( LO LO + OB + BM ) → → → 1→ = OL + λ LO LO + OB + — BC 2 → → → 1 → → = OL + λ LO LO + OB + —( BO BO + OC ) 2 1 1 1 = — a + λ – — a + b + —(–b + c) 2 2 2 1 1 1 1 = — a + λ – —a + b – —b + — c 2 2 2 2 1 1 1 1 = — a + λ – —a + —b + — c 2 2 2 2 1 1 = — a + — λ (– (–a + b + c) 2 2 1 1 = — a + — λ (b + c – a) [shown] 2 2 Hence, the position vector of any point on 1 1 the line LM is —a + —λ (b + c – a). [shown] 2 2 Let Y be a point on the line PQ. → → → OY = OP + PY → → = OP + µ PQ PQ, where PY = µ PQ PQ → → → → = OP + µ (PO + OA + AQ) → → → 1 → = OP + µ PO + OA + — AC 2 → → → → 1 → = OP + µ [PO + OA + — ( AO )] AO + OC )] 2 1 1 1 = — b + µ – — b + a + —(–a + c) 2 2 2 1 1 1 1 = — b + µ – —b + a – — a + — c 2 2 2 2 1 1 1 1 = —b + µ – —b + — a + — c 2 2 2 2 1 1 = —b + — µ (– (–b + a + c) 2 2 1 1 = —b + —µ (a + c – b) 2 2 Hence, the position vector of any point on 1 1 the line PQ is — b + — µ (a + c – b). 2 2
10
→ → → (b) YX = OX – OY 1 1 = —a + —λ (b + c – a) 2 2 1 1 – —b + — µ (a + c – b) 2 2 1 1 1 1 1 = —a + —λ b + —λ c – —λ a – —b 2 2 2 2 2 1 1 1 – — µ a – — µ c + — µ b 2 2 2 1 1 1 1 1 = — – — λ – —µ a + —λ – — 2 2 2 2 2 1 1 1 + —µ b + —λ – —µ c 2 2 2 Let T be the point of interesection of the line LM and the line PQ. → At point T , YX = 0. 1 1 1 1 1 1 — – —λ – — µ a + —λ – — + —µ b 2 2 2 2 2 2 1 1 + —λ – — µ c = 0 2 2 1 1 1 — – — λ – —µ = 0 2 2 2 1 – λ – µ = 0 λ + µ = 1 ...(1) 1 1 1 —λ – — + —µ = 0 2 2 2 λ – 1 + µ = 0 λ + µ = 1 1 1 —λ – — µ = 0 2 2 λ – µ = 0 ...(2) (1) + (2): 2λ = 1 1 λ = — 2 1 From (1): — + µ = 1 2 1 µ = — 2 → 1 Substituting µ = — into OY , 2
Actual 2008 STPM Mathematics Examination Paper
→
1 1 1 2 2 2 1 1 1 1 = —b + —a + —c – —b 2 4 4 4 1 1 1 = —a + —b + —c 4 4 4 1 = —(a + b + c) 4 Hence, the position vector of the point of intersection of the line LM and the line 1 PQ is —(a + b + c). 4 6. (a) Let V be the volume of solution in the tank at time t minutes.
OT = —b + — — (a + c – b)
Rate of change of volume of solution
dV dt
(b)
Change in volume of solution = —————————————– Change in time δ V dV V —– = —– dt δ t t
—– = 4 – k
O
∫ —————— 10 + (4 – k )t 0
– k dQ —– = ——– 20 Q 4 – k Q
∫
(4 – k ) dt —————— 0 10 + (4 – k )t t
∫
k
= ——–ln 10 + (4 – k )t k – 4
t
0
k
10 + (4 – k )t Q k ln —– = ——— ln ———–—— 20 10 k – 4
k
——
10 + (4 – k )t Q ln —– = ln ———–—–— 20 10
10 + (4 – k )t —– = —————— 20 10 Q
Take Note
This event does not follow a binomial distribution because the towels are chosen without replacement. Alternative method
k
——
t
k – 4
P(at least one red towel is chosen) = 1 – P(all the five towels chosen are not red)
k – 4
4 – k Q = 20 1 + ——— t 10
20
7. P(at least one red towel is chosen) = 1 – P(all the five towels chosen are not red) 10 9 8 7 6 = 1 – —– × —– × —– × — × — 12 11 10 9 8 7 = 1 – —– 22 15 = —– 22
ln |Q| – ln 20 = ——— [ln |10 + (4 – k )t | k – 4 – ln 10]
4
dt
t
—– = – k
20
k dt k dt
t
–1
20
0
Q
Q
= –∫ —————— ∫ —– 10 + (4 – k )t Q
[ln |Q|]
0
(c)
Q k
Q
20
100 When t = 20, Q = ——— = 4 5 + 20 Hence, at the instant overflow occurs, the amount of salt is 4 kg.
= – —————— [shown] 10 + (4 – k )t
20
dQ Q × (– k ) Hence, —— = ———— —————— —— 10 + (4 – k )t dt
∫
= (4 – k )t
1 = 20 ——–— 1 1 + — t 5 5 = 20 —–— 5 + t 100 = ——— 5 + t
Rate of change of amount of salt (kg/min) Amount of salt at time t minutes (kg) × Rate of change of volume of mixture leaving the tank (l /min) /min) = ————————————————– Volume of solution at time t minutes (l)
dQ
50
(4 – k ) dt
1 = 20 1 + —t 5
It is given that the amount of salt in the tank at time t minutes is Q.
Q
0
50 – 10 = (4 – k )(20 )(20 – 0) 2 = 4 – k k = 2 k = 4 [shown] When k = 4, 4 ——— 4– 4 4– 4 Q = 20 1 + ——–— t 10
Hence, the volume of solution in the tank at time t minutes = Initial volume of solution + δV = 10 + (4 – k )t
20
10
10
Rate of volume of distilled water pouring into the tank – Rate of volume of mixture leaving the tank
dQ
20
[V ]
dV δ V V = —– × δ t t dt = (4 – k )( )(t – 0) = (4 – k )t
Q
50
∫ dV = ∫
k
——
k – 4
C0 × 10C5 = 1 – ————— 12 C5 2
Actual 2008 STPM Mathematics Examination Paper
11
0.4 × 0.7 = ———— 0.64 = 0.4375 Hence, the percentage of ripe fruits which are mangoes = 0.4375 × 100 = 43.75% = 43.8% [ correct to three significant figures]
252 = 1 – ——– 792 15 = —– 22 8. X ~ B(n, p)
1 2
X ~ B 500, —
1 µ = E( X X ) = np = 500 × — = 250 2 1 1 2 σ = npq = 500 × — × — = 125 2 2 1 Since n > 50, p = — and np > 5, then the 2 normal approximation is used. X ~ N(250, 125) approximately P(| X )| ≤ 25) X – E( X X )| = P(| X X – 250| ≤ 25) = P(–25 ≤ X – 250 ≤ 25) = P(225 ≤ X ≤ 275) = P(225 – 0.5 < X < 275 + 0.5) [Taking continuity correction] = P(224.5 < X < 275.5) 224.5 – 250 X – 250 275.5 – 250 = P ————— < ———–– < ———–––— 125 125 125 = P(–2.281 < Z < 2.281) = 1 – 0.0113 – 0.0113 = 0.9774
0.0113
0.0113
z –2.281
2.281
9. M – Event that a mango is chosen P – Event that a papaya is chosen R – Event that a ripe fruit is chosen Outcomes 0.7
R
MR
— R
— MR
R
PR
— R
— PR
M 0.4
0.3 0.6
0.6
P 0.4
(a) P(fruits are ripe) = P( MR MR) + P(PR) = (0.4 × 0.7) + (0.6 × 0.6) = 0.64 Hence, the percentage of fruits which are ripe = 0.64 × 100 = 64% (b) P( M M / R R) P( M M ∩ R) = ———— R) P( R 12
Actual 2008 STPM Mathematics Examination Paper
10. Current Mid-point Cumulative Frequency x) Frequency f ) (C amperes) ( x ( f
fx
10 ≤ C < 11
10.5
8
8
84.0
11 ≤ C < 12
11.5
30
22
253.0
12 ≤ C < 13
12.5
63
33
412.5
13 ≤ C < 14
13.5
88
25
337.5
14 ≤ C < 15
14.5
97
9
130.5
15 ≤ C < 16
15.5
99
2
31.0
16 ≤ C < 17
16.5
100
1
16.5
Σ f =
Σ fx
100
= 1265
1265 – Σ fx Mean, x = —— = ——— = 12.65 Σ f 100 The median class is 12 ≤ C < 13. N
— – F 2 Median, M = Lm + ——–— c f m
100 —— – 30 2 = 12 + ————— (13 – 12) 33 = 12.61 The modal class is 12 ≤ C < 13.
d
1 Mode, mo = Lm + ——–— c o d 1 + d 2 11 = 12 + ——— (1) 11 + 8 = 12.58 Since mean > median > mode, the distribution is positively skewed.
11.
x < 0, ⎫0, 5 ⎪— – x, 0 ≤ x < 1, f ( x x) = ⎬ 4 1 ⎪——, x ≥ 1. ⎭ 4 x 2 x
∫ f ( x x) dx = ∫ 0 dx
(a) For x < 0, F ( x x) =
–∞
x
–∞
=0 x
∫ f ( x x) dx = ∫ f ( x x) dx + ∫ f ( x x) dx 5 = F (0) (0) + ∫ — – x dx 4
For 0 ≤ x < 1, F ( x x) =
–∞ 0
x
–∞
0
x
0
5 x2 = 0 + — x – — 4 2
x
= 1 – 0.0183 – 0.0733 – 0.1465 – 0.1954 = 0.567 [correct to three significant figures]
0
5 x2 = — x – — 4 2 For x ≥ 1, F ( x x) =
x
∫ f ( x x) dx
(b)
Even if the number of cars requested is more than 4, the number of cars rented out is still 4 because there are only 4 cars available to be rented out.
–∞
=
1
∫
–∞
x
∫ f ( x x) dx
f ( x x) dx +
1
x
= F (1) (1) +
1
dx ∫ —— 4 x 2
1
x
= F (1) (1) +
1
∫ —4 x
–2
dx
1
2
–1
5 1 1 x = —(1) – — + — —– 4 2 4 –1 3 1 = — – —— 4 4 x
(b) P( X X ≤ 3) =
2
3
≥4
0.0183 0.0733
0.1465
0.1954
0.5665
xP( X = x)
0
0.2930
0.5862
2.2660
0.0733
x
E( X X ) = Σ xP( X X = x) = 0 + 0.0733 + 0.2930 + 0.5862 + 2.2660 = 3.2185 Hence, the expected daily income from the rentals of cars = 3.2185 × 50 = RM160.93 e–4 44 X = 4) = ——— = 0.1954 (c) P( X 4! P( X X ≥ 5) = 1 – P( X X = 0) – P( X X = 1) – P( X X = 2) – P( X X = 3) – P( X X = 4) = 1 – 0.0183 – 0.0733 – 0.1465 – 0.1954 – 0.1954 = 0.3711
1
x
1
1 = 1 – —– 4 x
F ( x x) =
1
P( X = x)
3 1 1 = — – —– – —— 4 4 x 4(1) ⎫0, 5 x 2 ⎪— x – —–, 2 ⎬4 1 ⎪1 – —–, ⎭ 4 x
0
x
x < 0,
0 ≤ x < 1, x ≥ 1.
3
∫ f ( x x) dx –∞
= F (3) (3) 1 = 1 – —— 4(3) 11 = —– 12 P(at least one of two independent X is greater than 3) observed of X = 1– P(both the independent observed values of X X is less than or equal to 3) = 1 – P( X X ≤ 3) • P( X X ≤ 3) 11 11 = 1 – —– —– 12 12 23 = —— 144
12. Let X represent the number of cars requested. X ~ Po (4) in a day e–4 40 X = 0) = ——– = 0.0183 (a) (i) P( X 0! [correct to three significant figures] (ii) P( X X ≥ 4) = 1 – P( X X = 0) – P( X X = 1) – P( X X = 2) – P( X X = 3) e–4 40 e–4 41 e–4 42 e–443 = 1 – ——– – ——– – ——– – ——– 0! 1! 2! 3!
Even if the number of cars requested is more than 5, the number of cars rented out is still 5 because there are only 5 cars available to be rented out.
x
0
P( X X = x)
1
2
3
4
≥5
0.0183 0.0733 0.1465 0.1954 0.1954 0.3711
xP( X X = x) 0
0.0733 0.2930 0.5862 0.7816 1.8555
E( X X ) = Σ x P( X X = x) = 0 + 0.0733 + 0.2930 + 0.5862 + 0.7816 + 1.8555 = 3.5896 Hence, the expected daily income from the rentals of cars = 3.5896 × 50 = RM179.48 Additional daily income = RM179.48 – RM160.93 = RM18.55 Since the additional daily income (RM18.55) is less than the additional cost incurred per day (RM20.00), then the shop should not buy another car for rental. Actual 2008 STPM Mathematics Examination Paper
13