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Test - 4 (Code-G) (Answers) All India Aakash Test Series for JEE (Main)-2018
1/7
1. (4)
2. (1)
3. (4)
4. (1)
5. (1)
6. (1)
7. (3)
8. (1)
9. (1)
10. (2)
11. (3)
12. (3)
13. (1)
14. (1)
15. (4)
16. (1)
17. (3)
18. (3)
19. (2)
20. (1)
21. (3)
22. (1)
23. (2)
24. (1)
25. (1)
26. (3)
27. (2)
28. (3)
29. (2)
30. (3)
PHYSICS CHEMISTRY MATHEMATICS
31. (2)
32. (3)
33. (1)
34. (1)
35. (4)
36. (3)
37. (3)
38. (4)
39. (3)
40. (2)
41. (3)
42. (1)
43. (4)
44. (1)
45. (2)
46. (4)
47. (3)
48. (2)
49. (4)
50. (3)
51. (3)
52. (2)
53. (2)
54. (2)
55. (4)
56. (2)
57. (4)
58. (3)
59. (4)
60. (1)
61. (3)
62. (1)
63. (4)
64. (4)
65. (1)
66. (2)
67. (3)
68. (4)
69. (4)
70. (2)
71. (3)
72. (3)
73. (1)
74. (3)
75. (4)
76. (1)
77. (3)
78. (2)
79. (2)
80. (1)
81. (4)
82. (1)
83. (2)
84. (4)
85. (4)
86. (3)
87. (2)
88. (1)
89. (2)
90. (2)
Test Date : 14/01/2018
ANSWERS
TEST - 4 (Code-G)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)
2/7
1. Y = 4 sin2t sin2t
= 4(1 – cos2 ) sin2
2t t = 2 sin2t – sin4t
3. Vmax
= A = mg
K
= 10
40
g
= 1
4 m/s
4. Y = 2
8
( – 2 )x t
,
3
–16 2
( – 2 )
dy
dt x t
= 4 m/s
5.max
y
t
⎛ ⎞⎜ ⎟⎝ ⎠
= 2f
8. P0V
0 – V
0 = nRT
0
V0 =
0
0( – )
nRT
P , n = 1
w = P0V =
0 0
0( – )
P RT
P
9. U = 1
2 (stress × strain × volume)
10. Q = –dT
TAdx
0
lQ
dxA∫ =
2
1
–
T
T
T dT ∫
Q = 2 2
1 2( – )
2
AT T
l
12. = 2
13
= 120°
13. T = 20
21800
= 2s
3
t = 4
T =
1s
6
14.
45°
45°
A
A
A
Ar
= (2A cos45° + A) = ( 2 1)A15. Maximum possible will occur if it is around mean
position.
(Vavg
)max
= 2
4
A
T =
4 2A
T
PART - A (PHYSICS)
16.1
Ir
I A2
2
1
2
25
9
A
A
⎛ ⎞ ⎜ ⎟⎝ ⎠
1
2
5
3
A
A
17. T = 2m
K
T2 = 2 2
1 2T T 1 2
eq1 2
K KK
K K
⎛ ⎞⎜ ⎟⎝ ⎠18. (
1 –
2)T =
= 2 2 4
–5
TT T
⎡ ⎤⎢ ⎥⎣ ⎦
= 2
5
= 72°
19. dW∫ = P dv∫
= av2 dv = 3 3
2 1–
3
av v
W = 2 1( – )
3
RT T
21. WCD
= nRT2 ln
D
C
V
V
WAB
= nRT1 ln
B
A
V
V
VA = V
C, V
B = V
D
2WCD
= WAB
2T2 = T
1
1
2
T
T = 2
22.
4
A
B
T
T
⎛ ⎞⎜ ⎟⎝ ⎠
= 0.81
0.01 = 81
A
B
T
T = 3,
ATA =
BTB
23. 0 1 0 2–
mg mgP L P L
A A
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
1
101
99
L
L
24. T = Const.
4 4
2 1
1
16T T
T2 =
1
2
T
Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
3/7
25. TV r – 1 = Constant
–( –1)
dV V
dT T
–tan = 0–
( –1)V
T
( – 1) = 0
0tan
V
T
CV =
–1R
= 0
0
tanRT
V
26. f.L1 =
2
B
4.3
Lf =
1
B
1
4
3
L
L =
1
2
L1 =
1
2
4
3
L
27. 0.42
= 0.8 m
TV
= 8 m/s
f = 10 Hz
28. Q = U + W
W = 4
Q
3
4
Q = C
V dT
3
4
Q =
5
2R dT
C = 10
3R
29. On heating volume will increase. So, it will move
towards right.
30. TB
= B
A
A
VT
V = 600 K
WBC
= lnC
B
B
VnRT
V = 1200 R ln2
PART - B (CHEMISTRY)
35.
No rearrangement
36. CH CH + SeO2 CHO—CHO
3CH CH Red hot
iron tube
873 K
37. Heat of hydrogenation 1
stability of alkene
38. A = , B =
39.BH , THF
3
H
BH2
H
H
CH COOH3
(Alkane)
40.
OH
+ Zn (Dust)
(X)
CH3
+ CH Cl3
AlCl3
(Y)
CH3
alk. KMnO4
(Z)
COOH
NaOH
(U)
COONa
Sodalime
(X)
42. +I effect of D is more than of H.
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)
4/7
45. 2R COOK + 2H2O R—R + 2CO
2 + H
2 + 2KOH
46. % of P = 2 2 7
wt. of Mg P O62100
222 wt. of compound
= 62 0.75
100222 0.6
= 35%
50.
X
Li/NH3
X
[with EDG]
Y
Li/NH3
Y
[with EWG]
due to stability of carbanion intermediates.
56.Br
2
Br
Br(A)
(i) alc. KOH
(ii) NaNH2
CH –C C–CH3 3
(B)
H /Pd
BaCO
2
3
C=CCH
3CH
3
(C) HH
59.
PART - C (MATHEMATICS)
61. If is angle between tangents, then
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
= radius
distance between 3 3, 3 and (0, 0)
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
= 3
6 =
1
2
2
=
6
=
3
So, cot = cot3
⎛ ⎞⎜ ⎟⎝ ⎠
= 1
3
62. Let P(h, k) be the point of intersection of E1 and E
2
2
2
21
hk
a h2 = a2(1 – k2) ...(i)
and
2 2
21
h k
a = 1 k2 = a2 (1 – h2) ...(ii)
Eliminating a from equation (i) and (ii), we get
2 2
2 21– 1–
h k
k h
h2(1 – h2) = k2 (1 – k2)
(h – k) (h + k) (h2 + k2 – 1) = 0
Hence, the locus is a set of curves consisting of the
straight lines y = x, y = –x and circle x2 + y2 = 1.
63. Equation of hyperbola is
2 2
– 1.9 16
x y
Equation of tangent is
29 – 16y mx m
29 – 16 2 5m m = ±2
a + b = sum of roots = 0
64. Make homogeneous 3x2 – y2 –2x + 4y = 0 with the
line y = mx + c
i.e., 2 2 – –
3 – – 2 4y mx y mx
x y x yc c
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 0
According to question coeff. of x2 + coeff. of y2 = 0
2 43 –1 0
m
c c
m + c + 2 = 0
which gives fixed point (1, –2) p + q = –1
65. The point 6
P⎛ ⎞
⎜ ⎟⎝ ⎠
is sec , tan6 6
a b ⎛ ⎞
⎜ ⎟⎝ ⎠
or
2,
3 3
a bP⎛ ⎞⎜ ⎟⎝ ⎠
.
Equation of tangent at P is – 13 3
2
x y
a b
Area of the triangle = 1 3
32 2
ab = 3a2
b
a = 4
e2 =
2
21 17
b
a
66. The two circles are x2 + y2 – 4x – 6y – 3 = 0 and
x2 + y2 + 2x + 2y + 1 = 0
Centres : C1(2, 3) and C
2(–1, –1)
radii : r1 = 4 and r
2 = 1.
We have, C1C
2 = 5 = r
1 + r
2, circles touch externally
therefore there are three common tangents to the
given circles.
Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
5/7
67. Centre and radius of circle x2 + y2 + 8x – 10y – 40 = 0
are C(–4, 5) and 9 respectively. Distance of the centre
C(–4, 5) from the point P(–2, 3) is CP = 4 4 =
2 2 .
a = CP + r = 2 2 + 9
b = |CP – r| = 9 – 2 2
a – b = 4 2
68. If S1 = 0 and S
2 = 0 are the equations, then
S1+ S
2= 0 is a second degree curve passing through
the points of intersection of S1 = 0 and S
2 = 0.
( + 4)x2 + 2( + 1)y2 – 2(3 + 10)x
– 12( + 1)y + (23 + 35) = 0 ...(i)
For it to be a circle, choose such that the coefficient
of x2 and y2 are equal
+ 4 = 2 + 2
= 2
This gives the equation of the circle as
6(x2 + y2) – 32x – 36y + 81 = 0 {(using (i))}
2 2 16 27– – 6 0
3 2x y x y
Its centre is 8, 3
3C⎛ ⎞⎜ ⎟⎝ ⎠
and radius is
r = 64 27
9 –9 2
= 1 47
3 2
69. The circle and the ellipse do not meet at any point.
Length of the common chord = 0.
70. Putting y = 2t2 in the equation of the given ellipse
2 2
1,9 4
x y we get
2 44
19 4
x t x2 = 9(1 – t4)
= 9(1 – t2) (1 + t2)
This will give real values of x if 1 – t2 0, i.e., |t| 1.
71. Clearly ax + by = 1
i.e.,– 1a
y xb b
is tangent to cx2 + dy2 = 1
2 2
11 1
x y
c d
2 2
1 1 – 1a
b c b d
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 =
2 2a b
c d
72.
2 2
22
15 ( – 1) 2– 22
x y
aa a
2 2
22
13 ( – 1) 2( – 1)2
x y
aa
where, A2 = (a – 1)2 + 3
2, B2 = (a – 1)2 + 2
We know that,
A2 = B2 (1 – e2), (for vertical ellipse)
2 3( – 1)
2a = 2 1
( – 1) 2 1–4
a⎛ ⎞ ⎜ ⎟⎝ ⎠
4(a – 1)2 + 6 = 3(a – 1)2 + 6
(a – 1)2 = 0 a = 1
Equation of ellipse is
2 2
13 2
2
x y
So, length of L.R. =
22A
B =
32
2
2
=
9
2
73.
Pt
t(2 , 4 )
11
2
Q t t(2 , 4 )2 2
2
Let absciss of point P is 2
12 18t 2
19t
t1 = 3
We know that, If PQ is normal then 2 1
1
2– –t t
t
2
11–
3t
So, coordinates of P and Q are (18, 12) and
242 –44,
9 3
⎛ ⎞⎜ ⎟⎝ ⎠
respectively.
PQ =
2 2242 –44
– 18 – 129 3
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 80 10
9
74.
BR
P
X
AO
Y
Equation of tangent at P(a cos, b sin) is
cos sin 1x y
a b
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)
6/7
It meets the major axis at A(a sec, 0) and the
minor axis at B(0, b cosec).
Equation of the OR, perpendicular to the tangent at
P is ax sin – by cos = 0.
Now PR = 2 2 2 2
( cos )sin – ( sin )cos
sin cos
a a b b
a b
=
2 2
2 2 2 2
( – ) sin cos
sin cos
a b
a b
Also, AB =
2 2 2 2sin cos
sin cos
a b
So, (AB) (PR) = a2 – b2
Here, a = 4, b = 3
(AB)(PR) = 16 – 9 = 7
75. The equation ax3 + 3bx2y + 3cxy2 + dy3 = 0
3 2
3 3 0y y y
d c b ax x x
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Let y
mx
dm3 + 3cm2 + 3bm + a = 0
If the roots are m1, m
2 and m
3, then m
1 + m
2 + m
3
= –3
,
c
d m
1m
2 + m
2m
3 + m
3m
1 =
3b
d and m
1m
2m3
= –a
d.
For three coincident lines, we have m1 = m
2 = m
3,
so, m = –c
d, m2 =
b
d and m3 =
–a
d
So,
3– –c a
d d
⎛ ⎞ ⎜ ⎟⎝ ⎠
c3 = ad2
76.
XO
Y
PQ
A (1, 3)
Let P(2cos, 2sin). The mid point Q(h, k) of AP
will be given by
h = 1 2cos
2
cos =
1–2
h
and k = 3 2sin
2
sin =
3–2
k
Eliminating , we get
2 21 3
– – 12 2
h k⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Hence, the locus is
2 21 3
– – 12 2
x y⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
So, radius = 1.
77. Any point on the parabola is P(t2, 2t). The
circumcenter Q(h, k) of triangle PAB is midpoint of
P(t2, 2t) and the centre C(–3, 2) of the circle.
Hence,
2– 3
2
th and k = t + 1
Eliminating t, we get (k – 1)2 = 2h + 3
Locus of P(h, k) is (y – 1)2 = 2x + 3
Equation of tangent to parabola with slope 1 is
(y – 1) =
1
3 21
2 1x
⎛ ⎞ ⎜ ⎟⎝ ⎠
y = x + 3
78. The given parabolas 2y2 = 2x – 1 and 2x2 = 2y – 1
are symmetrical about the line y = x. Also, the
shortest distance occurs along the common normal
which is perpendicular to the line y = x. Then the
shortest distance is
1 1 1
16 16 2 2d
79. The locus of P is an ellipse, if K > AB
K > 50 K > 5 2
80. Ellipse 16x2 + 25y2 = 400
2 2
125 16
x y
Let the points on ellipse be P(5 cos1, 4 sin
1) and
Q(5cos2, 4sin
2).
Circle described on PQ as diameter touches x-axis
at (3, 0).
1 2cos cos
5 32
⎛ ⎞ ⎜ ⎟⎝ ⎠
and 1 2
sin sin4
2r
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 24tan
5 2 3
r ⎛ ⎞ ⎜ ⎟⎝ ⎠
1 2 5
tan2 12
r ⎛ ⎞ ⎜ ⎟⎝ ⎠
slope of PQ = 1 2–4
cot5 2
⎛ ⎞⎜ ⎟⎝ ⎠
= –48
25r = –1
96
4825
2 25r
⎛ ⎞⎜ ⎟
⎜ ⎟⎝ ⎠∵
81. Equation of tangent on parabola y2 = 4x is
1y mx
m ...(i)
Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018
7/7
It passes through point (–2, –1).
So, 1
–1 – 2mm
2m2 – m – 1 = 0
Roots of equation are m1, m
2.
Then, tan = 1 2
1 2
1 8
– 2
111–
2
m m
m m
tan = 3
82.
2 2( – 1) ( 3)1
4 16
x y
( 3) ,b
ye
a2 = b2 (1 – e2)
8
33
y
5
3y or
–11
3y
83. Foci of the parabolas are (0, a) and (2a, 0)
respectively, so equation of the circle described on
the line segment joining the foci of the parabolas as
diameter is
(x – 0) (x – 2a) + (y – a) (y – 0) = 0
x2 + y2 – 2ax – ay = 0
84. In slope form equation of tangent to 4x2 – 9y2 = 36 is
y = 2
9 – 4mx m ; m < 0
2
2
4 9 – 4
1
m m
m
= 4
–2
5
m
Required equation of tangent is
2 5 4 0x y 85. Equation of common chord on two circles is
S1 – S
2 = 0.
x2 + y2 – 6x – 4y + 9 – x2 – y2 + 8x + 6y – 23 = 0
2x + 2y – 14 = 0 x + y – 7 = 0
P
C2
(4, 3)
B
A
C1
(3, 2)
r1
C1P =
3 2 – 7
2
=
22
2
AB = 2AP = 2 2
1 12 – ( )r C P = 2 4 – 2 = 2 2
86. x2y2 – 9y2 – 6x2y + 54y = 0
y2 (x2 – 9) – 6y(x2 – 9) = 0
(x2 – 9) (y2 – 6y) = 0
(x + 3) (x – 3) (y) (y – 6) = 0
So, x = –3, x = 3, y = 0, y = 6
It will form a square, so
Area of quadrilateral = 36.
87. Let the direct common tangent intersect at P and
PN1 = l, PC
1 = .
b
C2
N2
a
C1
N1
Pl
PC1N
1 and PC
2N
2 are similar, so,
1 2
1 1 2 2
PC PC
C N C N
a
=
a b
b
=
2
–
a ab
b a
Now l2 = 2 – a2 =
22
2–
–
a aba
b a
⎛ ⎞⎜ ⎟⎝ ⎠
=
3
2
4
( – )
a b
a b
l =
3 1
2 22
( – )
a b
b a
89. Point of contact of y = mx + c is
2 2
2 2 2 2 2 2
,
a m b
a m b a m b
⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠
=
1 1–3 4,7 7
12 12
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
= –2 21 21
,21 14
⎛ ⎞⎜ ⎟⎝ ⎠
90. (x – 1)2 – (y – 2)2 – 1 + 4 – 7 = 0
2 2( – 1) ( – 2)– 1
4 4
x y
It is a rectangular hyperbola with eccentricity 2.
� � �
Test - 4 (Code-H) (Answers) All India Aakash Test Series for JEE (Main)-2018
1/7
1. (3)
2. (2)
3. (3)
4. (2)
5. (3)
6. (1)
7. (1)
8. (2)
9. (1)
10. (3)
11. (1)
12. (2)
13. (3)
14. (3)
15. (1)
16. (4)
17. (1)
18. (1)
19. (3)
20. (3)
21. (2)
22. (1)
23. (1)
24. (3)
25. (1)
26. (1)
27. (1)
28. (4)
29. (1)
30. (4)
PHYSICS CHEMISTRY MATHEMATICS
31. (1)
32. (4)
33. (3)
34. (4)
35. (2)
36. (4)
37. (2)
38. (2)
39. (2)
40. (3)
41. (3)
42. (4)
43. (2)
44. (3)
45. (4)
46. (2)
47. (1)
48. (4)
49. (1)
50. (3)
51. (2)
52. (3)
53. (4)
54. (3)
55. (3)
56. (4)
57. (1)
58. (1)
59. (3)
60. (2).
61. (2)
62. (2)
63. (1)
64. (2)
65. (3)
66. (4)
67. (4)
68. (2)
69. (1)
70. (4)
71. (1)
72. (2)
73. (2)
74. (3)
75. (1)
76. (4)
77. (3)
78. (1)
79. (3)
80. (3)
81. (2)
82. (4)
83. (4)
84. (3)
85. (2)
86. (1)
87. (4)
88. (4)
89. (1)
90. (3)
Test Date : 14/01/2018
ANSWERS
TEST - 4 (Code-H)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-H) (Hints)
2/7
1. TB
= B
A
A
VT
V = 600 K
WBC
= lnC
B
B
VnRT
V = 1200 R ln2
2. On heating volume will increase. So, it will move
towards right.
3. Q = U + W
W = 4
Q
3
4
Q = C
V dT =
5
2R dT
C = 10
3R
4. 0.42
= 0.8 m
TV
= 8 m/s
f = 10 Hz
5. f.L1 =
2
B
4.3
Lf =
1
B
1
4
3
L
L =
1
2
L1 =
1
2
4
3
L
6. TV r – 1 = Constant
–( –1)
dV V
dT T
–tan = 0–
( –1)V
T
( – 1) = 0
0tan
V
T
CV =
–1R
= 0
0
tanRT
V
7. T = Const.
4 4
2 1
1
16T T
T2 =
1
2
T
PART - A (PHYSICS)
8. 0 1 0 2–
mg mgP L P L
A A
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
1
101
99
L
L
9.
4
A
B
T
T
⎛ ⎞⎜ ⎟⎝ ⎠
= 0.81
0.01 = 81
A
B
T
T = 3,
ATA =
BTB
10. WCD
= nRT2 ln
D
C
V
V
WAB
= nRT1 ln
B
A
V
V
VA = V
C, V
B = V
D
2WCD
= WAB
2T2 = T
1
1
2
T
T = 2
12. dW∫ = P dv∫
= av2 dv = 3 3
2 1–
3
av v
W = 2 1( – )
3
RT T
13. (1 –
2)T =
= 2 2 4
–5
TT T
⎡ ⎤⎢ ⎥⎣ ⎦
= 2
5
= 72°
14. T = 2m
K
T2 = 2 2
1 2T T 1 2
eq1 2
K KK
K K
⎛ ⎞⎜ ⎟⎝ ⎠
15.1
Ir
I A2
2
1
2
25
9
A
A
⎛ ⎞ ⎜ ⎟⎝ ⎠
1
2
5
3
A
A
16. Maximum possible will occur if it is around mean
position.
(Vavg
)max
= 2
4
A
T =
4 2A
T
Test - 4 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
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PART - B (CHEMISTRY)
32.
35.Br
2
Br
Br(A)
(i) alc. KOH
(ii) NaNH2
CH –C C–CH3 3
(B)
H /Pd
BaCO
2
3
C=CCH
3CH
3
(C) HH
41.
X
Li/NH3
X
[with EDG]
Y
Li/NH3
Y
[with EWG]
due to stability of carbanion intermediates.
17.
45°
45°
A
A
A
Ar
= (2A cos45° + A) = ( 2 1)A
18. T = 20
21800
= 2s
3
t = 4
T =
1s
6
19. = 2
13
= 120°
21. Q = –dT
TAdx
0
lQ
dxA∫ =
2
1
–
T
T
T dT ∫
Q = 2 2
1 2( – )
2
AT T
l
22. U = 1
2 (stress × strain × volume)
23. P0V
0 – V
0 = nRT
0
V0 =
0
0( – )
nRT
P , n = 1
w = P0V =
0 0
0( – )
P RT
P
26.max
y
t
⎛ ⎞⎜ ⎟⎝ ⎠
= 2f
27. Y = 2
8
( – 2 )x t
,
3
–16 2
( – 2 )
dy
dt x t
= 4 m/s
28. Vmax
= A = mg
K
= 10
40
g
= 1
4 m/s
30. Y = 4 sin2t sin2t
= 4(1 – cos2 ) sin2
2t t = 2 sin2t – sin4t
45. % of P = 2 2 7
wt. of Mg P O62100
222 wt. of compound
= 62 0.75
100222 0.6
= 35%
46. 2R COOK + 2H2O R—R + 2CO
2 + H
2 + 2KOH
49. +I effect of D is more than of H.
51.
OH
+ Zn (Dust)
(X)
CH3
+ CH Cl3
AlCl3
(Y)CH
3
alk. KMnO4
(Z)
COOH
NaOH
(U)
COONa
Sodalime
(X)
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-H) (Hints)
4/7
PART - C (MATHEMATICS)
61. (x – 1)2 – (y – 2)2 – 1 + 4 – 7 = 0
2 2( – 1) ( – 2)– 1
4 4
x y
It is a rectangular hyperbola with eccentricity 2.
62. Point of contact of y = mx + c is
2 2
2 2 2 2 2 2
,
a m b
a m b a m b
⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠
=
1 1–3 4,7 7
12 12
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
= –2 21 21
,21 14
⎛ ⎞⎜ ⎟⎝ ⎠
64. Let the direct common tangent intersect at P and
PN1 = l, PC
1 = .
b
C2
N2
a
C1
N1
Pl
PC1N
1 and PC
2N
2 are similar, so,
1 2
1 1 2 2
PC PC
C N C N
a
=
a b
b
=
2
–
a ab
b a
Now l2 = 2 – a2 =
22
2–
–
a aba
b a
⎛ ⎞⎜ ⎟⎝ ⎠
=
3
2
4
( – )
a b
a b
l =
3 1
2 22
( – )
a b
b a
52.BH , THF
3
H
BH2
H
H
CH COOH3
(Alkane)
53. A = , B =
65. x2y2 – 9y2 – 6x2y + 54y = 0
y2 (x2 – 9) – 6y(x2 – 9) = 0
(x2 – 9) (y2 – 6y) = 0
(x + 3) (x – 3) (y) (y – 6) = 0
So, x = –3, x = 3, y = 0, y = 6
It will form a square, so
Area of quadrilateral = 36.
66. Equation of common chord on two circles is
S1 – S
2 = 0.
x2 + y2 – 6x – 4y + 9 – x2 – y2 + 8x + 6y – 23 = 0
2x + 2y – 14 = 0 x + y – 7 = 0
P
C2
(4, 3)
B
A
C1
(3, 2)
r1
C1P =
3 2 – 7
2
=
22
2
AB = 2AP = 2 2
1 12 – ( )r C P = 2 4 – 2 = 2 2
67. In slope form equation of tangent to 4x2 – 9y2 = 36 is
y = 2
9 – 4mx m ; m < 0
2
2
4 9 – 4
1
m m
m
= 4
–2
5m
Required equation of tangent is
2 5 4 0x y 68. Foci of the parabolas are (0, a) and (2a, 0)
respectively, so equation of the circle described on
the line segment joining the foci of the parabolas as
diameter is
(x – 0) (x – 2a) + (y – a) (y – 0) = 0
x2 + y2 – 2ax – ay = 0
54. Heat of hydrogenation 1
stability of alkene
55. CH CH + SeO2 CHO—CHO
3CH CH Red hot
iron tube
873 K
56.
No rearrangement
Test - 4 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
5/7
69.
2 2( – 1) ( 3)1
4 16
x y
( 3) ,b
ye
a2 = b2 (1 – e2)
8
33
y
5
3y or
–11
3y
70. Equation of tangent on parabola y2 = 4x is
1y mx
m ...(i)
It passes through point (–2, –1).
So, 1
–1 – 2mm
2m2 – m – 1 = 0
Roots of equation are m1, m
2.
Then, tan = 1 2
1 2
1 8
– 2
111–
2
m m
m m
tan = 3
71. Ellipse 16x2 + 25y2 = 400
2 2
125 16
x y
Let the points on ellipse be P(5 cos1, 4 sin
1) and
Q(5cos2, 4sin
2).
Circle described on PQ as diameter touches x-axis
at (3, 0).
1 2cos cos
5 32
⎛ ⎞ ⎜ ⎟⎝ ⎠
and 1 2
sin sin4
2r
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 24tan
5 2 3
r ⎛ ⎞ ⎜ ⎟⎝ ⎠
1 2 5
tan2 12
r ⎛ ⎞ ⎜ ⎟⎝ ⎠
slope of PQ = 1 2–4
cot5 2
⎛ ⎞⎜ ⎟⎝ ⎠
= –48
25r = –1
96
4825
2 25r
⎛ ⎞⎜ ⎟
⎜ ⎟⎝ ⎠∵
72. The locus of P is an ellipse, if K > AB
K > 50 K > 5 2
73. The given parabolas 2y2 = 2x – 1 and 2x2 = 2y – 1
are symmetrical about the line y = x. Also, the
shortest distance occurs along the common normal
which is perpendicular to the line y = x. Then the
shortest distance is
1 1 1
16 16 2 2d
74. Any point on the parabola is P(t2, 2t). The
circumcenter Q(h, k) of triangle PAB is midpoint of
P(t2, 2t) and the centre C(–3, 2) of the circle.
Hence,
2– 3
2
th and k = t + 1
Eliminating t, we get (k – 1)2 = 2h + 3
Locus of P(h, k) is (y – 1)2 = 2x + 3
Equation of tangent to parabola with slope 1 is
(y – 1) =
1
3 21
2 1x
⎛ ⎞ ⎜ ⎟⎝ ⎠
y = x + 3
75.
XO
Y
PQ
A (1, 3)
Let P(2cos, 2sin). The mid point Q(h, k) of AP
will be given by
h = 1 2cos
2
cos =
1–2
h
and k = 3 2sin
2
sin =
3–2
k
Eliminating , we get
2 21 3
– – 12 2
h k⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Hence, the locus is
2 21 3
– – 12 2
x y⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
So, radius = 1.
76. The equation ax3 + 3bx2y + 3cxy2 + dy3 = 0
3 2
3 3 0y y y
d c b ax x x
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Let y
mx
dm3 + 3cm2 + 3bm + a = 0
If the roots are m1, m
2 and m
3, then m
1 + m
2 + m
3
= –3
,
c
d m
1m
2 + m
2m
3 + m
3m
1 =
3b
d and m
1m
2m3
= –a
d.
For three coincident lines, we have m1 = m
2 = m
3,
so, m = –c
d, m2 =
b
d and m3 =
–a
d
So,
3– –c a
d d
⎛ ⎞ ⎜ ⎟⎝ ⎠
c3 = ad2
All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-H) (Hints)
6/7
77.
BR
P
X
AO
Y
Equation of tangent at P(a cos, b sin) is
cos sin 1x y
a b
It meets the major axis at A(a sec, 0) and the
minor axis at B(0, b cosec).
Equation of the OR, perpendicular to the tangent at
P is ax sin – by cos = 0.
Now PR = 2 2 2 2
( cos )sin – ( sin )cos
sin cos
a a b b
a b
=
2 2
2 2 2 2
( – ) sin cos
sin cos
a b
a b
Also, AB =
2 2 2 2sin cos
sin cos
a b
So, (AB) (PR) = a2 – b2
Here, a = 4, b = 3
(AB)(PR) = 16 – 9 = 7
78.
Pt
t(2 , 4 )
11
2
Q t t(2 , 4 )2 2
2
Let absciss of point P is 2
12 18t 2
19t
t1 = 3
We know that, If PQ is normal then 2 1
1
2– –t t
t
2
11–
3t
So, coordinates of P and Q are (18, 12) and
242 –44,
9 3
⎛ ⎞⎜ ⎟⎝ ⎠
respectively.
PQ =
2 2242 –44
– 18 – 129 3
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 80 10
9
79.
2 2
22
15 ( – 1) 2– 22
x y
aa a
2 2
22
13 ( – 1) 2( – 1)2
x y
aa
where, A2 = (a – 1)2 + 3
2, B2 = (a – 1)2 + 2
We know that,
A2 = B2 (1 – e2), (for vertical ellipse)
2 3( – 1)
2a = 2 1
( – 1) 2 1–4
a⎛ ⎞ ⎜ ⎟⎝ ⎠
4(a – 1)2 + 6 = 3(a – 1)2 + 6
(a – 1)2 = 0 a = 1
Equation of ellipse is
2 2
13 2
2
x y
So, length of L.R. =
22A
B =
32
2
2
=
9
2
80. Clearly ax + by = 1
i.e.,– 1a
y xb b
is tangent to cx2 + dy2 = 1
2 2
11 1
x y
c d
2 2
1 1 – 1a
b c b d
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 =
2 2a b
c d
81. Putting y = 2t2 in the equation of the given ellipse
2 2
1,9 4
x y we get
2 44
19 4
x t
x2 = 9(1 – t4) = 9(1 – t2) (1 + t2)
This will give real values of x if 1 – t2 0, i.e., |t| 1.
82. The circle and the ellipse do not meet at any point.
Length of the common chord = 0.
83. If S1 = 0 and S
2 = 0 are the equations, then
S1+ S
2= 0 is a second degree curve passing through
the points of intersection of S1 = 0 and S
2 = 0.
( + 4)x2 + 2( + 1)y2 – 2(3 + 10)x
– 12( + 1)y + (23 + 35) = 0 ...(i)
For it to be a circle, choose such that the coefficient
of x2 and y2 are equal
+ 4 = 2 + 2
= 2
Test - 4 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018
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� � �
This gives the equation of the circle as
6(x2 + y2) – 32x – 36y + 81 = 0 {(using (i))}
2 2 16 27– – 6 0
3 2x y x y
Its centre is 8, 3
3C⎛ ⎞⎜ ⎟⎝ ⎠
and radius is
r = 64 27
9 –9 2
= 1 47
3 2
84. Centre and radius of circle x2 + y2 + 8x – 10y – 40 = 0
are C(–4, 5) and 9 respectively. Distance of the centre
C(–4, 5) from the point P(–2, 3) is CP = 4 4 =
2 2 .
a = CP + r = 2 2 + 9
b = |CP – r| = 9 – 2 2
a – b = 4 2
85. The two circles are x2 + y2 – 4x – 6y – 3 = 0 and
x2 + y2 + 2x + 2y + 1 = 0
Centres : C1(2, 3) and C
2(–1, –1)
radii : r1 = 4 and r
2 = 1.
We have, C1C
2 = 5 = r
1 + r
2, circles touch externally
therefore there are three common tangents to the
given circles.
86. The point 6
P⎛ ⎞
⎜ ⎟⎝ ⎠
is sec , tan6 6
a b ⎛ ⎞
⎜ ⎟⎝ ⎠
or
2,
3 3
a bP⎛ ⎞⎜ ⎟⎝ ⎠
.
Equation of tangent at P is – 13 3
2
x y
a b
Area of the triangle = 1 3
32 2
ab = 3a2
b
a = 4
e2 =
2
21 17
b
a
87. Make homogeneous 3x2 – y2 –2x + 4y = 0 with the
line y = mx + c
i.e., 2 2 – –
3 – – 2 4y mx y mx
x y x yc c
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 0
According to question coeff. of x2 + coeff. of y2 = 0
2 43 –1 0
m
c c
m + c + 2 = 0
which gives fixed point (1, –2) p + q = –1
88. Equation of hyperbola is
2 2
– 1.9 16
x y
Equation of tangent is
29 – 16y mx m
29 – 16 2 5m m = ±2
a + b = sum of roots = 0
89. Let P(h, k) be the point of intersection of E1 and E
2
2
2
21
hk
a h2 = a2(1 – k2) ...(i)
and
2 2
21
h k
a = 1 k2 = a2 (1 – h2) ...(ii)
Eliminating a from equation (i) and (ii), we get
2 2
2 21– 1–
h k
k h
h2(1 – h2) = k2 (1 – k2)
(h – k) (h + k) (h2 + k2 – 1) = 0
Hence, the locus is a set of curves consisting of the
straight lines y = x, y = –x and circle x2 + y2 = 1.
90. If is angle between tangents, then
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
= radius
distance between 3 3, 3 and (0, 0)
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
= 3
6 =
1
2
2
=
6
=
3
So, cot = cot3
⎛ ⎞⎜ ⎟⎝ ⎠
= 1
3
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