Test - 2 (Paper-I) (Code-A) (Answers & Hints)
All In India Aakash Test Se Series fo for JE JEE (M (Main)-2018
All India Aakash Test Series for JEE (Main)-2018
TEST - 2 (Code-A) Test Date : 27/08/2017
ANSWERS PHYSICS
CHEMISTRY
MATHEMATICS
1.
(4)
31.
(3)
61.
(2)
2.
(3)
32.
(2)
62.
(4)
3.
(2)
33.
(4)
63.
(2)
4.
(4)
34.
(4)
64.
(4)
5.
(1)
35.
(1)
65.
(1)
6.
(1)
36.
(2)
66.
(1)
7.
(3)
37.
(2)
67.
(4)
8.
(4)
38.
(4)
68.
(1)
9.
(4)
39.
(4)
69.
(2)
10.
(4)
40.
(4)
70.
(4)
11.
(2)
41.
(3)
71.
(4)
12.
(4)
42.
(3)
72.
(1)
13.
(2)
43.
(1)
73.
(2)
14.
(4)
44.
(1)
74.
(4)
15.
(4)
45.
(4)
75.
(4)
16.
(2)
46.
(3)
76.
(2)
17.
(4)
47.
(4)
77.
(1)
18.
(1)
48.
(4)
78.
(4)
19.
(3)
49.
(1)
79.
(1)
20.
(3)
50.
(1)
80.
(1)
21.
(1)
51.
(3)
81.
(3)
22.
(2)
52.
(3)
82.
(1)
23.
(2)
53.
(2)
83.
(1)
24.
(4)
54.
(3)
84.
(1)
25.
(3)
55.
(2)
85.
(3)
26.
(3)
56.
(1)
86.
(3)
27.
(4)
57.
(2)
87.
(4)
28.
(2)
58.
(3)
88.
(1)
29.
(2)
59.
(3)
89.
(4)
30.
(2)
60.
(3)
90.
(4)
1/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Code-A) (Answers & Hints)
PART - A (PHYSICS) 1.
Answer (4)
6.
Answer (1)
dB 2 R B0 dt
2 As E × 2 l = R
2
E
R
2l
B0 and
for equilibrium of charge
B
x
2 1 qR B0 mg x K 2l
2.
B
Answer (3)
As F q v B
∫
dx
0i 2 x
B.dl
2
a2
0idx cos 90 –
a
∫
a
2 x
3
Force on electron will be along BO direction 3.
2
a2
ia 1 –1 x 0 tan 2 a a
Answer (2)
a
a
3
v y
v
30°
B v x
r
Mv y qB
and p
qB
v
0i 0i – 6 24 2 4
7.
Answer (3)
8.
Answer (4)
9.
Answer (4) As
2M
θ 9 0 –
θ
qE + mg = Kx
E B
V speed of wave and
V
K
E = V × B
x
3
108 4 10–6 600 V/m
Mv y v v sin30 qB y 2v x 2 v cos 30 p 2Mv x qB r
2
10. Answer (4) A i
r
2 3
4.
Answer (4)
5.
Answer (1) As time constant
i
(2, 0)
L R
– t 1 – e 6 1 – R
V
4 1
B
4 seconds
F
1
3.8 A
i dl B
2 2
e
Power P = i 2R = 14.4 W
2/8
y 6
p
F
8
6 2 – i ˆ N
6 – i ˆ N
30 10
9
200
Test - 2 (Code-A) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018 15. Answer (4)
11. Answer (2) 2
M
L
Q ⇒
2m0
M
LQ
2m0
5
m0 R
2
Emf induced between O and A
Q
2m0
1 5
R
2
Q
1
E
2
12. Answer (4)
final
π
α
–
α=δ
2
i
so
in loop
Ba2 a 2 2R 2
a
1
=π –
Total deviation = 1 2 2 –
B dl
∫
2
qB
=
2
2m
2
2
2R
a 4
4R
0 i i d
0 i 0
d E dt
d E 0 i 0 0 dt
18. Answer (1)
–
u avg =
2
=
2 –
2m
B
Ba
17. Answer (4)
β
Correct equation is
T
R
Due to the flux change eddy currents will develop and retard the motion till a constant speed is achieved.
Here =
As t =
E
16. Answer (2)
β δ
2
Bia B 2
initial
α
Ba
1 2
0 E 2 =
0 2
10 4
–
Volume of cylinder = 10 cm2 × 50 cm = 5 × 10 –4 m3
qB
U = u avg × Volume
13. Answer (2) V L
0
=
104 5 10–4
2
50 2
19. Answer (3) i
V R
20. Answer (3) 21. Answer (1)
V C
22. Answer (2)
At resonance V L = V C
= 0°
Electric field between the plates of capacitor
So power factor cos = 1
If electron moves on straight line
14. Answer (4)
F e = F m
Emf in square of side ‘a’ e1
d dt
Ba B a 2
2
0
In small square
e2
cos t
Net emf = e1 – e2 = i
i
e1 – e2 R
2
B0 a – b
2
b2)
cos t
4 a b
a – b B0 cos t 4
eE = evB
B0 b2 cos t B0(a2 –
E
t
l v
lB E
cos t
v
lB
0
E B
0lB
23. Answer (2) For BNet = 0
0 i 2 R
–
0i 0 4 R
= 2 rad
3/8
0
All India Aakash Test Series for JEE (Main)-2018 24. Answer (4)
Test - 2 (Code-A) (Answers & Hints) 29. Answer (2)
VL
– V
A
V
c
F 1
F 2 F
φ V R
i
B
As after resonant frequency X L dominates
Net force per unit length on A
25. Answer (3) N
F
e
n i s
v
v
2R
R
45°
M
v cos
2
F1
60 F22 2F1F2 cos
0 i 2 F 2 where F1 2 a F = 2 F 1 cos 30 =
0 i 2 F 2 a
2R v cos
F 1 3
3
30. Answer (2)
B 2Rv cos 45
=
v
R
B
C
1
e = BRv
α α
With N at high potential
– 0 9
26. Answer (3)
2
UC UL initial UC UL final
⇒
1 Total initial energy Energy in capacitor 2
1 1Q
2 2 Q
2
2
C
Q
1 Q
tan 1
2 C
⇒
r
B
0H
B cos H
2
tan 2
8 4 10
–7
BV
tan cos
In second plane
Q
Q
27. Answer (4)
2
2
2
In first plane
2 103
BV BH sin
tan sin
cos = tancot1, sin = tan cot2
104
cos2 + sin2 = tan2 [cot21 + cot22] cot2 = cot21 + cot22
28. Answer (2)
PART - B (CHEMISTRY) 31. Answer (3)
34. Answer (4)
In electrolytic refining impure metal act as anode. 32. Answer (2)
35. Answer (1)
33. Answer (4) CaO + SiO2 Impurity
CaSiO3
(Slag)
In blast furnace reduction of iron oxides takes places in different temperature range.
4/8
Zone refining is used for refining of high pure metal that are used in semiconductor. Siderite FeCO3. 36. Answer (2) 37. Answer (2) AgCl, Cu(OH)2 and Zn(OH)2 are soluble in aqueous NH3 solution.
Test - 2 (Code-A) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018 48. Answer (4)
38. Answer (4) NiSO4 + 4Py + 2NaNO 2 Na2SO4 + [Ni(Py)4](NO2)2 39. Answer (4) Bauxite contains the impurity of SiO2, iron oxide and TiO2.
Compound
E.A.N
Ni(CO)4
36
Mn(CO)5
35
Fe(CO)5
36
49. Answer (1)
40. Answer (4)
50. Answer (1)
Mn can show +7 oxidation state.
Fact.
41. Answer (3) [Cr(NH3)4(CN)Cl]Br is optically inactive.
51. Answer (3)
42. Answer (3) NH3 2– Cr2O7
+ 4H2O2 +
2H+
2CrO5 + 5H2O
43. Answer (1) 2+
Pb
Co NH3
+ 2OH Pb(OH)2 –
Pb(OH)2 + 2OH
[Pb(OH)4]2–
Cis (optically active). 52. Answer (3)
Soluble in excess of reagent. Cd2+ + 2OH
[Cr(en)3]Cl3 cannot show geometrical isomerism.
Cd(OH)2
Precipitate is insoluble in excess of reagent Ni2+ + 2OH Ni(OH)2
Fact. 54. Answer (3)
The precipitate is insoluble in excess reagent. Mg 2+ + 2OH reagent.
53. Answer (2)
Mg(OH2) insoluble in excess
Hg2+ does not give precipitate with dil. HCl. 55. Answer (2) Ag2S2O3 + H2O Ag2 S + H2SO4
44. Answer (1)
(Black) Y
56. Answer (1) 6.35
Moles of Cu =
63.5
In lanthanoids as atomic number increase size decrease.
= 0.1 mol
57. Answer (2)
Formula = [Cu(H2O)4]SO4·H2O Moles of water = 0.4
Rotten egg smell gas is H2S.
Mass of water = 0.4 × 18 = 7.2 g
Na2S + dil. H2SO4 H2S + Na 2SO4 H2S + Pb2+ PbS + 2H+
45. Answer (4)
Blackening of filter paper moistened with lead acetate solution.
46. Answer (3) Spin only magnetic moment() =
n(n 2) B.M.
n Number of unpaired electron. Cu2+, n = 1, = 1.73 B.M. Cu , n = 0, = 0 +
Cr , n = 5, = 5.9 B.M. +
Zn , n = 0, = 0 2+
47. Answer (4) Name of positively charged ligands end with ium.
58. Answer (3) NO3
+ 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O
2+ [Fe(H2O)6]2+ + NO [Fe(H2 O)5NO] + H2O Brown
59. Answer (3) HCl give white fumes with NH3. 60. Answer (3) [Ag(NH3)2][Ag(Cl)2] Diamminesilver(I) dichloridoargentate(I)
5/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Code-A) (Answers & Hints)
PART - C (MATHEMATICS) 61. Answer (2) As
lim
0
70. Answer (4)
the function approaches to
x
3 e
0 hence
e R ~ {0}.
62. Answer (4) As lim f ( x ) –1, lim f ( x ) 1 and f (1) = 0 1–
1
x
x
Range = {–1, 0, 1}
x 2 – x 5 2 – x x 5 f ( x ) = x 2 – x 5 9 – x
; 0
x
1
; 1
x
;
2
x
2
72. Answer (1)
1 1 – x – 1 x –
y = max f
n
2
y = x
1 –1 1 1 h0 – – h – – 2 2 lim
h
) 1 , 0 (
–π
π
–3π 2
2 2 – 1 2h 1 2h h0
=
0
f ( x ) is continuous everywhere but non-differentiable at x = 0, 1 and 2.
lim
=
x
71. Answer (4)
63. Answer (2)
x
;
lim
–π
π
2
2
2π
x
Always continuous.
2 = lim – h0 1 2h
73. Answer (2) y = f(x )
= –2 64. Answer (4)
2e e e– x
Simplifying lim loge
x
x
x
log 2 Q
x
c
e
65. Answer (1) 66. Answer (1) 1
d 2 y
2 As lim 3 1 3 h ⇒ 3 3
For
The required limit value is 1.
d 2y If 2 dx
x
x
dx 2
=0
at x = a
x
67. Answer (4)
d 2 y 2 dx x
2
lim
h .sin h
h 0
h
3
1
68. Answer (1) lim f x
1
4 (from less than 4) hence G.I.F. is 3
x
69. Answer (2) The function |sin x | is not differentiable at x = n for all n being integer.
6/8
a–
x
0 a
Then x = a is a point of inflection. 74. Answer (4) As f ( x ) will intersect at least once in each x (a, b) and (b, c ) and (c , d ) hence three zeros atleast. 75. Answer (4) For f ( x ) = 3ax 2 + 2bx + c the extremum will exist if 4 b2 – 4 (3ac ) > 0 but not in all cases.
Test - 2 (Code-A) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018 81. Answer (3)
76. Answer (2) For f ( x ) = x x (1 + loge x )
1 f 0, as well as e
g ( x ) will not be differentiable at x
lim
x
0
1
0,
–2 5
,
–2 3e 5
82. Answer (1)
x
A
y = f(x ) (0, 1)
(1, 1)
13
5
x
O
x
1 x = e
B
1 ⇒ , is longest bijective interval with e
1 b e
1 e
f ( x ) = ⇒
2 ⇒ a b 0 e
C
12
–1
sin
tan
e
12 sin x 5 cos x –1 sin sin x 13 5 ⇒
0
12
4
As x varies in real the slope of f ( x ) will be –1 or 1 only.
77. Answer (1)
83. Answer (1)
78. Answer (4)
y = f(x )
y = f(x )
f (x ) > 0
y = 1
y = x y = 0 O
x –1
y = f (x ) Always continuous
x
79. Answer (1) Let x 0
g ( x ) g (0)
f ( x ) > 0
f (g ( x )) f (g (0))
concave upwards
h( x ) 0
But h( x ) 0
R
f (1) = 9 and f (1) = 5
1 1
x
f ( x ) is bijective function
80. Answer (1)
dx
(f –1(d )) 0
As f ( x ) = 6 x 2 + 3
h( x ) is a zero function.
0
2
84. Answer (1)
h( x ) = 0
As
2
d
2
, x 0 2 1 x 1 x 2 1
1 hence
x
Therefore, f ( x ) =
x – 1
and f ( x ) = 1 at x = 0
1
1
x R ~ 0 2
1 x
Therefore, f –1(5) = 1 and
(y – 1) =
1 9
f –1 5
1 f 1
x – 5
9y – x = 4
7/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Code-A) (Answers & Hints)
85. Answer (3)
e – ; y = e ; x
x
88. Answer (1) x
0
x
0
y (b, f(b))
y = e|x | is symmetric about the y -axis.
} v , u {
y
2a 3b 2a 3 b , f 5 5
(a, f(a))
(, e )
x
=
– tan
v
2a
e
x
3b
2f a 3f b 5
As evident
Angle between tangents is – 2tan–1e. 86. Answer (3)
v
As f ( x ) = f –1(0) = Roots of f ( x ) ⇒
b
5
2
Hence f ( x ) = x 1, x 2, x 3
3b
Here
= tan–1e –1
2a
5
u
a
O
(0, 0)
2a 3b 5
f
89. Answer (4)
9 intersection points
As f {f ( x )} < f (3 – 4 x ) is required y = f(x ) (–2,19) (–1, 14) (0,3) –3 –2 –1 1 2 x 0 x
–4
1
and f ( x ) = –2 – 6 x 2
(4,19)
f ( x ) f ( x ) > 3 – 4 x 3 – 2 x – 2 x 3 > 3 – 4 x
3
4 x
2 x – 2 x 3 > 0 x ( x – 1) ( x + 1) < 0
x
3
2
(1, –8)
(3,–6)
(–4,–13)
(2, –13)
87. Answer (4) b
2a
⇒
a –1
f a
ab
2
a –1
4a a – 1 – 2a
a – 12
0
–
1
90. Answer (4)
f a , now 2
As g {f ( x )} = x 1
g {f ( x )} = f x
2
2a – 4 a
a – 12
Now f (1) = 2 hence g 2
f (a) = 0 a = 2 f (2) is minimum
8/8
– –1
2a
+
+
1
f 1
1 4
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018
TEST - 2 (Paper-I) - Code-B Test Date : 27/08/2017
ANSWERS PHYSICS
CHEMISTRY
MATHEMATICS
1.
(2)
31.
(3)
61.
(4)
2.
(2)
32.
(3)
62.
(4)
3.
(2)
33.
(3)
63.
(1)
4.
(4)
34.
(2)
64.
(4)
5.
(3)
35.
(1)
65.
(3)
6.
(3)
36.
(2)
66.
(3)
7.
(4)
37.
(3)
67.
(1)
8.
(2)
38.
(2)
68.
(1)
9.
(2)
39.
(3)
69.
(1)
10.
(1)
40.
(3)
70.
(3)
11.
(3)
41.
(1)
71.
(1)
12.
(3)
42.
(1)
72.
(1)
13.
(1)
43.
(4)
73.
(4)
14.
(4)
44.
(4)
74.
(1)
15.
(2)
45.
(3)
75.
(2)
16.
(4)
46.
(4)
76.
(4)
17.
(4)
47.
(1)
77.
(4)
18.
(2)
48.
(1)
78.
(2)
19.
(4)
49.
(3)
79.
(1)
20.
(2)
50.
(3)
80.
(4)
21.
(4)
51.
(4)
81.
(4)
22.
(4)
52.
(4)
82.
(2)
23.
(4)
53.
(4)
83.
(1)
24.
(3)
54.
(2)
84.
(4)
25.
(1)
55.
(2)
85.
(1)
26.
(1)
56.
(1)
86.
(1)
27.
(4)
57.
(4)
87.
(4)
28.
(2)
58.
(4)
88.
(2)
29.
(3)
59.
(2)
89.
(4)
30.
(4)
60.
(3)
90.
(2)
1/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
PART - A (PHYSICS) 1.
1 1 Q
Answer (2)
2 2
1 α α
– 0 9
2 Q
C Q
1 Q
2
2 C
2
⇒
2
2
6.
2
Q
Q
2
Answer (3) N
In first plane
cos B H
tan 1
tan
BV
tan 2
BV BH sin
tan
e
sin
2.
2R v cos
B 2Rv cos 45
Answer (4)
A
VL
F 1
– V
V
c
F 2
φ
F
V R
B
C
2
F1
Answer (2) For BNet = 0
60 F22 2F1F2 cos
0i 2 R
0 i 2 F 2 F where 1 2 a F = 2 F 1 cos 30 =
i
As after resonant frequency X L dominates 8.
Net force per unit length on A F
v
With N at high potential 7.
Answer (2)
v cos
e = BRv
cos2 + sin2 = tan2 [cot21 + cot22] cot2 = cot21 + cot22
R
B =
cos = tancot1, sin = tan cot2
R
45°
M
In second plane
v
v
2R
cos
n i s
–
0 i 0 4 R
= 2 rad
F 1 3
9.
Answer (2)
2
F
0 i 2 a
Electric field between the plates of capacitor E
3
3.
Answer (2)
If electron moves on straight line
4.
Answer (4)
F e = F m
r
5.
B
0H
8 4 10
–7
2 10
3
104
Answer (3) UC UL initial UC UL final ⇒
1 2
2/8
Total initial energy Energy in capacitor
eE = evB t
l v
lB E
10. Answer (1)
v
lB
0
E B
0lB
0
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018
11. Answer (3)
At resonance V L = V C
12. Answer (3)
= 0°
13. Answer (1)
So power factor cos = 1
1
u avg =
2
0
0E 2 =
2
19. Answer (4)
10 4
Volume of cylinder = 10 cm2 × 50 cm = 5 × 10 –4 m3
final
U = u avg × Volume
=
0
104 5 10 –4
2
50
∫ B dl i i
2
=π –
d E dt
d E 0 i 0 0 dt
As t =
T
2 –
2
2m
15. Answer (2) Due to the flux change eddy currents will develop and retard the motion till a constant speed is achieved.
qB
=
2
– =
2
Emf induced between O and A E
2
Ba
Bia
2m
– qB
20. Answer (2)
16. Answer (4)
1
β
Total deviation = 1 2 2 –
d
1
Here =
0 i 0
α=δ
2
δ
0
–
β
Correct equation is
π
α
14. Answer (4)
initial
α
2
i
so
in loop
Ba 2 a 2 B R 2 2
a 2
E R B
2
2
Ba
2
M
L
2R
Q ⇒
2m0
M
LQ 2m0
5
m0R
2
Q
2m0
1 5
R
2
Q
21. Answer (4)
a 4
A
4R
i
17. Answer (4)
y 6
Emf in square of side ‘a’ e1
d dt
Ba
2
B0a2 cos t e2
In small square
(2, 0)
B0 b2 cos t
Net emf = e1 – e2 = B0(a2 – b2) cos t i
i
e1 – e2 R
2
B0 a – b
2
cos t
B
F
2 2
4 a b
a – b B0 cos t
i dl B
F
4
8
6 2 – i ˆ N
6 – i ˆ N
22. Answer (4)
18. Answer (2) V L
As
E B
V speed of wave and
V
K
30 10
9
200
E = V × B i V C
V R
3
108 4 10–6 600 V/m 2
3/8
All India Aakash Test Series for JEE (Main)-2018 23. Answer (4)
Test - 2 (Paper-I) (Code-B) (Answers & Hints) 28. Answer (2) v y
24. Answer (3) 25. Answer (1)
v
30°
B v x
B
θ 9 0 –
θ
x
B
2
a
∫ B.dl ∫
qB
dx
and p
0i 2 x
Mv y
r
a2
2 x
3
2
a
ia 1 –1 x 0 tan 2 a a
r p
2
r
a
x
v v sin30 qB y 2Mv x 2 v x 2 v cos 30 qB p 2 3
29. Answer (3)
As F
0i 0i – 6 24 2 4
q v B
Force on electron will be along BO direction
26. Answer (1)
30. Answer (4)
As time constant
i
v
a
3
qB
Mv y
0idx cos 90 –
a
2M
L R
4 1
4 seconds
dB 2 R B0 dt
2 As E × 2 l = R
2
– t 1 1 – e 6 1 – 3.8 A R e
E
V
R
2l
B0 and
for equilibrium of charge
qE + mg = Kx
2
Power P = i R = 14.4 W
qR 2B0 mg x K 2l 1
27. Answer (4)
PART - B (CHEMISTRY) 31. Answer (3)
34. Answer (2)
[Ag(NH3)2][Ag(Cl)2] Diamminesilver(I) dichloridoargentate(I) HCl give white fumes with NH3. 33. Answer (3) + 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O
2+ [Fe(H2O)6]2+ + NO [Fe(H2O)5NO] + H2O Brown
4/8
Na2S + dil. H2SO4 H2S + Na 2SO4 H2S + Pb2+ PbS + 2H+
32. Answer (3)
NO3
Rotten egg smell gas is H2S.
Blackening of filter paper moistened with lead acetate solution. 35. Answer (1) In lanthanoids as atomic number increase size decrease.
Test - 2 (Paper-I) (Code-B) (Answers & Hints) 36. Answer (2)
All India Aakash Test Series for JEE (Main)-2018 48. Answer (1)
Ag2S2O3 + H2O Ag2 S + H2SO4 (Black) Y
Pb2+ + 2OH– Pb(OH)2 Pb(OH)2 + 2OH
37. Answer (3) Hg2+ does not give precipitate with dil. HCl.
Soluble in excess of reagent. Cd2+ + 2OH
38. Answer (2) Fact.
[Pb(OH)4]2–
Cd(OH)2
Precipitate is insoluble in excess of reagent
39. Answer (3) [Cr(en)3]Cl3 cannot show geometrical isomerism. 40. Answer (3)
Ni2+ + 2OH Ni(OH)2 The precipitate is insoluble in excess reagent. Mg 2+ + 2OH reagent.
NH3
Mg(OH 2) insoluble in excess
49. Answer (3) Co 2–
Cr2O7
NH3
+ 4H2O2 + 2H+ 2CrO5 + 5H2O
50. Answer (3) Cis (optically active).
[Cr(NH3)4(CN)Cl]Br is optically inactive.
41. Answer (1)
51. Answer (4)
Fact.
Mn can show +7 oxidation state.
42. Answer (1)
52. Answer (4)
43. Answer (4) Compound
E.A.N
Ni(CO)4
36
Mn(CO)5
35
Fe(CO)5
36
Bauxite contains the impurity of SiO2, iron oxide and TiO2. 53. Answer (4) NiSO4 + 4Py + 2NaNO 2 Na2SO4 + [Ni(Py)4](NO2)2 54. Answer (2)
44. Answer (4) Name of positively charged ligands end with ium. 45. Answer (3)
55. Answer (2)
Spin only magnetic moment() = n Number of unpaired electron. Cu2+, n = 1, = 1.73 B.M. Cu+,
n = 0, = 0
Cr +, n = 5, = 5.9 B.M. Zn2+, n = 0, = 0
n(n 2) B.M.
56. Answer (1) Siderite FeCO3. 57. Answer (4) Zone refining is used for refining of high pure metal that are used in semiconductor. 58. Answer (4)
46. Answer (4)
CaO + SiO2
47. Answer (1)
Impurity
6.35
Moles of Cu =
AgCl, Cu(OH)2 and Zn(OH)2 are soluble in aqueous NH3 solution.
63.5
= 0.1 mol
(Slag)
In blast furnace reduction of iron oxides takes places in different temperature range.
Formula = [Cu(H2O)4]SO4·H2O
59. Answer (2)
Moles of water = 0.4
60. Answer (3)
Mass of water = 0.4 × 18 = 7.2 g
CaSiO3
In electrolytic refining impure metal act as anode.
5/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
PART - C (MATHEMATICS) f a
61. Answer (4) As g {f ( x )} = x
4a a – 1 – 2a
2
a – 12
2
2a – 4 a
a – 12
f (a) = 0 a = 2 f (2) is minimum
1
g {f ( x )} = f x
65. Answer (3)
Now f (1) = 2 hence g 2
1
f 1
1 4
As f ( x ) = f –1(0) = Roots of f ( x ) Hence f ( x ) = x 1, x 2, x 3
9 intersection points
⇒
y = f(x )
62. Answer (4) As f {f ( x )} < f (3 – 4 x ) is required
(4,19)
(–2,19) (–1, 14)
and f ( x ) = –2 – 6 x 2
f ( x ) f ( x ) > 3 – 4 x –4
3
3 – 2 x – 2 x > 3 – 4 x
(0,3) –3 –2 –1 1 2 x 0 x 1
3
2
(1, –8)
2 x – 2 x 3 > 0 x ( x – 1) ( x + 1) < 0
(–4,–13)
4
x
x
3
(3,–6) (2, –13)
66. Answer (3) +
+
– –1
–
0
1
63. Answer (1)
e – ; y = e ; x
x
x
0
x
0
y = e|x | is symmetric about the y -axis.
y
y (b, f(b))
(, e ) } v , u {
(a, f(a))
a
O
2a
3b
b
= tan–1e = – tan–1 e
Here
v
2a 3b
2
Angle between tangents is – 2tan–1e.
5 2f a 3f b
67. Answer (1)
5
As f ( x ) = 6 x 2 + 3
As evident v
f ( x ) is bijective function
2a 3b 5
f (1) = 9 and f (1) = 5
f
Therefore, f –1(5) = 1 and
64. Answer (4) b
6/8
2a a –1
x
x
5
u
(0, 0)
2a 3b 2a 3 b , f 5 5
⇒
ab
2a
2
a –1
f a , now
(y – 1) =
1 9
9y – x = 4
x – 5
f –1 5
1 f 1
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
All India Aakash Test Series for JEE (Main)-2018
68. Answer (1)
72. Answer (1) Let x 0
y = f(x )
g ( x ) g (0)
f (x ) > 0 y = x
f (g ( x )) f (g (0)) h( x ) 0 But h( x ) 0
–1
y = f (x )
h( x ) = 0 x
h( x ) is a zero function. 73. Answer (4) y = f(x ) y = 1
f ( x ) > 0 concave upwards 2
d
2
dx
(f –1(d )) 0
y = 0 x
R
O
x
Always continuous
69. Answer (1)
74. Answer (1)
A
75. Answer (2) For f ( x ) = x x (1 + loge x )
13
5
1 e
f 0, as well as
B
⇒
–1
sin
tan
x
x
1
y = f(x )
12 sin x 5 cos x –1 sin sin x 13 5
0
C
12
f ( x ) =
lim x
⇒
12
0
(0, 1)
(1, 1)
4
O
As x varies in real the slope of f ( x ) will be –1 or 1
x 1 x = e
only.
1 ⇒ , is longest bijective interval with e
70. Answer (3) g ( x ) will not be differentiable at
x
0,
–2 5
,
As
0
1
2
x
3e 5
1 1 , x 0 1 hence 2 1 x 1 x 2
Therefore, f ( x ) =
x – 1
and f ( x ) = 1 at x = 0
1 2
1 x
1 e
–2
71. Answer (1) 1
1 b e
x R ~ 0
2 ⇒ a b 0 e e
76. Answer (4) For f ( x ) = 3ax 2 + 2bx + c the extremum will exist if 4 b2 – 4 (3ac ) > 0 but not in all cases. 77. Answer (4) As f ( x ) will intersect at least once in each x (a, b) and (b, c ) and (c , d ) hence three zeros atleast.
7/8
All India Aakash Test Series for JEE (Main)-2018
Test - 2 (Paper-I) (Code-B) (Answers & Hints)
78. Answer (2)
83. Answer (1)
y = f( x )
lim f x
1
4 (from less than 4) hence G.I.F. is 3
x
84. Answer (4) 2
x
h .sin h
lim h 0
h
1
3
85. Answer (1) 1
2 As lim 3 1 3 h ⇒ 3 3 x
For
d 2 y 2
=0
dx
x
at x = a
The required limit value is 1.
d2y If 2 dx
d 2 y 2 dx x
a–
x
86. Answer (1)
0
87. Answer (4)
a
Simplifying lim loge
2e x
e e
x
Then x = a is a point of inflection.
x
– x
log 2 Q
c
e
88. Answer (2)
79. Answer (1) y = max f
n
lim
y = x
x
) 1 0, (
–π
π
–3π
–π
π
2
2
2
=
2π
x
x 2 – x 5 2 – x x 5 f ( x ) = x 2 – x 5 9 – x
;
x
; 0
0 x x
;
2
x
2
1 –1 1 1 h 0 – – h – – 2 2 lim
h
=
2 – 1 2h h 0
lim
lim
= –2
1
; 1
1
1 1 x – x
2 2 – 1 2h 1 2h h 0
80. Answer (4) 81. Answer (4)
–
–
= Always continuous.
89. Answer (4)
2
As lim f ( x ) –1, lim f ( x ) 1 and f (1) = 0 1–
x
90. Answer (2)
82. Answer (2)
As
The function |sin x | is not differentiable at x = n for all n being integer.
1
x
Range = {–1, 0, 1}
f ( x ) is continuous everywhere but non-differentiable at x = 0, 1 and 2.
8/8
x
lim
0
the function approaches to
x
e R ~ {0}.
3 e
0 hence