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Test - 7 (Code-G) (Answers) All India Aakash Test Series for JEE (Main)-2018
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1. (1)
2. (2)
3. (1)
4. (2)
5. (2)
6. (3)
7. (4)
8. (3)
9. (2)
10. (2)
11. (1)
12. (1)
13. (4)
14. (3)
15. (2)
16. (1)
17. (4)
18. (2)
19. (3)
20. (3)
21. (2)
22. (2)
23. (1)
24. (3)
25. (4)
26. (3)
27. (3)
28. (1)
29. (4)
30. (1)
PHYSICS CHEMISTRY MATHEMATICS
31. (2)
32. (2)
33. (1)
34. (2)
35. (4)
36. (2)
37. (3)
38. (4)
39. (2)
40. (1)
41. (3)
42. (2)
43. (1)
44. (1)
45. (3)
46. (3)
47. (2)
48. (3)
49. (3)
50. (4)
51. (3)
52. (4)
53. (1)
54. (4)
55. (4)
56. (2)
57. (4)
58. (3)
59. (1)
60. (3)
61. (1)
62. (3)
63. (4)
64. (4)
65. (3)
66. (4)
67. (3)
68. (2)
69. (4)
70. (1)
71. (1)
72. (4)
73. (2)
74. (3)
75. (3)
76. (3)
77. (3)
78. (3)
79. (2)
80. (2)
81. (2)
82. (1)
83. (1)
84. (4)
85. (4)
86. (1)
87. (2)
88. (3)
89. (1)
90. (4)
Test Date : 25/02/2018
ANSWERS
TEST - 7 (Code-G)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-G) (Hints)
2/8
1. ∵ Component of incident ray along normal is
ˆ ˆ( )i j .
to express reflected ray, ˆ ˆ( )i j reverses in
expression of incident ray.
ˆ ˆ ˆ– – –R i j k
�
= ˆ ˆ ˆ–( )i j k
ˆ ˆ ˆ–( )ˆ3
i j kR⎧ ⎫ ⎨ ⎬⎩ ⎭
2.
h = 2 m
i 90° O x
y
If is the angle of refraction for the horizontal rays.
(1 +
)
ay
1/2
1
ay
0
sin90
sin
sin = 1
1 ay, tan =
1
ay
1dx
dy ay
1 –12–2 2
0 0
x
dx a y dy∫ ∫
x =
2–1 1
2 20
2
⎡ ⎤⎢ ⎥⎣ ⎦a y =
1–1
–6 222 2 10 2 = 2000 m
4. The parallel rays will focus at focal point of concave
lens so after refraction from it, they become parallel.
5. When lenses are in contact
1 2
1 1 1
12f f ...(i)
When lenses are at separation
1 2 1 2
1 1 50– 0
f f f f ...(ii)
On solving equations (i) and (ii), we get
f1 = 20 cm, f
2 = 30 cm
PART - A (PHYSICS)
7. The focal length of the glass (lens) used 100
m,1.5
f
if y is the distance of near point, then
1 1 1.5––25 100y
{y = –40 cm}
The focal length of lens of microscope
fe =
1 100
20P = 5 cm
Magnifying power of microscope together with glass
M = 1
e
D
f =
251
5 = 6
In case of without glass, D = 40 cm = y
M = 1
e
D
f
= 40
15
= 9
8.
O
d
d
S
S
y
∵ Due to reflection. A phase difference of occurs.
Let the distance of first maxima from O is y
2
2 2
D
dy
⎛ ⎞⎜ ⎟ ⎝ ⎠
4
Dy
d
where, = fringe width.
9.
2r1
r2
e
i + e – A = 2, e =
1
sin2sinr
, 1
sinsin( – )r
on solving we get,
2–1
cos8
10. n11 = n
22
∵ n
11 = n
22
12 × 600 = n2 × 400
{n2 = 18}
Test - 7 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
3/8
11.
r1
r2
i2
i1
sin
sin
i
r
⎡ ⎤ ⎢ ⎥⎣ ⎦
= (i1 – r
1) + (i
2 – r
2)
For minimum deviation, i1 = i
2, r1 = r
2
sin60
sin30
= , = 3
12.
x
ir
y
∵ i + r = 90° and i = r
i = r = 45°, i + = 90°
= 90° – i = 90° – 45° = 45°
∵ Given equation of surface, y2 = 2x
2y
dy
dx = 2
1dy
dx y = tan45°
y = 1
x =
21
2 2
y
Coordinates are 1,1
2
⎛ ⎞⎜ ⎟⎝ ⎠
13. ∵ A1 = 4°,
1 = 1.54,
2 = 1.72
Condition for no deviation
A2 =
1
1
2
( – 1)–( – 1)
A
1.54 – 1
– 41.72 – 1
⎛ ⎞ ⎜ ⎟⎝ ⎠
A2 = –3° |A
2| = 3°
15.
O
36 c
m
24 c
m
x
C
For lens:
∵ v = –24 cm, u = –x
1 1 1–24 x f
, 1 1 1–24 40x
x = 15 cm
17. 3 cos = 2
cos = 2
3
3
tanX
D
X = 5
2D
18. According to diagram given in question
OP = d sec, CO = OP cos2 = d sec cos2 Path difference, x = CO + OP
x = d sec + d sec cos2Effective path difference
xeff
= d(sec + seccos2) + 2
For constructive interference, xeff
=
d sec (1 + cos2) + 2
=
or cos4d
19. For A:
2.
2 4
⎡ ⎤ ⎢ ⎥ ⎣ ⎦IA = I + I + 2I cos = 0
For B:
–4
x PB QB⎛ ⎞ ⎜ ⎟
⎝ ⎠ = (PB – QB) +
4
= 04 4
2
IB = I + I + 2I cos
2
= 2I
For C:
2– 0
2 4
⎛ ⎞ ⎜ ⎟ ⎝ ⎠IC = I + I + 2I cos0° = 4I
20. ∵ ,
h
mv so with the increase in velocity of
electron, wavelength decreases and so fringe width
decreases.
22. I1 = 4I
0, I2 = I
0, Imax
= 9I0, Imin
= I0
23. The deviation produced by the prism
= ( – 1)A = (1.5 – 1) × 4° = 2°
Thus the angle through which mirror should be rotated
is 1°.
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-G) (Hints)
4/8
PART - B (CHEMISTRY)
32. LiAlH4 reduce ester group and H
2/Ni reduce C C
also.
33.
+ CH —CH=CH3 2
H+
CH
CH3
CH3
(A)
O2
CH —C—O—OH3
CH3
(B)
H+
/H O2
CH —C—CH3 3
+
O
OH
(C)
24. If P is equivalent power of the combination, then
2 1
15
P ,
1–
m
Pf
By using mirror equation
1 1 1
–20 –7.5v
– 12 cmv
25. For first lens,
1
1 1 1––30 10v
v1 = 15 cm at I
1
For second lens, u2 = 15 – 5 = 10 cm
5 cm 10 cm 30 cm
I1
I
O
The focal length of second lens is also 10 cm, thus
it forms the image at infinity.
Finally the third lens forms the image at its focal point
i.e., 30 cm from it.
26. Image formed by lens should be on the pole of mirror.
28. ∵ Fringe width
D
d
and new fringe width
= (1.1 )
(1.1 )
D D
d d
= 0.2 mm
29.
d = 10 cm
I1 I
3
I2
For refraction through concave lens:
1 1 1–
v u f
, 1 1 –1–– 10v
v = –10 cm
I2 is 20 cm behind the plane mirror.
1 1 –1––(30) 10v
1 –1 1 –4–
10 30 30v
= –7.5 cm
I3 is formed at 2.5 cm in front of mirror.
30. I =2 20
cos cos (90 – )2
I⎡ ⎤ ⎢ ⎥⎣ ⎦
I = 2 20
cos sin2
I =
20sin 2
8
I
34.CH MgBr
3
OMgBr+
CH3
H O2
OHO
41. CH —CH—CH—CH3 3
CH3
OH
H+
CH —CH—CH—CH3 3
CH3
(+)
CH —C—CH CH3 2 3
CH3
+
Rearrange
CH —C—CHCH3 3
CH3
–H O2
42. CH3—CH
2Br
LiAlH4 CH3—CH
3 + LiAlH
3Br
47. The structure that can form the more stable
carbocation will undergoes SN1 faster.
Test - 7 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
5/8
PART - C (MATHEMATICS)
61. I = cos8 cos7
1 2cos5
x x
x
∫ dx
= 2
152sin sin
2 2
53 4sin
2
x x
x
∫ dx
=
3
2
5 52 3sin 4sin sin
2 2 2
53 4sin
2
x x x
x
⎛ ⎞ ⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
∫ dx
= 5
2sin sin2 2
x xdx ∫
= cos3 cos2x x dx∫
= 1 1sin3 sin2
3 2x x c
62. Let 1
sec tx
63. I = 3 5
4 5
x x
x x
e edx
e e
∫ =
2
2
3 5
4 5
x
x
edx
e
∫
=
2
2
3 154 5 5
4 4
4 5
x
x
e
dxe
∫
49.
Cl
, Cl,
Cl
,
Cl
,
Cl
,
Cl
51. X
Br /h2
Racemic mixture Presence of
alkyl part⇒
Br /Fe2 Product Presence of benzene ring⇒
Br , CCl2 4 No reaction No C=C is present in
side chain⇒
CH CH2 3
X =
53.
O
CH CH2 3
HBr
OH
+ C H Br2 5
54.
OH
SOCl2
S iN
Cl
(Retention ofconfiguration)
58. NH2
OH HNO2
N2
OH
+
–N2
OH
+
ring
expansion
:OH
+
–H+
O
= 23 35 1
ln 4 54 4 10
xx e c
= 23 7 7
ln(4 5)4 8 4
xx e x c
= –x + 7
8 ln| 4e
2x – 5 | + c
64. Given equation is
2
1 3dy
dx
⎛ ⎞⎜ ⎟⎝ ⎠ =
33
3
34
d y
dx
⎛ ⎞ ⎜ ⎟⎝ ⎠
Degree = 3
65. Required area =
12 2
0
1212
xx dx
⎛ ⎞⎜ ⎟⎝ ⎠∫
= 48 sq. units
66. The tangent at p(x, y) is : Y – y = (X – x)dy
dx
It meet x-axis at A = , 0
dxx y
dy
⎛ ⎞⎜ ⎟⎝ ⎠
∵ OP2 = PA
2 x2 + y2 = y2
2dx
dy
⎛ ⎞⎜ ⎟⎝ ⎠
+ y2
dy
xdx
= ±y
xy = c2 or y = cx, where c is constant
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-G) (Hints)
6/8
67.
( 5, 0)
2(1,0)1
( 5 )
0,
Required area
=
2
2
1
1 1(5 ) 2 2 1 1
2 2x dx
⎧ ⎫⎪ ⎪ ⎨ ⎬⎪ ⎪⎩ ⎭∫ sq. units
=
2
2 1
1
5 55 sin
2 2 25
x xx
⎡ ⎤ ⎢ ⎥⎣ ⎦
= 5 5 5 2
24 2 4
⎛ ⎞ ⎜ ⎟⎝ ⎠ sq. units
68. I =
ln2 1
y
x
dx
e ∫
Let ex –1 = t2
exdx = 2tdt
I =
1
2
1
2
( 1)
ye
tdt
t t
∫
= 11
12 tan
ye
t
= 1
2 tan 14
ye
⎛ ⎞ ⎜ ⎟⎝ ⎠
12tan 1
ye
= 6 2
1tan 1
ye
= 3
ey = 3 + 1
y = ln4
69. I = 21
dx
x x x ∫
Let x = sin2 dx = 2sin cos d
I = 21 sin
d ∫
= 22(sec sec tan )d ∫
= 2(tan – sec) + c
= 2 1
1
xc
x
70. Let ln x = t dx = etdt
I =
2
2
1
1
tte dt
t
⎛ ⎞ ⎜ ⎟⎝ ⎠∫ = 2 2 2
1 2
1 (1 )
tdt
t t
⎛ ⎞⎜ ⎟⎝ ⎠ ∫
= 2 21 1 (ln )
te x
c ct x
71. I = 3
tan 2 sec2x xdx∫Let sec2x = t
tan 2x sec2xdx = 1
2dt
I = 21
tan 22
xdt∫
= 211
2t dt∫
= 31 1
sec 2 sec26 2
x x c
72. Let t = tanx
I = 2
2 2 2
2 4 2 4
0
( )
(1 )( )
∫a t b dt
t a t b
= 2 2
2
a b
2 2
2 4 2 4
0
1
1
a bdt
t a t b
⎛ ⎞⎜ ⎟
⎝ ⎠∫
=
2
1 1
2 2 20
2tan tan
a tt
a b b
⎛ ⎞
⎜ ⎟⎝ ⎠= 2 2
2
a b
73. 2 2 2 2cos
x xx y d x y d
y y
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
2 2sin
xx y C
y
⎛ ⎞ ⎜ ⎟⎝ ⎠
74. I =
2017
1
sin( ln )e
xdx
x
∫
Let lnx = t dx
x
= dt
I =
2017
0
sin 1 cos(2017 )tdt
∫
= 2 = even integer
75. |cos x| is periodic function of period and
0
| cos | 2x
∫
Test - 7 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
7/8
76. ∵ I1 =
1 1
0
tan xdx
x
∫
Let x = tan dx = sec2d
I1
= 24 4
0 0
sec 2
tan sin2
d d
∫ ∫
Now let 2 = t d = 1
2dt
I1 =
2
0
1
2 sin
tdt
t
∫
I1 = 2
1
2I
1
2
1
2
I
I
77. 4(x2 – 2x) + y2 + 4y – 4 = 0
2( 1)
3
x +
2( 2)
12
y = 1
This is an ellipse and its area = 3 12 = 6
78. ∵
4
( ) sin cos 24
f x dx
∫
On differentiating both sides w.r.t we get
f() = sin + cos – 4
sin + 2
1 1 1
24 4 42 2 2
f ⎛ ⎞ ⎜ ⎟⎝ ⎠
= 1
22
= 3
2=3 2
2
80. The equation of circle is
x2 + y2 –2hx = 0 ..... (i)
where h R– is a parameter
On differentiating : x + y dy
dx= h ..... (ii)
On eliminating h from equation (i) and (ii) we get
2dy
x x ydx
⎛ ⎞⎜ ⎟⎝ ⎠ = x2 + y2
x2 – y2 = – 2xy
dy
dx
81. f (x) = 6(x – 1) f (x) = 3(x – 1)2 + a
The slope of tangent at (2, 1) is 3
f (2) = 3 a = 0
f (x) = 3(x – 1)2
f(x) = (x – 1)3 + b
But f(2) = 1 b = 0
f(x) = (x – 1)3
82.2 2
lnxdy ydx xc
x x
2
lny xdxd
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 lny x
x x x
f (x) = 2
1
x
83. A.T.Q: dx
y x ady
dy
y=
dx
a x lny = lnc – ln|(x – a)|
y = ,| |
c
x a where x = 2a, y = a c = a2
y =
2
| |a
x a
84. ∵ y = cx2
dy
dx=
2y
x
Now replace dy
dx by
dx
dy we get
2ydy + xdx = 0
y2 +
2
2
x
= d1, i.e., x2 + 2y
2 = d
86. ∵
1
1lim
⎛ ⎞⎜ ⎟⎝ ⎠∑
n
nr
rf
n n =
1
0
( )f x dx∫
=
1
0
| sin2 |x dx∫
=
1
0
cos2
2
x⎡ ⎤⎢ ⎥⎣ ⎦
= 1 cos2
2
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-G) (Hints)
8/8
� � �
87. f(x) = ex
0
( )
x
te f t dt ∫ (x
2 – x + 1)ex
Again on differentiating w.r.t. x
f (x) = ex(2x + 1)
f(x) = (2x + 1) ex – 2
xe dx∫
f(x) = (2x – 1)ex
1
2f⎛ ⎞⎜ ⎟⎝ ⎠ = 0
88. ∵ f(x) = 3
4
cos( )
x
t dt∫
f (x) = cos(x3)
f (4) = cos 64 = 1
f(4) =
4
3
4
cos( ) 0t dt ∫
Equation of tangents : y – 0 = 1 (x – 4)
x – y = 4
89. I = 8 7( 2)
dx
x x ∫
= 7 8
7
21
dx
x xx
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
Let 7
21
x
= t
8
dx
x =
14
dt
I = 1
28
11 dt
t
⎛ ⎞⎜ ⎟⎝ ⎠∫
= 1
287 7
2 2ln 1 1
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠ + c
= 1
287 7
2 2ln 1
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠ + c
90. I =
534 ( 1) 2
dx
x x ∫
= 3
42 1( 2)
2
dx
xx
x
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
Let 1
2
x
x
= t4
2( 2)
dx
x =
4
3t3dt
I =
3
3
4 4
4
3( )
t dt
t
∫ =4
3t + c
Test - 7 (Code-H) (Answers) All India Aakash Test Series for JEE (Main)-2018
1/8
1. (1)
2. (4)
3. (1)
4. (3)
5. (3)
6. (4)
7. (3)
8. (1)
9. (2)
10. (2)
11. (3)
12. (3)
13. (2)
14. (4)
15. (1)
16. (2)
17. (3)
18. (4)
19. (1)
20. (1)
21. (2)
22. (2)
23. (3)
24. (4)
25. (3)
26. (2)
27. (2)
28. (1)
29. (2)
30. (1)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (1)
33. (3)
34. (4)
35. (2)
36. (4)
37. (4)
38. (1)
39. (4)
40. (3)
41. (4)
42. (3)
43. (3)
44. (2)
45. (3)
46. (3)
47. (1)
48. (1)
49. (2)
50. (3)
51. (1)
52. (2)
53. (4)
54. (3)
55. (2)
56. (4)
57. (2)
58. (1)
59. (2)
60. (2)
61. (4)
62. (1)
63. (3)
64. (2)
65. (1)
66. (4)
67. (4)
68. (1)
69. (1)
70. (2)
71. (2)
72. (2)
73. (3)
74. (3)
75. (3)
76. (3)
77. (3)
78. (2)
79. (4)
80. (1)
81. (1)
82. (4)
83. (2)
84. (3)
85. (4)
86. (3)
87. (4)
88. (4)
89. (3)
90. (1)
Test Date : 25/02/2018
ANSWERS
TEST - 7 (Code-H)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-H) (Hints)
2/8
1. I =2 20
cos cos (90 – )2
I⎡ ⎤ ⎢ ⎥⎣ ⎦
I = 2 20
cos sin2
I =
20sin 2
8
I
2.
d = 10 cm
I1 I
3
I2
For refraction through concave lens:
1 1 1–
v u f
, 1 1 –1–– 10v
v = –10 cm
I2 is 20 cm behind the plane mirror.
1 1 –1––(30) 10v
1 –1 1 –4–
10 30 30v
= –7.5 cm
I3 is formed at 2.5 cm in front of mirror.
3. ∵ Fringe width
D
d
and new fringe width
= (1.1 )
(1.1 )
D D
d d
= 0.2 mm
5. Image formed by lens should be on the pole of mirror.
6. For first lens,
1
1 1 1––30 10v
v1 = 15 cm at I
1
For second lens, u2 = 15 – 5 = 10 cm
5 cm 10 cm 30 cm
I1
I
O
PART - A (PHYSICS)
The focal length of second lens is also 10 cm, thus
it forms the image at infinity.
Finally the third lens forms the image at its focal point
i.e., 30 cm from it.
7. If P is equivalent power of the combination, then
2 1
15
P ,
1–
m
Pf
By using mirror equation
1 1 1
–20 –7.5v
– 12 cmv
8. The deviation produced by the prism
= ( – 1)A = (1.5 – 1) × 4° = 2°
Thus the angle through which mirror should be rotated
is 1°.
9. I1 = 4I
0, I2 = I
0, Imax
= 9I0, Imin
= I0
11. ∵ ,
h
mv so with the increase in velocity of
electron, wavelength decreases and so fringe width
decreases.
12. For A:
2.
2 4
⎡ ⎤ ⎢ ⎥ ⎣ ⎦IA = I + I + 2I cos = 0
For B:
–4
x PB QB⎛ ⎞ ⎜ ⎟
⎝ ⎠ = (PB – QB) +
4
= 04 4
2
IB = I + I + 2I cos
2
= 2I
For C:
2– 0
2 4
⎛ ⎞ ⎜ ⎟ ⎝ ⎠IC = I + I + 2I cos0° = 4I
13. According to diagram given in question
OP = d sec, CO = OP cos2 = d sec cos2 Path difference, x = CO + OP
x = d sec + d sec cos2Effective path difference
xeff
= d(sec + seccos2) + 2
For constructive interference, xeff
=
d sec (1 + cos2) + 2
=
or cos4d
Test - 7 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
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14. 3 cos = 2
cos = 2
3
3
tanX
D
X = 5
2D
16.
O36 c
m
24 c
m
x
C
For lens:
∵ v = –24 cm, u = –x
1 1 1–24 x f
, 1 1 1–24 40x
x = 15 cm
18. ∵ A1 = 4°,
1 = 1.54,
2 = 1.72
Condition for no deviation
A2 =
1
1
2
( – 1)–( – 1)
A
1.54 – 1
– 41.72 – 1
⎛ ⎞ ⎜ ⎟⎝ ⎠
A2 = –3° |A
2| = 3°
19.
x
ir
y
∵ i + r = 90° and i = r
i = r = 45°, i + = 90°
= 90° – i = 90° – 45° = 45°
∵ Given equation of surface, y2 = 2x
2ydy
dx = 2
1dy
dx y = tan45°
y = 1
x =
21
2 2
y
Coordinates are 1,1
2
⎛ ⎞⎜ ⎟⎝ ⎠
20.
r1
r2
i2
i1
sin
sin
i
r
⎡ ⎤ ⎢ ⎥⎣ ⎦
= (i1 – r
1) + (i
2 – r
2)
For minimum deviation, i1 = i
2, r1 = r
2
sin60
sin30
= , = 3
21. n11 = n
22
∵ n
11 = n
22
12 × 600 = n2 × 400
{n2 = 18}
22.
2r1
r2
e
i + e – A = 2, e =
1
sin2sinr
, 1
sinsin( – )r
on solving we get,
2–1
cos8
23.
O
d
d
S
S
y
∵ Due to reflection. A phase difference of occurs.
Let the distance of first maxima from O is y
2
2 2
D
dy
⎛ ⎞⎜ ⎟ ⎝ ⎠
4
Dy
d
where, = fringe width.
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-H) (Hints)
4/8
PART - B (CHEMISTRY)
33. NH2
OH HNO2
N2
OH
+
–N2
OH
+
ring
expansion
:OH
+
–H+
O
24. The focal length of the glass (lens) used 100
m,1.5
f
if y is the distance of near point, then
1 1 1.5––25 100y
{y = –40 cm}
The focal length of lens of microscope
fe =
1 100
20P = 5 cm
Magnifying power of microscope together with glass
M = 1
e
D
f =
251
5 = 6
In case of without glass, D = 40 cm = y
M = 1
e
D
f
= 40
15
= 9
26. When lenses are in contact
1 2
1 1 1
12f f ...(i)
When lenses are at separation
1 2 1 2
1 1 50– 0
f f f f ...(ii)
On solving equations (i) and (ii), we get
f1 = 20 cm, f
2 = 30 cm
27. The parallel rays will focus at focal point of concave
lens so after refraction from it, they become parallel.
29.
h = 2 m
i 90° O x
y
If is the angle of refraction for the horizontal rays.
(1 +
)
ay
1/2
1
ay
0
sin90
sin
sin = 1
1 ay, tan =
1
ay
1dx
dy ay
1 –12–2 2
0 0
x
dx a y dy∫ ∫
x =
2–1 1
2 20
2
⎡ ⎤⎢ ⎥⎣ ⎦a y =
1–1
–6 222 2 10 2 = 2000 m
30. ∵ Component of incident ray along normal is
ˆ ˆ( )i j .
to express reflected ray, ˆ ˆ( )i j reverses in
expression of incident ray.
ˆ ˆ ˆ– – –R i j k
�
= ˆ ˆ ˆ–( )i j k
ˆ ˆ ˆ–( )ˆ3
i j kR⎧ ⎫ ⎨ ⎬⎩ ⎭
37.
OH
SOCl2
S iN
Cl
(Retention ofconfiguration)
38.
O
CH CH2 3
HBr
OH
+ C H Br2 5
40. X
Br /h2
Racemic mixture Presence of
alkyl part⇒
Br /Fe2 Product Presence of benzene ring⇒
Br , CCl2 4 No reaction No C=C is present in
side chain⇒
CH CH2 3
X =
Test - 7 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
5/8
PART - C (MATHEMATICS)
61. I =
534 ( 1) 2
dx
x x ∫
= 3
42 1( 2)
2
dx
xx
x
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
Let 1
2
x
x
= t4
2( 2)
dx
x =
4
3t3dt
I =
3
3
4 4
4
3( )
t dt
t
∫ =4
3t + c
62. I = 8 7( 2)
dx
x x ∫
= 7 8
7
21
dx
x xx
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
42.
Cl
, Cl,
Cl
,
Cl
,
Cl
,
Cl
44. The structure that can form the more stable
carbocation will undergoes SN1 faster.
49. CH3—CH
2Br
LiAlH4 CH3—CH
3 + LiAlH
3Br
50. CH —CH—CH—CH3 3
CH3
OH
H+
CH —CH—CH—CH3 3
CH3
(+)
CH —C—CH CH3 2 3
CH3
+
Rearrange
CH —C—CHCH3 3
CH3
–H O2
57.CH MgBr
3
OMgBr+
CH3
H O2
OHO
58.
+ CH —CH=CH3 2
H+
CH
CH3
CH3
(A)
O2
CH —C—O—OH3
CH3
(B)
H+
/H O2
CH —C—CH3 3
+
O
OH
(C)
59. LiAlH4 reduce ester group and H
2/Ni reduce C C
also.
Let 7
21
x
= t
8
dx
x =
14
dt
I = 1
28
11 dt
t
⎛ ⎞⎜ ⎟⎝ ⎠∫
= 1
287 7
2 2ln 1 1
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠
+ c
= 1
287 7
2 2ln 1
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠
+ c
63. ∵ f(x) = 3
4
cos( )
x
t dt∫
f (x) = cos(x3)
f (4) = cos 64 = 1
f(4) =
4
3
4
cos( ) 0t dt ∫
Equation of tangents : y – 0 = 1 (x – 4)
x – y = 4
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-H) (Hints)
6/8
64. f(x) = ex
0
( )
x
te f t dt ∫ (x2 – x + 1)ex
Again on differentiating w.r.t. x
f (x) = ex(2x + 1)
f(x) = (2x + 1) ex – 2x
e dx∫f(x) = (2x – 1)ex
1
2f⎛ ⎞⎜ ⎟⎝ ⎠ = 0
65. ∵
1
1lim
⎛ ⎞⎜ ⎟⎝ ⎠∑
n
nr
rf
n n =
1
0
( )f x dx∫
=
1
0
| sin2 |x dx∫ =
1
0
cos2
2
x⎡ ⎤⎢ ⎥⎣ ⎦ =
1 cos2
2
67. ∵ y = cx2 dy
dx=
2y
x
Now replace dy
dx by
dx
dy we get
2ydy + xdx = 0
y2 +
2
2
x
= d1, i.e., x2 + 2y2 = d
68. A.T.Q: dx
y x ady
dy
y=
dx
a x
lny = lnc – ln|(x – a)|
y = ,| |
c
x a where x = 2a, y = a c = a2
y =
2
| |a
x a
69.2 2
lnxdy ydx xc
x x
2
lny xdxd
x x
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 lny x
x x x
f (x) = 2
1
x
70. f (x) = 6(x – 1) f (x) = 3(x – 1)2 + a
The slope of tangent at (2, 1) is 3
f (2) = 3 a = 0
f (x) = 3(x – 1)2
f(x) = (x – 1)3 + b
But f(2) = 1 b = 0
f(x) = (x – 1)3
71. The equation of circle is
x2 + y2 –2hx = 0 ..... (i)
where h R– is a parameter
On differentiating : x + y dy
dx= h ..... (ii)
On eliminating h from equation (i) and (ii) we get
2dy
x x ydx
⎛ ⎞⎜ ⎟⎝ ⎠ = x2 + y2
x2 – y2 = – 2xydy
dx
73. ∵
4
( ) sin cos 24
f x dx
∫
On differentiating both sides w.r.t we get
f() = sin + cos – 4
sin + 2
1 1 1
24 4 42 2 2
f ⎛ ⎞ ⎜ ⎟⎝ ⎠
= 1
22
= 3
2=3 2
2
74. 4(x2 – 2x) + y2 + 4y – 4 = 0
2( 1)
3
x +
2( 2)
12
y = 1
This is an ellipse and its area = 3 12 = 6
75. ∵ I1 =
1 1
0
tan xdx
x
∫
Let x = tan dx = sec2d
I1
= 24 4
0 0
sec 2
tan sin2
d d
∫ ∫
Now let 2 = t d = 1
2dt
Test - 7 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
7/8
I1 =
2
0
1
2 sin
tdt
t
∫
I1 = 2
1
2I
1
2
1
2
I
I
76. |cos x| is periodic function of period and
0
| cos | 2x
∫
77. I =
2017
1
sin( ln )e
xdx
x
∫
Let lnx = t dx
x
= dt
I =
2017
0
sin 1 cos(2017 )tdt
∫
= 2 = even integer
78. 2 2 2 2cos
x xx y d x y d
y y
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
2 2sin
xx y C
y
⎛ ⎞ ⎜ ⎟⎝ ⎠
79. Let t = tanx
I = 2
2 2 2
2 4 2 4
0
( )
(1 )( )
∫a t b dt
t a t b
= 2 2
2
a b
2 2
2 4 2 4
0
1
1
a bdt
t a t b
⎛ ⎞⎜ ⎟
⎝ ⎠∫
=
2
1 1
2 2 20
2tan tan
a tt
a b b
⎛ ⎞
⎜ ⎟⎝ ⎠= 2 2
2
a b
80. I = 3
tan 2 sec2x xdx∫Let sec2x = t
tan 2x sec2xdx = 1
2dt
I = 21
tan 22
xdt∫
= 211
2t dt∫
= 31 1
sec 2 sec26 2
x x c
81. Let ln x = t dx = etdt
I =
2
2
1
1
tte dt
t
⎛ ⎞ ⎜ ⎟⎝ ⎠∫ = 2 2 2
1 2
1 (1 )
tdt
t t
⎛ ⎞⎜ ⎟⎝ ⎠ ∫
= 2 21 1 (ln )
te x
c ct x
82. I = 21
dx
x x x ∫
Let x = sin2 dx = 2sin cos d
I = 21 sin
d ∫
= 22(sec sec tan )d ∫
= 2(tan – sec) + c
= 2 1
1
xc
x
83. I =
ln2 1
y
x
dx
e ∫
Let ex –1 = t2
exdx = 2tdt
I =
1
2
1
2
( 1)
ye
tdt
t t
∫
= 11
12 tan
ye
t
= 1
2 tan 14
ye
⎛ ⎞ ⎜ ⎟⎝ ⎠
12tan 1
ye
= 6 2
1tan 1
ye
= 3
ey = 3 + 1
y = ln4
84.
( 5, 0)
2(1,0)1
( 5 )
0,
Required area
=
2
2
1
1 1(5 ) 2 2 1 1
2 2x dx
⎧ ⎫⎪ ⎪ ⎨ ⎬⎪ ⎪⎩ ⎭∫ sq. units
All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-H) (Hints)
8/8
� � �
=
2
2 1
1
5 55 sin
2 2 25
x xx
⎡ ⎤ ⎢ ⎥⎣ ⎦
= 5 5 5 2
24 2 4
⎛ ⎞ ⎜ ⎟⎝ ⎠ sq. units
85. The tangent at p(x, y) is : Y – y = (X – x)dy
dx
It meet x-axis at A = , 0
dxx y
dy
⎛ ⎞⎜ ⎟⎝ ⎠
∵ OP2 = PA2 x2 + y2 = y2
2dx
dy
⎛ ⎞⎜ ⎟⎝ ⎠
+ y2
dy
xdx
= ±y
xy = c2 or y = cx, where c is constant
86. Required area =
12 2
0
1212
xx dx
⎛ ⎞⎜ ⎟⎝ ⎠∫
= 48 sq. units
87. Given equation is
2
1 3dy
dx
⎛ ⎞⎜ ⎟⎝ ⎠ =
33
3
34
d y
dx
⎛ ⎞ ⎜ ⎟⎝ ⎠
Degree = 3
88. I = 3 5
4 5
x x
x x
e edx
e e
∫
=
2
2
3 5
4 5
x
x
edx
e
∫
=
2
2
3 154 5 5
4 4
4 5
x
x
e
dxe
∫
= 23 35 1
ln 4 54 4 10
xx e c
= 23 7 7
ln(4 5)4 8 4
xx e x c
= –x + 7
8 ln| 4e2x – 5 | + c
89. Let 1
sec tx
90. I = cos8 cos7
1 2cos5
x x
x
∫ dx
= 2
152sin sin
2 2
53 4sin
2
x x
x
∫ dx
=
3
2
5 52 3sin 4sin sin
2 2 2
53 4sin
2
x x x
x
⎛ ⎞ ⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
∫ dx
= 5
2sin sin2 2
x xdx ∫
= cos3 cos2x x dx∫
= 1 1sin3 sin2
3 2x x c