Test - 5 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/15
1. (2)
2. (3)
3. (3)
4. (2)
5. (3)
6. (4)
7. (4)
8. (3)
9. (1)
10. (2)
11. (3)
12. (4)
13. (4)
14. (2)
15. (2)
16. (2)
17. (3)
18. (3)
19. (3)
20. (3)
21. (2)
22. (3)
23. (4)
24. (3)
25. (2)
26. (1)
27. (2)
28. (4)
29. (3)
30. (3)
PHYSICS CHEMISTRY MATHEMATICS
31. (1)
32. (2)
33. (3)
34. (3)
35. (2)
36. (1)
37. (3)
38. (4)
39. (4)
40. (4)
41. (2)
42. (2)
43. (4)
44. (3)
45. (3)
46. (3)
47. (2)
48. (3)
49. (1)
50. (2)
51. (4)
52. (2)
53. (3)
54. (2)
55. (1)
56. (2)
57. (3)
58. (1)
59. (1)
60. (3)
61. (3)
62. (4)
63. (3)
64. (3)
65. (1)
66. (3)
67. (4)
68. (3)
69. (2)
70. (2)
71. (3)
72. (3)
73. (4)
74. (2)
75. (4)
76. (1)
77. (1)
78. (3)
79. (2)
80. (1)
81. (1)
82. (1)
83. (3)
84. (2)
85. (3)
86. (2)
87. (2)
88. (2)
89. (2)
90. (3)
Test Date : 16/12/2018
ANSWERS
TEST - 5 - Code-A
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
2/15
1. Answer (2)
Hint : Vc =
2
0
14
RG
r drr r
Sol. : M =
22
0
14 4
2
RR
r drr
Vc =
2
0
1 2(4 )
RG GM
r drr r R
2. Answer (3)
Hint : Two surfaces are formed in the capillary
Sol. : Since force due to surface tension becomes
twice
h = 2h
3. Answer (3)
Hint : Net force on particle = Centripetal force
Sol. :
2
3
mv
a
=
2
23Gm
a
v = Gm
a
T =
32
23
r a
v Gm
4. Answer (2)
Hint : p = T
r for cylindrical surface
Sol. : p = 2T T
r d
A = V m
d d
F = (p) × A = 2
2
Tm
d
5. Answer (3)
Hint : dr
dt =
2gR
r
Sol. : u = 2gR
21
2mv –
GMm
r =
21
2GMm
muR
PART - A (PHYSICS)
v = 2g
Rr
dr
dt =
2gR
r
4
R
R
r dr =
0
2 t
R g dt
t = 7 2
3
R
g
6. Answer (4)
Hint : Imagine hemisphere.
Sol. : Assume upper hemisphere
B = 32
3 R g
F2 = R2 × g(2R)
F1 = F
2 – B =
34
3 R g
7. Answer (4)
Hint : Torque = ( ) p dA r
Sol. : d = 45
( )3
g H x adx x ga
8. Answer (3)
Hint : dA
dt =
2
L ab
m T
Sol. : Areal speed, vA =
abT
Area(A B)
= 1
22 2
abb ae
t = ( )Area 1
12
A B
A
T
v
9. Answer (1)
Hint : Loss in PE = gain in KE
Sol. :21
2mv =
3 2 GMm GMm
R R
v = 3
GM
R
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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10. Answer (2)
Hint : R = time of fall × speed of efflux then 0dR
dh
Sol. : hreq
= 2
2
HH
= 3
4
H
height from bottom of vessel = 3
4 2 4 H H H
11. Answer (3)
Hint : = p
B
Sol. : p = gh
= pB
� 2 kg/m3
12. Answer (4)
Hint : Force = Pav
× Projection area.
Sol. : Fx= P
av × Area
= ( )2
�gRR
=
2
2
�g R
13. Answer (4)
Hint : T = Vg, � = �T
YA
Sol. : T = × Vg = 34
3 R g
� =
34
3
� �T R g
YA YA
14. Answer (2)
Hint : Use equation of ellipse.
a = 1 2
2
r r
Sol. :E
S
r1r2
2 2
2 2x y
a b = 1
|x| = ae = a – r2,
a = 1 2
2
r r
y2 =
2
2
21
xb
a
y = 1 2
1 2
2
( )r r
r r
15. Answer (2)
Hint : W = 2( ) a x g dx
Sol. : F = (a2x)g
W =
0.1
2
0
( ) a
a x g dx
=
4
200
a g
16. Answer (2)
Hint : dv
dt = av
Sol. : av =
dv
dt
1 × 10–4 × 2gh =
330 10
60
h = 1.25 m
17. Answer (3)
Hint : E = U + K = 21
2kA
Sol. : v =
2 2
2 3
4 4 A A
A
v = 2v = 22
32
A
E = U + K =
2 2
21 1 34
2 2 2 4
A Ak m
=
221 1
32 4 2
Ak k A
2 2113
2 2 A
kA A
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
4/15
18. Answer (3)
Hint :
2
2
d
dt
= –2 for angular SHM
Sol. : = 6 6 3 3 L L L L
k k
2 2
2
2
2
3
mL dmL
dt = –
2 1 1
36 9
kL
2 2
2
5
3
mL d
dt =
25
36 kL
2
2
d
dt =
12 k
m
= 12
k
m
f = 1
2 12k
m =
1
4 3k
m
19. Answer (3)
Hint : 2R = R
R3R
Sol. : 2R = R
= 2
mg sinR =
22
2
7
5
dmRdt
=
22
2
14
5
dmRdt
2
2
d
dt =
5
14
g
R
= 5 14
, 214 5
g RT
R g
20. Answer (3)
Hint : fstring
= 2 µ
n T
L
fc =
4
V
L
Sol. : f0 =
4
V
L =
320100 Hz
4 0.8
100 = 2
50 cm2
TLL
L m
21. Answer (2)
Hint : g(x, t) = f((x – v(t – t0)), t)
Sol. : g(x, t) = f((x – v(t – t0)), t)
22. Answer (3)
Hint : Power, P = Area under E versus
Sol. : Power, P = Area under E versus
4
1/41 11 2 2
2 2
4 4 2
P TT T T
P T
23. Answer (4)
Hint : Work done by a gas in a cyclic process is
negative if P-V graph is in anticlockwise
sequence.
Sol. : Wby gas
=1
– 1 40 –20 J2
24. Answer (3)
Hint : For polytropic process, PVx = constant
C = 1
v
RC
x
Sol. : C = 5 2 1
13
R R =
3
2 2 R RR
Q = 3
,2 2 R R
n T U n T
W = –nRT
25. Answer (2)
Hint : For process PVx = constant
[ ]
1
i fnR T T
Wx
Sol. PT = constant P2V = constant
PV 1/2 = constant
W = [ ]
11
2
i fnR T T
= [600 300]
1
2
R = –600R
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/15
26. Answer (1)
Hint : n1CV1
+ n2CV2
= (n1 + n
2)CV
Sol. :3 3 5
2 2
R R = 4C
v
Cv =
18 9
2 4 4
R R
= 13
9
27. Answer (2)
Hint : 1 dV
V dT
Sol. :1 dV
V dT , PT2 = Constant
T3 = kV
2
2 33
dV T dVT k
dT k dT
2
3
3 1T dVk
V dTkT
28. Answer (4)
Hint : For maximum intensity, path difference = n
Sol. : Path difference = 2 × 8.5 cm = 17 cm
29. Answer (3)
Hint : For maximum, path difference = n
In a quadrant path difference varies continuously
from 3 to 0 4.
Sol. : For maximum, path difference = n
In a quadrant path difference varies continuously
from 3 to 0 4.
30. Answer (3)
Hint : Work done by external pressure = Pex
[Vi – V
f]
Sol. : Work done by external pressure
= 2P0[V
0 – V] =
0
3[ ]
2R
n T T [–W = U]
Also 0 0 0 0
0 0
2 2PV PV VTT
T T V
P0V
0 = nRT
0
4[V0 – V] = 3[2V – V
0]
7V0 = 10V V =
07
10
V
31. Answer (1)
Hint : 3 2 2 2LiNO Li O NO O
Sol. : Lithium cannot form alkynide on reaction with
ethyne
32. Answer (2)
Hint : K form superoxide.
Sol. : KO2, Na
2O
2, Li
2O
33. Answer (3)
Hint : Concentrated solution is diamagnetic.
Sol. : Alkali metal dissolve in ammonia.
34. Answer (3)
Hint : Change in pH of buffer is minimum
PART - B (CHEMISTRY)
Sol. :
CH3COO– + H+ CH
3COOH
t = 0 1 0.1 1
t = eq 0.9 — 1.1
(pH)f = pK
a + log
0.9
1.1
(pH)i = pK
a + log
1
1
pH decreases
As concentration of buffer increase buffer
capacity increase
NaOH + H+ H2O
0.1 0.1
Salt is formed, pH = 7 and change in pH is
maximum
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
6/15
35. Answer (2)
Hint : Sulphate of 2nd group are white
Sol. : Solubility 2nd group sulphate or carbonate
decrease as atomic number increase
36. Answer (1)
Hint : Mg2+ has 6 or 8 molecule of water
Sol. : Ba(NO3)28H
2O does not exist.
37. Answer (3)
Hint : For ppt
Kip > K
sp
Sol. : Experiment I :
For saturated solution of AgCl
[Ag+] = [Cl–] = 10–5 M
If 10–3 mol NaBr is added it means,
[Br–] = 10–3 M.
(Kip)AgBr
= 10–5 × 10–3 = 10–8 > 10–14
(Ksp
of AgBr)
So AgBr form ppt.
Experiment II :
For saturated solution of AgBr
[Ag+] = [Br–] = 10–7
If 10–5 mol of NaCl is added in 100 mL solution so
[Cl–] = 10–4
Kip of AgCl = 10–7 × 10–4 = 10–11
Kip < K
sp so no ppt
38. Answer (4)
Hint : A = B3N
3H
3Cl
Y = B3N
3H
6
Y + 3HCl = B3N
3H
9Cl
3
Sol. :
YHCl
B N
B N
BN
—
—
Cl
H
H
H HH
H
H
HH
Cl
Cl
+
+
+
—
39. Answer (4)
Hint : Al2Cl
6 has 3c – 4e– as well as 2c – 2e– bond.
Sol. :
Cl
Cl
Cl
Cl
Cl
Cl
Al Al
40. Answer (4)
Hint : Buckminster fullerene is aromatic in nature.
Sol. : Stability: Graphite > Diamond > C60
41. Answer (2)
Hint : R3SiCl
Sol. : R3SiCl terminate the chain
42. Answer (2)
Hint :f
Distance moved by the substance
from base lineR
Distance moved by the solvent from
base line
Sol. :4
0.410
43. Answer (4)
Hint : H2O
2 is miscible with water
Sol. :3 2 2 2
Mg(HCO ) Mg(OH) 2CO
3 2 3 2 2Ca(HCO ) CaCO H O CO (g)
44. Answer (3)
Hint : Lassaigne's test for nitrogen
Sol. : Fe2+ + CN– [Fe(CN)6]4–
Blue coloured compound is prussian blue
45. Answer (3)
Hint : I effect act on O > M > P while hyperconjugation
effect act on O and P with equal intensity
Sol. : +I power CD3 > CH
3
+H power CH3 > CD
3
Stability order
A > D > C > B
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/15
46. Answer (3)
Hint : 2 mol
H H
O
O O
2 mol
H
O
H
Means P must contain 8 carbon
Sol. :
H
O
OH
O
H
(I) O 3
(II) Zn/H O2
+
1 mol
2 mol
H
O
2 mol
47. Answer (2)
Hint : A is 3° Alcohol
Sol. :
H+
Ring expansion
(+)
Ring expansion
(+)
H O2
OH
+
48. Answer (3)
Hint : In F – CH2 – CH
2 – OH, gauche form is more
stable than anti.
Sol. :H
H
F
H
H
O
H
Most stable
49. Answer (1)
Hint : 2Cr+6 2Cr+3
+6e–
2N3– N2 + 6e–
Sol. : So a total of 6e– are involved in the above
intramolecular redox
n factor = 6
50. Answer (2)
Hint : Let m moles of Na2CO
3 be x 2x meq
and NaOH be y y meq
Sol. : For phenolphthalein end point
2xy
2 = 30 × 0.5
x + y = 15
For methyl orange end point
2x
2 = 10 × 0.5
x = 5
y = 10
weight of Na2CO
3 = 5 × 10–3 × 106 = 0.53 g
weight of NaOH = 10 × 10–3 × 40 = 0.4 g
Impurity = 0.07 g (7% impurity is present)
51. Answer (4)
Hint : Zn + 4HNO3 4Zn(NO
3)2 + 2NO
2 + 2H
2O
Sol. : The H+ required for the conversion will be
provided by HNO3
Zn + 2NO + 4H Zn +2NO + 2H O3 2 2
– + 2++ 2NO3
–
+ 2NO3
–
Zn + 4HNO Zn(NO ) +2NO + 2H O3 3 2 2 2
(0.2)
0.8 moles of HNO3 will be consumed
1.6 litres of 0.5 M HNO3 is required
52. Answer (2)
Hint : M + H2SO
4 Ma+ + H
2
(8.96 L)
Ma+ + KMnO4 Mb+ + Mn2+
Sol. : Reaction (1)
g meq of H2 = g meq of metal
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
8/15
8.962
22.4
= 0.4 × (x)
0.8 = 0.4 × x
x = 2 a 2
Reaction (2)
eq of KMnO4 = eq of metal
40 × 2 × 5 × 10–3 = 0.4 × x
40 × 10 × 10–3 = 0.4x
x = 1
b – a = 1
b = 3 a + b = 5
53. Answer (3)
Hint : Conc. of solid is constant
Sol. : Addition of inert at constant pressure move
towards more no. of gaseous mole.
54. Answer (2)
Hint : Nitro benzene is formed which is m-directing
Sol. : 3Zn, CH OH
2 2 2 2 2Cl CH CHCl C H 2ZnCl
CH2 2
Red hot
Cu
HNO3
H SO2 4
NO2
Br /Fe2
Br
NO2
55. Answer (1)
Hint : As for polluted water B.O.D is higher.
Sol. : B.O.D: A < C < B < D
Pollution level in water A < C < B < D
56. Answer (2)
Hint : E2 elimination
Sol. : Anti H is removed
57. Answer (3)
Hint : AI
Sol. : AI has highest value of oxidation potential
58. Answer (1)
Hint : [H+] = 1 2
a 1 a 2K C K C
Sol. : Case of simultaneous equilibrium
4O
4O
HA A H , 1.7 10
(0.1 x) x (x y)
HB B H , 4.7 10
(0.1 y) y (y x)
���⇀↽���
���⇀↽���
x(x + y) = 1.7 × 10–5
y(x + y) = 4.7 × 10–5
(x + y)2 = 6.4 × 10–5
(x + y)2 = 64 × 10–6
x + y = 8 × 10–3
[H+] = 8 × 10–3
59. Answer (1)
Hint : Correct order is a > b > c
Sol. : As s character increases bond length decreases
60. Answer (3)
Hint :
AB(g) A(g) B(g)
1 p p p' p p ''
���⇀↽���
A(g) C(g) D(g)
p p' 1 p' p ' p ''
���⇀↽���
D B E
p' p '' p p '' p ''
���⇀↽���
Given [A][B] p p
0.15[AB] 1 p
Sol. : Given p = 0.2
1 – p = p – p
1 p''
2
= p, p = 0.6 atm
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
9/15
PART - C (MATHEMATICS)
61. Answer (3)
Hint :1
loglog
a
b
ba
Sol. :cos
cos
3log sin 4
log sin
logcossin = 1 or log
cossin = 3
There are two solutions.
62. Answer (4)
Hint : Max. value of sinnx = 1
Sol. : sinx + sin4x = 2
sinx = 1 and sin4x = 1
(4 1)2
x n and 4 (4 1)
2x m
(4 1)2
x n and (4 1)
8x m
There is no common solution.
63. Answer (3)
Hint : Max. value of sinx + cosx is 2 at x = 4
Sol. : 5 + 7sin2x + 3 3
sin cos2 22 2
x x
Max. value of 3sin 4cos2 2
x x is 3 at 2
x
Max. value of given expression = 15
64. Answer (3)
Hint : In ax2 + bx + c = 0
sum of roots = –b
a
Product of roots = c
a
From equilibrium of reaction I
(0.6 p')(0.4)
0.4
=
15
100
p = 0.45
pAB
= 0.4 atm
pA = 0.15 atm
pB = 0.4 atm
pC = 0.55
pD = 0.25
pE = 0.2
Sol. : cosx1 + cosx
2 = –a and sinx
1 + sinx
2 = –c
1 2 1 2–
2cos cos –2 2
x x x xa
and1 2 1 2
–2sin cos –
2 2
x x x xc
1 2tan
2
x x c
a
sec(x1 + x
2) =
2
2 22
2 2 2
2
1
–1–
c
a ca
c a c
a
65. Answer (1)
Hint : In a triangle
cot cot cot2 2 2
A B C = cot .cot .cot2 2 2
A B C
Sol. : In a triangle we know that
cot cot cot2 2 2
A B C = cot cot cot
2 2 2
A B C
cot cot cot 3cot2 2 2 2
A B C B
cot cot 32 2
A C
cot cot 2 cot cot
2 2 2 2
A C A B
2cot 2 32
B cot 32
B
1tan
2 3B
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
10/15
66. Answer (3)
Hint :
–1
–1
sin2
cos2 .cos2
r
r r rT
Sol. :
–1
–1
1
sin2
cos2 .cos2
rn
r r
r
=
–1
–1
1
sin(2 – 2 )
cos2 .cos2
r rn
r r
r
= –1
1
(tan2 – tan2 )n
r r
r
= (tan2n – tan)
67. Answer (4)
Hint : 2sinA sinB = cos(A – B) – cos(A + B)
Sol. :–
4 sin .sin2 2
A C A CR
= 2R(cosC – cosA)
= 2 2 2 2
2 –a c
R
a c a c
= a – c
68. Answer (3)
Hint :
2 2 2–
cos2
a c bB
ac
Sol. :
2 2 2 2 2 2– cos –
cos ,2 2
a c b b B bB
ac ac
=
2 2(cos –1)0
2
b B
ac
cosB < 0 2
B
69. Answer (2)
Hint : Speed = distance
time
Sol. :30°
45°
x
xy
1200 m
A B
P
1200tan45
y y = 1200 ...(1)
1200tan30
x y
1200 1
3x y
1200 3 = x + y
x = 1200 3 – 1200 x = 1200( 3 – 1)
Speed = 1200( 3 – 1)
5 = 240 3 –1 m/s
70. Answer (2)
Hint : Lines are y = x + 3; y = 1 – x, x = 0, y = 0
Sol. : Let the equation of tangent be
y = 25 5
3 2mx m
2t passes through (1, 2)
2 = 25 5
3 2m m
4m2 + 24m – 9 = 0
tan = 12
5
71. Answer (3)
Hint : Use property-vertex of parabola bisects the
subtangent and median through the point
passes through the vertex.
Sol. :
V(1, 2)
P(2, 1)
x
y
NT
∵ Vertex bisects the subtangent TN
Median through the point passes through
the vertex V(1, 2)
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
11/15
Equation of median of PTN, through P(2, 1)
is y – 1 = 1– 2
( – 2)2 –1
x x + y = 3
72. Answer (3)
Hint : Equation of normal at 1
,A tt
to the hyperbola
xy = 1.
is 1
–yt
= t2(x – t)
Sol. : Equations of normals at 1
1
1,A tt
and
2
2
1,B tt
to the hyperbola xy = 1 are
2
1 1
1
1– – ,y t x tt ...(i)
and 2
2 2
2
1– –y t x tt
...(ii) respectively.
(i) and (ii) will represent the same normal if
2 2
1 2t t and
3 3
1 2
1 2
1 1– –t t
t t
t1 = t
2 (Rejected) 3 3
1 2
1 2
1 1– –t t
t t
2 1
1 2
1 2
2 2
1 2 1 2
( )
( )
t tt t
t t
t t t t
∵ t
2 = –t
1
or t1 = –t
2 ...(iii) 4 2
1 11 1t t
t1 = ± 1 ...(iv)
From (iii) and (iv)
t1
= +1, t2
= –1 or t1 = –1, t
2 = 1
Points A and B are (1, 1) and (–1, –1)
2 2AB [AB] = 2
73. Answer (4)
Hint : Concept— If the normal at t1, again cuts the
parabola y2 = 4x at t2, then t
2 =
1
1
2– –t
t
Sol. : Camparing it with (y – m, x)(y – m, x) = 0 we
get
2
1
sin =
2 2
1 2
1 2
( ) tan cos
2tan
m m
mm
m1 + m
2 = 2
2tan 2
sin cossin
m1 + m
2 =
2 2 2 4
2 2 2
tan cos sin cos
sin sin (cos )
(m1 – m
2)2 = (m
1 + m
2)2 – 4m
1m2
2 2
4
sin cos =
2 2
2 2
4cos sin
cos sin
m1 – m
2 = ±2
tan1 – tan
2 = 2
74. Answer (2)
Hint : Solve equations of conics S1 and S
2, we find
values of x2 and y2 and non-substitute in the
equation of conic S3 = 0
Sol. : Solving conics (1) and (2), we get
2 2 2
2 1 1 1 1
2 2 2 2
1 1
( )a a a b a bx
a b a b
and
2 2 2
2 1 1 1 1
2 2 2 2
1 1
( – )b b a b a by
a b a b
Putting these values of x2 and y2 in the
equation of conic S3 = 0.
We get the requried result as in option (2).
75. Answer (4)
Hint : If points O(0, 0) and P(, 1) will lie on opposite
sides of line x + y = 2
(0 + 0 – 2) ( + 1 – 2) < 0 > 1 and
P(, 1) lies inside of circumcentre
< 2
Sol. :
O
(0, 0)(2, 0)
(0, 2)
Y
P( , 1)(0, 1)
(1, 1)(1 2,1)
(1 2,1)Y = 1
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
12/15
76. Answer (1)
Hint : Equation of circle touching the line x – 3y – 4 = 0
at the point (1, –1) is (x – 1)2 + (y + 1)2 +
(x – 3y – 4) = 0
Sol. : The equation of circle touching the line
x – 3y – 4 = 0 at the point (1, –1) is
y
O
x y – 3 – 4 = 0
(1, –1)
2 – 3 – 2,
2 2
y
(x – 1)2 + (y + 1)2 + (x – 3y – 4) = 0
x2 + y2 + x( – 2) + y(–3 + 2) + 2 – 4 = 0
whose centre is
C 2 – 3 – 2
,2 2
From figure, 2 –
2
> 0,
3 – 2
2
< 0 and
2 – 2 – 3
2 2
= 0
(If the circle touches both axes)
Circle is (x – 1)2 + (y + 1)2 = 0
Here, this circle does not touch the coordinate
axes.
Number of required circles is zero.
77. Answer (1)
Hint : If the line atx + 2ay + 1 = 0 touches the
parabola x2 = 4ay then at2x + 2atx
2
04
xt
a
will have equal roots and then D = 0
Sol. : For intersection point of the parabola
x2 = 4ay and the line atx + 2ay + 1 = 0,
Here equation (i) must have only one solution
for touching case
(, ) 1
– 2,2a
12 4 2 – 2 4 .
2a a
a
= –2 + 2 = 0
78. Answer (3)
Hint : (bsin, acos) lies on the ellipse
2 2
2 21
x y
a b ,
2 2 2 2
2 2
sin cos1
b a
a b
2 2 2 2
sin , cosa b
a b a b
Sol. : (bsin, acos) lies on the ellipse
2 2
2 21
x y
a b
2 2 2 2
2 2
sin cos1
b a
a b
2 2 2 2
sin , cosa b
a b a b
sin2 = 2sincos = 2 2 2
2
22
1
b
ab a
a b b
a
=
2 2
2 2
2 1– 2 1–
1 1– 2 –
e e
e e
2
2
21–
be
a
∵
79. Answer (2)
Hint :
P
A BD
C
CP = r3, CA = r
2, CD = r
1 = 2
2 1
3 2
sinr r
r r
2
2 1 3r r r
Sol. :
P
A BD
C(1, –2)
r2
r1
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
13/15
Here, CP = r3, CA = r
2,
CD = r1 = 2 2(–1) 2 – 3 = 2
From PAC, 2
3
sinr
r
and from ACD, 1
2
sinr
r
2 1
3 2
r r
r r
2
2 1 3r r r
where r2 = CA = 1 4 – ,
2 2
3(–1) 2 –1r = 2
(1 + 4 – ) = 2 2
= 5 – 2 2
80. Answer (1)
Hint :1 1 1 1
2 2
– – –2( )x y ax by c
a b a b
P x y( , )1 1
Q( , ) image of point
ax by c + + = 0
Sol.:
Let ( )M h, k
x y n + =
1,t
t
1
–(–1) –1
–
kt
h t and
1
2 2
kh t t
n
1– –t h kt ...(i)
12 – –t n h k
t ...(ii)
From (i) and (ii)
t = (n – k) and 1
t = (n – h) and
11t
t
(n – k)(n – h) = 1, locus of (h, k) is
(x – n)(y – n) = 1 whose centre is (n, n)
Sum of coordinates of
Ci = (1 + 1) + (2 + 2) + ... + (20 + 20)
= 20(21)
2 4202
81. Answer (1)
Hint : Equation of tangent at (acos, bsin) to the
ellipse
2 2
2 21
x y
a b is cos sin 1
x y
a b
Sol. : The equation of tangents at the points
(acos, bsin) and (asin, –bcos) to the given
ellipse are
cos sin 1x y
a b ...(i)
and sin – cos 1x y
a b ...(ii)
Let P(h, k) be the intersection point of tangents
(i) and (ii)
cos sinh k
a b = 1 ...(iii)
sin – cosh k
a b = 1 ...(iv)
Now squaring and adding (iii) and (iv), we get
2 2
2 22
h k
a b
locus of P(h, k) is
2 2
2 21
2 2
x y
a b
...(v)
Let e be the eccentricity of ellipse (v)
2b2 = 2a2(1 – e2)
2
2
21–
be
a
1 – e2 = 1 – e2
e = e.
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)
14/15
82. Answer (1)
Hint : G denotes OH in the ratio 1 : 2 intenally.
Centroid
O (Circumcentre)(3, 0)
6,
3 3
h kG
H h k ( , )orthocentre
Sol. : Centroid is
G 11 2sin 2cos 2sin 2cos
,3 3
Clearly circumcentre is O = (3, 0) so that
OA = OB = OC = 2. Let orthocentre be
H(h, k).
1
2
G
O
(3, 0)
H
h k( , )
6,
3 3
h k
G divides OH in the ratio 1 : 2.
G 1 2 1 2 0
,3 3
h b k
11 2sin 2cos 6
3 3
h ,
2sin 2cos
3 3
k
h – k = 5
locus of orthocentre H(h, k) is x – y = 5
(a straight line)
83. Answer (3)
Hint : Equation of required circle is s + 2 = 0
Sol. : Here the equation of the required circle is
x2 + y2 – 2x + 2y + 1 + (x + y – 1) = 0 ...(i)
Centre C 1– , – 1 –2 2
of circle (i) lies
on the line x + y = 1
1– –1– 12 2
= –1
Equation of circle is x2 + y2 – 3x + y + 2 = 0
84. Answer (2)
Hint : Tangents are y – 1 = ±x and y + 1 = ±x
Centre is C (0, 0)
Sol. : Tangents are
x2 – y2 + 2y – 1 = 0 and x2 – y2 – 2y – 1 = 0
x2 = y2 – 2y + 1 y + 1 = ±x
(y – 1)2 = x2
y – 1 = ±x
Equation of tangents are y = ±x + 1 and
y = ±x – 1
Radius is r = 2 1
2 2 , C (0, 0) centre
of circle
(∵ diameter of circle = 2 2
–1–12
1 1
= Distance between parallel tangents).
85. Answer (3)
Hint : Tangents at the ends of the minor axis of the
ellipse S1 = 0 are the chords of minimum
length of ellipse S2 = 0
Sol. : S1 =
2 2
2 2–1 0,
x y
a b
S2 =
2 2
2 2 2 2–1 0
4( )
x y
a b a b
Y
X
AB
X
Y
y b =
2 2
2 21
x y
a b
2 2
2 2 2 21
4( )
x y
a b a b
O
Tangents at the ends of the minor axis
of the ellipse
2 2
2 21
x y
a b are the
chords of maximum length of ellipse
2 2
2 2 2 21
4( )
x y
a b a b
Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
15/15
�����
Points A and B are intersection points of line
y = b and ellipse S2 = 0
A(2a, b) and B(–2a, b)
max.
= AB = 4a
86. Answer (2)
Hint : For common tangent, C1C
2 = |r
1 – r
2|
Sol. : For 1st circle
C1 (0, 0), r
1 = 1
A O
y
x(1, 0)
C2
(2, 0)
For 2nd circle,
C2 (2, 0), r
2 = 4 –
for exactly one common tangent
C1C
2 = |r
1 – r
2|
= –5
87. Answer (2)
Hint : Take images of points due to given line.
Sol. : Here, P(1, 1)
Q(1, 1) mirror image of P due to line y = x
1 7,
5 5R
17 31,
25 25S
lies on nx – y = 0 31
17n
88. Answer (2)
Hint : Make truth table
Sol. :
p q p q ~ p q( ) p q ~ p q( ) ~ p q ~p q( ) ( ) p p q ( ) ( ) p q p q p q ( )
T T T F T F T T T T
T F F T F T T F T T
F T T F F T F T F T
F F T F F T F T F T
89. Answer (2)
Hint : ~(p q) = p ~q
Sol. : ~((p q) (p q)) = (p q) (~(p q))
= (p q) (~p ~q)
90. Answer (3)
Hint : Make truth table
Sol. :
~ ~ ~ ~ ~ ~ ~ ~
T T T F F T T T F
T F F F T T T F T
F T T T F F T T T
F F T T T T F T T
p q p q p q p q p q q p q p
Test - 5 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/15
1. (3)
2. (3)
3. (4)
4. (2)
5. (1)
6. (2)
7. (3)
8. (4)
9. (3)
10. (2)
11. (3)
12. (3)
13. (3)
14. (3)
15. (2)
16. (2)
17. (2)
18. (4)
19. (4)
20. (3)
21. (2)
22. (1)
23. (3)
24. (4)
25. (4)
26. (3)
27. (2)
28. (3)
29. (3)
30. (2)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (1)
33. (1)
34. (3)
35. (2)
36. (1)
37. (2)
38. (3)
39. (2)
40. (4)
41. (2)
42. (1)
43. (3)
44. (2)
45. (3)
46. (3)
47. (3)
48. (4)
49. (2)
50. (2)
51. (4)
52. (4)
53. (4)
54. (3)
55. (1)
56. (2)
57. (3)
58. (3)
59. (2)
60. (1)
61. (3)
62. (2)
63. (2)
64. (2)
65. (2)
66. (3)
67. (2)
68. (3)
69. (1)
70. (1)
71. (1)
72. (2)
73. (3)
74. (1)
75. (1)
76. (4)
77. (2)
78. (4)
79. (3)
80. (3)
81. (2)
82. (2)
83. (3)
84. (4)
85. (3)
86. (1)
87. (3)
88. (3)
89. (4)
90. (3)
Test Date : 16/12/2018
ANSWERS
TEST - 5 - Code-B
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-B) (Hints & Solutions)
2/15
1. Answer (3)
Hint : Work done by external pressure = Pex
[Vi – V
f]
Sol. : Work done by external pressure
= 2P0[V
0 – V] =
0
3[ ]
2R
n T T [–W = U]
Also 0 0 0 0
0 0
2 2PV PV VTT
T T V
P0V
0 = nRT
0
4[V0 – V] = 3[2V – V
0]
7V0 = 10V V =
07
10
V
2. Answer (3)
Hint : For maximum, path difference = n
In a quadrant path difference varies continuously
from 3 to 0 4.
Sol. : For maximum, path difference = n
In a quadrant path difference varies continuously
from 3 to 0 4.
3. Answer (4)
Hint : For maximum intensity, path difference = n
Sol. : Path difference = 2 × 8.5 cm = 17 cm
4. Answer (2)
Hint : 1 dV
V dT
Sol. :1 dV
V dT , PT2 = Constant
T3 = kV
2
2 33
dV T dVT k
dT k dT
2
3
3 1T dVk
V dTkT
5. Answer (1)
Hint : n1CV
1 + n
2CV
2 = (n
1 + n
2)CV
PART - A (PHYSICS)
Sol. :3 3 5
2 2
R R = 4C
v
Cv =
18 9
2 4 4
R R
= 13
9
6. Answer (2)
Hint : For process PVx = constant
[ ]
1
i fnR T T
Wx
Sol. : PT = constant P2V = constant
PV1/2 = constant
W = [ ]
11
2
i fnR T T
= [600 300]
1
2
R = –600R
7. Answer (3)
Hint : For polytropic process, PVx = constant
C = 1
v
RC
x
Sol. : C = 5 2 1
13
R R =
3
2 2 R RR
Q = 3
,2 2 R R
n T U n T
W = –nRT
8. Answer (4)
Hint : Work done by a gas in a cyclic process is
negative if P-V graph is in anticlockwise
sequence.
Sol. : Wby gas
=1
– 1 40 –20 J2
9. Answer (3)
Hint : Power, P = Area under E versus
Sol. : Power, P = Area under E versus
4
1/41 11 2 2
2 2
4 4 2
P TT T T
P T
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
3/15
10. Answer (2)
Hint : g(x, t) = f((x – v(t – t0)), t)
Sol. : g(x, t) = f((x – v(t – t0)), t)
11. Answer (3)
Hint : fstring
= 2 µ
n T
L
fc =
4
V
L
Sol. : f0 =
4
V
L =
320100 Hz
4 0.8
100 = 2
50 cm2
TLL
L m
12. Answer (3)
Hint : 2R = R
R3R
Sol. : 2R = R
= 2
mg sinR =
22
2
7
5
dmRdt
=
22
2
14
5
dmRdt
2
2
d
dt =
5
14
g
R
= 5 14
, 214 5
g RT
R g
13. Answer (3)
Hint :
2
2
d
dt
= –2 for angular SHM
Sol. : = 6 6 3 3 L L L L
k k
2 2
2
2
2
3
mL dmL
dt = –
2 1 1
36 9
kL
2 2
2
5
3
mL d
dt =
25
36 kL
2
2
d
dt =
12 k
m
= 12
k
m
f = 1
2 12k
m =
1
4 3k
m
14. Answer (3)
Hint : E = U + K = 21
2kA
Sol. : v =
2 2
2 3
4 4 A A
A
v = 2v = 22
32
A
E = U + K =
2 2
21 1 34
2 2 2 4
A Ak m
=
221 1
32 4 2
Ak k A
2 2113
2 2 A
kA A
15. Answer (2)
Hint : dv
dt = av
Sol. : av =
dv
dt
1 × 10–4 × 2gh =
330 10
60
h = 1.25 m
16. Answer (2)
Hint : W = 2( ) a x g dx
Sol. : F = (a2x)g
W =
0.1
2
0
( ) a
a x g dx
=
4
200
a g
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-B) (Hints & Solutions)
4/15
17. Answer (2)
Hint : Use equation of ellipse.
a = 1 2
2
r r
Sol. :E
S
r1r2
2 2
2 2x y
a b = 1
|x| = ae = a – r2,
a = 1 2
2
r r
y2 =
2
2
21
xb
a
y = 1 2
1 2
2
( )r r
r r
18. Answer (4)
Hint : T = Vg, � = �T
YA
Sol. : T = × Vg = 34
3 R g
� =
34
3
� �T R g
YA YA
19. Answer (4)
Hint : Force = Pav
× Projection area.
Sol. : Fx= P
av × Area
= ( )2
�gRR
=
2
2
�g R
20. Answer (3)
Hint : = p
B
Sol. : p = gh
= pB
� 2 kg/m3
21. Answer (2)
Hint : R = time of fall × speed of efflux then 0dR
dh
Sol. : hreq
= 2
2
HH
= 3
4
H
height from bottom of vessel = 3
4 2 4 H H H
22. Answer (1)
Hint : Loss in PE = gain in KE
Sol. :21
2mv =
3 2 GMm GMm
R R
v = 3
GM
R
23. Answer (3)
Hint : dA
dt =
2
L ab
m T
Sol. : Areal speed, vA =
abT
Area(A B)
= 1
22 2
abb ae
t = ( )Area 1
12
A B
A
T
v
24. Answer (4)
Hint : Torque = ( ) p dA r
Sol. : d = 45
( )3
g H x adx x ga
25. Answer (4)
Hint : Imagine hemisphere.
Sol. : Assume upper hemisphere
B = 32
3 R g
F2 = R2 × g(2R)
F1 = F
2 – B =
34
3 R g
26. Answer (3)
Hint : dr
dt =
2gR
r
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/15
31. Answer (3)
Hint : AB(g) A(g) B(g)
1 p p p' p p ''
���⇀↽���
A(g) C(g) D(g)
p p' 1 p' p ' p ''
���⇀↽���
D B E
p' p '' p p '' p ''
���⇀↽���
Given [A][B] p p
0.15[AB] 1 p
Sol. : Given p = 0.2
1 – p = p – p
1 p''
2
= p, p = 0.6 atm
PART - B (CHEMISTRY)
From equilibrium of reaction I
(0.6 p')(0.4)
0.4
=
15
100
p = 0.45
pAB
= 0.4 atm
pA = 0.15 atm
pB = 0.4 atm
pC = 0.55
pD = 0.25
pE = 0.2
32. Answer (1)
Hint : Correct order is a > b > c
Sol. : As s character increases bond length decreases
Sol. : u = 2gR
21
2mv –
GMm
r =
21
2GMm
muR
v = 2g
Rr
dr
dt =
2gR
r
4
R
R
r dr =
0
2 t
R g dt
t = 7 2
3
R
g
27. Answer (2)
Hint : p = T
r for cylindrical surface
Sol. : p = 2T T
r d
A = V m
d d
F = (p) × A = 2
2
Tm
d
28. Answer (3)
Hint : Net force on particle = Centripetal force
Sol. :
2
3
mv
a
=
2
23Gm
a
v = Gm
a
T =
32
23
r a
v Gm
29. Answer (3)
Hint : Two surfaces are formed in the capillary
Sol. : Since force due to surface tension becomes twice
h = 2h
30. Answer (2)
Hint : Vc =
2
0
14
RG
r drr r
Sol. : M =
22
0
14 4
2
RR
r drr
Vc =
2
0
1 2(4 )
RG GM
r drr r R
All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-B) (Hints & Solutions)
6/15
33. Answer (1)
Hint : [H+] = 1 2
a 1 a 2K C K C
Sol. : Case of simultaneous equilibrium
4O
4O
HA A H , 1.7 10
(0.1 x) x (x y)
HB B H , 4.7 10
(0.1 y) y (y x)
���⇀↽���
���⇀↽���
x(x + y) = 1.7 × 10–5
y(x + y) = 4.7 × 10–5
(x + y)2 = 6.4 × 10–5
(x + y)2 = 64 × 10–6
x + y = 8 × 10–3
[H+] = 8 × 10–3
34. Answer (3)
Hint : AI
Sol. : AI has highest value of oxidation potential
35. Answer (2)
Hint : E2 elimination
Sol. : Anti H is removed
36. Answer (1)
Hint : As for polluted water B.O.D is higher.
Sol. : B.O.D: A < C < B < D
Pollution level in water A < C < B < D
37. Answer (2)
Hint : Nitro benzene is formed which is m-directing
Sol. : 3Zn, CH OH
2 2 2 2 2Cl CH CHCl C H 2ZnCl
CH2 2
Red hot
Cu
HNO3
H SO2 4
NO2
Br /Fe2
Br
NO2
38. Answer (3)
Hint : Conc. of solid is constant
Sol. : Addition of inert at constant pressure move
towards more no. of gaseous mole.
39. Answer (2)
Hint : M + H2SO
4 Ma+ + H
2
(8.96 L)
Ma+ + KMnO4 Mb+ + Mn2+
Sol. : Reaction (1)
g meq of H2 = g meq of metal
8.962
22.4
= 0.4 × (x)
0.8 = 0.4 × x
x = 2 a 2
Reaction (2)
eq of KMnO4 = eq of metal
40 × 2 × 5 × 10–3 = 0.4 × x
40 × 10 × 10–3 = 0.4x
x = 1
b – a = 1
b = 3 a + b = 5
40. Answer (4)
Hint : Zn + 4HNO3 4Zn(NO
3)2 + 2NO
2 + 2H
2O
Sol. : The H+ required for the conversion will be
provided by HNO3
Zn + 2NO + 4H Zn +2NO + 2H O3 2 2
– + 2++ 2NO3
–
+ 2NO3
–
Zn + 4HNO Zn(NO ) +2NO + 2H O3 3 2 2 2
(0.2)
0.8 moles of HNO3 will be consumed
1.6 litres of 0.5 M HNO3 is required
41. Answer (2)
Hint : Let m moles of Na2CO
3 be x 2x meq
and NaOH be y y meq
Sol. : For phenolphthalein end point
2xy
2 = 30 × 0.5
x + y = 15
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/15
For methyl orange end point
2x
2 = 10 × 0.5
x = 5
y = 10
Weight of Na2CO
3 = 5 × 10–3 × 106 = 0.53 g
Weight of NaOH = 10 × 10–3 × 40 = 0.4 g
Impurity = 0.07 g (7% impurity is present)
42. Answer (1)
Hint : 2Cr+6 2Cr+3
+6e–
2N3– N2 + 6e–
Sol. : So a total of 6e– are involved in the above
intramolecular redox
n factor = 6
43. Answer (3)
Hint : In F – CH2 – CH
2 – OH, gauche form is more
stable than anti.
Sol. :H
H
F
H
H
O
H
Most stable
44. Answer (2)
Hint : A is 3° Alcohol
Sol. :
H+
Ring expansion
(+)
Ring expansion
(+)
H O2
OH
+
45. Answer (3)
Hint : 2 mol
H H
O
O O
2 mol
H
O
H
Means P must contain 8 carbon
Sol. :
H
O
OH
O
H
(I) O 3
(II) Zn/H O2
+
1 mol
2 mol
H
O
2 mol
46. Answer (3)
Hint : I effect act on O > M > P while hyperconjugation
effect act on O and P with equal intensity
Sol. : +I power CD3 > CH
3
+H power CH3 > CD
3
Stability order
A > D > C > B
47. Answer (3)
Hint : Lassaigne's test for nitrogen
Sol. : Fe2+ + CN– [Fe(CN)6]4–
Blue coloured compound is prussian blue
48. Answer (4)
Hint : H2O
2 is miscible with water
Sol. :3 2 2 2
Mg(HCO ) Mg(OH) 2CO
3 2 3 2 2Ca(HCO ) CaCO H O CO (g)
49. Answer (2)
Hint : f
Distance moved by the substance
from base lineR
Distance moved by the solvent from
base line
Sol. :4
0.410
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50. Answer (2)
Hint : R3SiCl
Sol. : R3SiCl terminate the chain
51. Answer (4)
Hint : Buckminster fullerene is aromatic in nature.
Sol. : Stability: Graphite > Diamond > C60
52. Answer (4)
Hint : Al2Cl
6 has 3c – 4e– as well as 2c – 2e– bond.
Sol. :
Cl
Cl
Cl
Cl
Cl
Cl
Al Al
53. Answer (4)
Hint : A = B3N
3H
3Cl
Y = B3N
3H
6
Y + 3HCl = B3N
3H
9Cl
3
Sol. :
YHCl
B N
B N
BN
—
—
Cl
H
H
H HH
H
H
HH
Cl
Cl
+
+
+
—
54. Answer (3)
Hint : For ppt
Kip > K
sp
Sol. : Experiment I :
For saturated solution of AgCl
[Ag+] = [Cl–] = 10–5 M
If 10–3 mol NaBr is added it means,
[Br–] = 10–3 M.
(Kip)AgBr
= 10–5 × 10–3 = 10–8 > 10–14
(Ksp
of AgBr)
So AgBr form ppt.
Experiment II :
For saturated solution of AgBr
[Ag+] = [Br–] = 10–7
If 10–5 mol of NaCl is added in 100 mL solution so
[Cl–] = 10–4
Kip of AgCl = 10–7 × 10–4 = 10–11
Kip < K
sp so no ppt
55. Answer (1)
Hint : Mg2+ has 6 or 8 molecule of water
Sol. : Ba(NO3)28H
2O does not exist.
56. Answer (2)
Hint : Sulphate of 2nd group are white
Sol. : Solubility 2nd group sulphate or carbonate
decrease as atomic number increase
57. Answer (3)
Hint : Change in pH of buffer is minimum
Sol. :
CH3COO– + H+ CH
3COOH
t = 0 1 0.1 1
t = eq 0.9 — 1.1
(pH)f = pK
a + lo g
0.9
1.1
(pH)i = pK
a + log
1
1
pH decreases
As concentration of buffer increase buffer
capacity increase
NaOH + H+ H2O
0.1 0.1
Salt is formed, pH = 7 and change in pH is
maximum
58. Answer (3)
Hint : Concentrated solution is diamagnetic.
Sol. : Alkali metal dissolve in ammonia.
59. Answer (2)
Hint : K form superoxide.
Sol. : KO2, Na
2O
2, Li
2O
60. Answer (1)
Hint : 3 2 2 2LiNO Li O NO + O
Sol. : Lithium cannot form alkynide on reaction with
ethyne
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PART - C (MATHEMATICS)
61. Answer (3)
Hint : Make truth table
Sol. :
~ ~ ~ ~ ~ ~ ~ ~
T T T F F T T T F
T F F F T T T F T
F T T T F F T T T
F F T T T T F T T
p q p q p q p q p q q p q p
62. Answer (2)
Hint : ~(p q) = p ~q
Sol. : ~((p q) (p q)) = (p q) (~(p q))
= (p q) (~p ~q)
63. Answer (2)
Hint : Make truth table
Sol. :
p q p q ~ p q( ) p q ~ p q( ) ~ p q ~p q( ) ( ) p p q ( ) ( ) p q p q p q ( )
T T T F T F T T T T
T F F T F T T F T T
F T T F F T F T F T
F F T F F T F T F T
64. Answer (2)
Hint : Take images of points due to given line.
Sol. : Here, P(1, 1)
Q(1, 1) mirror image of P due to line y = x
1 7,
5 5R
17 31,
25 25S
lies on nx – y = 0 31
17n
65. Answer (2)
Hint : For common tangent, C1C
2 = |r
1 – r
2|
Sol. : For 1st circle
C1 (0, 0), r
1 = 1
A O
y
x(1, 0)
C2 (2, 0)
For 2nd circle,
C2 (2, 0), r
2 = 4 –
for exactly one common tangent
C1C
2 = |r
1 – r
2|
= –5
66 .Answer (3)
Hint : Tangents at the ends of the minor axis of the
ellipse S1 = 0 are the chords of minimum
length of ellipse S2 = 0
Sol. : S1 =
2 2
2 2–1 0,
x y
a b
S2 =
2 2
2 2 2 2–1 0
4( )
x y
a b a b
Y
X
AB
X
Y
y b =
2 2
2 21
x y
a b
2 2
2 2 2 21
4( )
x y
a b a b
O
Tangents at the ends of the minor axis of the
ellipse
2 2
2 21
x y
a b are the chords
of maximum length of ellipse
2 2
2 2 2 21
4( )
x y
a b a b
Points A and B are intersection points of line
y = b and ellipse S2 = 0
A(2a, b) and B(–2a, b)
max.
= AB = 4a
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67 Answer (2)
Hint : Tangents are y – 1 = ±x and y + 1 = ±x
Centre is C (0, 0)
Sol. : Tangents are
x2 – y2 + 2y – 1 = 0 and x2 – y2 – 2y – 1 = 0
x2 = y2 – 2y + 1 y + 1 = ±x
(y – 1)2 = x2
y – 1 = ±x
Equation of tangents are y = ±x + 1 and
y = ±x – 1
Radius is r = 2 1
2 2
, C (0, 0) centre
of circle
(∵ diameter of circle = 2 2
–1–12
1 1
= Distance between parallel tangents).
68. Answer (3)
Hint : Equation of required circle is s + 2 = 0
Sol. : Here the equation of the required circle is
x2 + y2 – 2x + 2y + 1 + (x + y – 1) = 0 ...(i)
Centre C 1– , – 1 –2 2
of circle (i) lies
on the line x + y = 1
1– –1– 12 2
= –1
Equation of circle is x2 + y2 – 3x + y + 2 = 0
69. Answer (1)
Hint : G denotes OH in the ratio 1 : 2 intenally.
Centroid
O (Circumcentre)(3, 0)
6,
3 3
h kG
H h k ( , )orthocentre
Sol. : Centroid is
G 11 2sin 2cos 2sin 2cos
,3 3
Clearly circumcentre is O = (3, 0) so that
OA = OB = OC = 2. Let orthocentre be
H(h, k).
1
2
G
O
(3, 0)
H
h k( , )
6,
3 3
h k
G divides OH in the ratio 1 : 2.
G 1 2 1 2 0
,3 3
h b k
11 2sin 2cos 6
3 3
h ,
2sin 2cos
3 3
k
h – k = 5
locus of orthocentre H(h, k) is x – y = 5
(a straight line)
70. Answer (1)
Hint : Equation of tangent at (acos, bsin) to the
ellipse
2 2
2 21
x y
a b is cos sin 1
x y
a b
Sol. : The equation of tangents at the points
(acos, bsin) and (asin, –bcos) to the given
ellipse are
cos sin 1x y
a b ...(i)
and sin – cos 1x y
a b ...(ii)
Let P(h, k) be the intersection point of tangents
(i) and (ii)
cos sinh k
a b = 1 ...(iii)
sin – cosh k
a b = 1 ...(iv)
Now squaring and adding (iii) and (iv), we get
2 2
2 22
h k
a b
locus of P(h, k) is
2 2
2 21
2 2
x y
a b ...(v)
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
11/15
Let e be the eccentricity of ellipse (v)
2b2 = 2a2(1 – e2)
2
2
21–
be
a
1 – e2 = 1 – e2
e = e.
71. Answer (1)
Hint :1 1 1 1
2 2
– – –2( )x y ax by c
a b a b
P x y( , )1 1
Q( , ) image of point
ax by c + + = 0
Sol.:
Let ( )M h, k
x y n + =
1,t
t
1
–(–1) –1
–
kt
h t and
1
2 2
kh t t
n
1– –t h kt ...(i)
12 – –t n h k
t ...(ii)
From (i) and (ii)
t = (n – k) and 1
t = (n – h) and
11t
t
(n – k)(n – h) = 1, locus of (h, k) is
(x – n)(y – n) = 1 whose centre is (n, n)
Sum of coordinates of
Ci = (1 + 1) + (2 + 2) + ... + (20 + 20)
= 20(21)
2 4202
72. Answer (2)
Hint :
P
A BD
C
CP = r3, CA = r
2, CD = r
1 = 2
2 1
3 2
sinr r
r r
2
2 1 3r r r
Sol. :
P
A BD
C(1, –2)
r2
r1
Here, CP = r3, CA = r
2,
CD = r1 = 2 2(–1) 2 – 3 = 2
From PAC, 2
3
sinr
r
and from ACD, 1
2
sinr
r
2 1
3 2
r r
r r
2
2 1 3r r r
where r2 = CA = 1 4 – ,
2 2
3(–1) 2 –1r = 2
(1 + 4 – ) = 2 2
= 5 – 2 2
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73. Answer (3)
Hint : (bsin, acos) lies on the ellipse
2 2
2 21
x y
a b ,
2 2 2 2
2 2
sin cos1
b a
a b
2 2 2 2
sin , cosa b
a b a b
Sol. : (bsin, acos) lies on the ellipse
2 2
2 21
x y
a b
2 2 2 2
2 2
sin cos1
b a
a b
2 2 2 2
sin , cosa b
a b a b
sin2 = 2sincos = 2 2 2
2
22
1
b
ab a
a b b
a
=
2 2
2 2
2 1– 2 1–
1 1– 2 –
e e
e e
2
2
21–
be
a
∵
74. Answer (1)
Hint : If the line atx + 2ay + 1 = 0 touches the
parabola x2 = 4ay then at2x + 2atx
2
04
xt
a
will have equal roots and then D = 0
Sol. : For intersection point of the parabola
x2 = 4ay and the line atx + 2ay + 1 = 0,
Here equation (i) must have only one solution
for touching case
(, ) 1
– 2,2a
12 4 2 – 2 4 .
2a a
a
= –2 + 2 = 0
75. Answer (1)
Hint : Equation of circle touching the line x – 3y – 4 = 0
at the point (1, –1) is (x – 1)2 + (y + 1)2 +
(x – 3y – 4) = 0
Sol. : The equation of circle touching the line
x – 3y – 4 = 0 at the point (1, –1) is
y
O
x y – 3 – 4 = 0
(1, –1)
2 – 3 – 2,
2 2
y
(x – 1)2 + (y + 1)2 + (x – 3y – 4) = 0
x2 + y2 + x( – 2) + y(–3 + 2) + 2 – 4 = 0
whose centre is
C 2 – 3 – 2
,2 2
From figure, 2 –
2
> 0,
3 – 2
2
< 0 and
2 – 2 – 3
2 2
= 0
(If the circle touches both axes)
Circle is (x – 1)2 + (y + 1)2 = 0
Here, this circle does not touch the coordinate
axes.
Number of required circles is zero.
76. Answer (4)
Hint : If points O(0, 0) and P(, 1) will lie on opposite
sides of line x + y = 2
(0 + 0 – 2) ( + 1 – 2) < 0 > 1 and
P(, 1) lies inside of circumcentre
< 2
Sol. :
O
(0, 0)(2, 0)
(0, 2)
Y
P( , 1)(0, 1)
(1, 1)(1 2,1)
(1 2,1)Y = 1
77. Answer (2)
Hint : Solve equations of conics S1 and S
2, we find
values of x2 and y2 and non-substitute in the
equation of conic S3 = 0
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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Sol. : Solving conics (1) and (2), we get
2 2 2
2 1 1 1 1
2 2 2 2
1 1
( )a a a b a bx
a b a b
and
2 2 2
2 1 1 1 1
2 2 2 2
1 1
( – )b b a b a by
a b a b
Putting these values of x2 and y2 in the
equation of conic S3 = 0.
We get the requried result as in option (2).
78. Answer (4)
Hint : Concept— If the normal at t1, again cuts the
parabola y2 = 4x at t2, then t
2 =
1
1
2– –t
t
Sol. : Camparing it with (y – m, x)(y – m, x) = 0 we
get
2
1
sin =
2 2
1 2
1 2
( ) tan cos
2tan
m m
mm
m1 + m
2 = 2
2tan 2
sin cossin
m1 + m
2 =
2 2 2 4
2 2 2
tan cos sin cos
sin sin (cos )
(m1 – m
2)2 = (m
1 + m
2)2 – 4m
1m2
2 2
4
sin cos =
2 2
2 2
4cos sin
cos sin
m1 – m
2 = ±2
tan1 – tan
2 = 2
79. Answer (3)
Hint : Equation of normal at 1
,A tt
to the hyperbola
xy = 1.
is 1
–yt
= t2(x – t)
Sol. : Equations of normals at 1
1
1,A tt
and
2
2
1,B tt
to the hyperbola xy = 1 are
2
1 1
1
1– – ,y t x tt ...(i)
and 2
2 2
2
1– –y t x tt
...(ii) respectively.
(i) and (ii) will represent the same normal if
2 2
1 2t t and
3 3
1 2
1 2
1 1– –t t
t t
t1 = t
2 (Rejected) 3 3
1 2
1 2
1 1– –t t
t t
2 1
1 2
1 2
2 2
1 2 1 2
( )
( )
t tt t
t t
t t t t
∵ t
2 = –t
1
or t1 = –t
2 ...(iii) 4 2
1 11 1t t
t1 = ± 1 ...(iv)
From (iii) and (iv)
t1
= +1, t2
= –1 or t1 = –1, t
2 = 1
Points A and B are (1, 1) and (–1, –1)
2 2AB [AB] = 2
80. Answer (3)
Hint : Use property-vertex of parabola bisects the
subtangent and median through the point
passes through the vertex.
Sol. :
V(1, 2)
P(2, 1)
x
y
NT
∵ Vertex bisects the subtangent TN
Median through the point passes through
the vertex V(1, 2)
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Equation of median of PTN, through P(2, 1)
is y – 1 = 1– 2
( – 2)2 –1
x x + y = 3
81. Answer (2)
Hint : Lines are y = x + 3; y = 1 – x, x = 0, y = 0
Sol. : Let the equation of tangent be
y = 25 5
3 2mx m
2t passes through (1, 2)
2 = 25 5
3 2m m
4m2 + 24m – 9 = 0
tan = 12
5
82. Answer (2)
Hint : Speed = distance
time
Sol. :30°
45°
x
xy
1200 m
A B
P
1200tan45
y y = 1200 ...(1)
1200tan30
x y
1200 1
3x y
1200 3 = x + y
x = 1200 3 – 1200 x = 1200( 3 – 1)
Speed = 1200( 3 – 1)
5 = 240 3 –1 m/s
83. Answer (3)
Hint :
2 2 2–
cos2
a c bB
ac
Sol. :
2 2 2 2 2 2– cos –
cos ,2 2
a c b b B bB
ac ac
=
2 2(cos –1)0
2
b B
ac
cosB < 0 2
B
84. Answer (4)
Hint : 2sinA sinB = cos(A – B) – cos(A + B)
Sol. :–
4 sin .sin2 2
A C A CR
= 2R(cosC – cosA)
= 2 2 2 2
2 –a c
R
a c a c
= a – c
85. Answer (3)
Hint :
–1
–1
sin2
cos2 .cos2
r
r r rT
Sol. :
–1
–1
1
sin2
cos2 .cos2
rn
r r
r
=
–1
–1
1
sin(2 – 2 )
cos2 .cos2
r rn
r r
r
= –1
1
(tan2 – tan2 )n
r r
r
= (tan2n – tan)
86. Answer (1)
Hint : In a triangle
cot cot cot2 2 2
A B C = cot .cot .cot2 2 2
A B C
Sol. : In a triangle we know that
cot cot cot2 2 2
A B C = cot cot cot
2 2 2
A B C
cot cot cot 3cot2 2 2 2
A B C B
cot cot 32 2
A C
Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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�����
cot cot 2 cot cot
2 2 2 2
A C A B
2cot 2 32
B cot 32
B
1tan
2 3B
87. Answer (3)
Hint : In ax2 + bx + c = 0
sum of roots = –b
a
Product of roots = c
a
Sol. : cosx1 + cosx
2 = –a and sinx
1 + sinx
2 = –c
1 2 1 2–
2cos cos –2 2
x x x xa
and1 2 1 2
–2sin cos –
2 2
x x x xc
1 2tan
2
x x c
a
sec(x1 + x
2) =
2
2 22
2 2 2
2
1
–1–
c
a ca
c a c
a
88. Answer (3)
Hint : Max. value of sinx + cosx is 2 at x = 4
Sol. : 5 + 7sin2x + 3 3
sin cos2 22 2
x x
Max. value of 3sin 4cos2 2
x x is 3 at 2
x
Max. value of given expression = 15
89. Answer (4)
Hint : Max. value of sinnx = 1
Sol. : sinx + sin4x = 2
sinx = 1 and sin4x = 1
(4 1)2
x n and 4 (4 1)
2x m
(4 1)2
x n and (4 1)
8x m
There is no common solution.
90. Answer (3)
Hint :1
loglog
a
b
ba
Sol. :cos
cos
3log sin 4
log sin
logcossin = 1 or log
cossin = 3
There are two solutions.