Post on 25-Jan-2020
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
1/14
1. (2)
2. (1)
3. (2)
4. (3)
5. (3)
6. (5)
7. (7)
8. (6)
9. (A, D)
10. (A, C)
11. (A, C)
12. (A, B, C)
13. (A, C)
14. (B, D)
15. (B, C)
16. (C, D)
17. (A, B, D)
18. (A, B, D)
19. (A, C)
20. (A, D)
21. (3)
22. (6)
23. (6)
24. (4)
25. (4)
26. (2)
27. (6)
28. (7)
29. (A, C)
30. (A, B, D)
31. (A)
32. (A, C)
33. (B, D)
34. (A, B, D)
35. (A, C, D)
36. (A, B, C)
37. (D)
38. (D)
39. (A, D)
40. (A, B)
41. (7)
42. (4)
43. (3)
44. (8)
45. (1)
46. (5)
47. (2)
48. (6)
49. (A, B, C)
50. (A, B)
51. (A, B, C, D)
52. (A, D)
53. (A, C, D)
54. (C, D)
55. (A, C, D)
56. (A, B, C, D)
57. (A, B, D)
58. (C)
59. (A, D)
60. (B)
ANSWERS
Test Date: 17/02/2019
PHYSICS CHEMISTRY MATHEMATICS
All India Aakash Test Series for JEE (Advanced)-2019
MOCK TEST - 1 (Paper-2) - Code-C
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
2/14
PART - I (PHYSICS)
ANSWERS & HINTS
1. Answer (2)
Hint : O B A BT T
1
2
Solution :
mgx = fR mg
f xR
g ga x
R R
R T R
T tg g
2 ,4 2
2. Answer (1)
Hint : E A2 2
0
1
4 2
Solution :
x1
7cm
8 x
2
17cm
8
2 212.5
4 2U E A
2 2
2
12.5
4 2
TU E A
v
2 2
2
5 1 1 4 100100 4
2 4 2 100 100U E
25U E 2
15
U
3. Answer (2)
L
H
Hint : 2dpx
dx
Solution :
22
LAL ghA
Lh
g
2 2
2
4. Answer (3)
Hint : dP
B VdV
–
Solution :
3 24, 4
3V a dV a da
–
–/
dP dPB V
dV V dV
MgdP
A
3
2
4–
3 4
a MgB
A a da
da Mg
a BA–3
5. Answer (3)
Hint : pr v
RC C
n–
–1
Solution :
T2 P3 = C
2
3 5 2
PVP C P V C
nR
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
3/14
2/5pV C
3 19–
22 6–1
5
pr
R R RC
Q = n Cpr T R
Q19
2 1006
RQ
1900
3
32 100
2
Ru
u = 300 R
1900 1000– – 300
3 3
R RW Q u R
6. Answer (5)
Hint : hc
Emin
max
Solution :
1 hC
ev
If V increase 1 decreases
7. Answer (7)
Hint : Length of string will remain constant
Solution :
V0 cos 37 = V
0
4
5 V V
VV0
5
4
8. Answer (6)
Hint :
Fmax
= 2 mAg
Solution :
Fmax
= 2 mAg
9. Answer (A, D)
Hint :
Apply Ampere’s law
Solution :
Amplere’s law
0 enclosed. .B dl I
�� ���
�
0–
ˆ –2
��
B dxi
10. Answer (A, C)
Hint : Ea
02
Solution :
( , )a a
( )p
3
1
x
y2
31 2
0 0 0
1 1
2 . 2 . 2 . 2 2 2pE j i i j
a a a
��� � � �
3 3
2 1
0 0
1 1
2 . 2 2 . 2
��� �
pE i ja a
11. Answer (A, C)
Hint : Apply energy conservation and momentum
conservation
Solution :
T
2ma0
Ring
T cos 37° = ma
For 2 m
T cos 37° + 2 ma0 = (2 m) g sin 37°
We get the result
12. Answer (A, B, C)
Hint : Use R – C circuit analysis.
Solution :
K Q KQV
d a
1–
0 0
1
QaQ
d
t
RcQ Q e–
1 t = Rc ln3 = 4
0 .aR ln3
1
3
2 2 2
22 200
8 .2 4 .
VQ a Q a
dd d
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
4/14
13. Answer (A, C)
Hint : Apply I ��
Solution :
(2 2mR + mg2
)N
fO
0 = I
0
RmR mg mR mR2 2 2
2 2 2 22
2
4
g R
R
14. Answer (B, D)
Hint : ax
VI
Z
max
m
Solution :
15
I2
60
30 v
I1
I
I t1
100sin 50
2 3
I t2
100sin 50 –
2 6
I t50 2 sin 5012
Pam
= Vrm
Irm
cos
100 5 2cos15
2 2
15. Answer (B, C)
Hint : Apply superposition principle
Solution :
00
1
4
R
C
x dxV
x
0
4
R
02
P
RV
16. Answer (C, D)
Hint : After reflection from mirror there will be phase
change of Solution :
D 2D
y
O
Screen
d
D
y
P
Q
OP dOP d
D D2 2
d OROR d
D D3
3
y = 2d = 1 mm
also = yd
nD
1–
2 2
or y d n 5.5, n 15.5
17. Answer (A, B, D )
18. Answer (A, B, D)
Hint : f Nmax
1 1 1
Solution for Q. Nos. 17 and 18
4 kg
2 kg
1 kg f1
f2
f3
14 N
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
5/14
ca
14 – 7
7
= 1 m/s2
max
37 Nf
max
12 Nf
max
23 Nf
max
1a = 2 m/s2
max
2a = 1 m/s2
19. Answer (A, C)
20. Answer (A, D)
Hint :
Rt
LI I e–
max1–
in R - L circuit
Solution for Q. Nos. 19 and Q. Nos. 20
1
2
1 V
1
2
I
4H
3
Circuit can be reduced as
t
LI I e
R–
max1–
4R
3
4 2 4
4 2 3
R
t
I e
–4
31 4
1–4 3
Imax
1A
4
I = 0.25 (1 – e–t) A
also L1I1 = L
2I2
I t I I t I1 2
2 1,
3 3
tI t e–
1
11–
6 , tI t e
–
2
11–
12
Total
1 4 1 1J
2 3 16 24
U
PART - II (CHEMISTRY)
21. Answer (3)
Hint: Kp remain constant for a reaction, even in
simultaneous equilibria.
Solution :
a 2b a
PQ(s) P(g) Q(g)
���⇀↽��� K
p = 4.8 × 10–6
2
2b a
RQ (s) R(s) 2Q(g)
���⇀↽��� K
p = 5.76 × 10–6
(2b + a)2 = 5.76 × 10–6
(2b + a) = 2.4 × 10–3
(2.4 × 10–3) × a = 4.8 × 10–6
a = 2 × 10–3 = 2 × 10–x
x = 3
22. Answer (6)
Hint: To obtain aromatic character R can lose a
hydride H–, and become sp2 hybridised.
Solution :
C C
H
+
+
H
+ C(Ph)3
+ H – C(Ph)3
(R)
–
BF4
–
BF4H
a a
Number of electrons in ring R is 6.
23. Answer (6)
Hint: First convert molecules tetrahedral geometry into
Fisher form.
Solution :
(A) CH3
H
OH
2
4
3
1
(R)
(1)C H2 5
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
6/14
(B)
HH
Cl
24
3
1
HO C H2 5
(R)
(1)
(C)
HH
Cl2
4
3
Br (S)(2)
CH3
1
(D)
HH
C H2 5
Br
4
3
2
(S)(2)
CH3
1
24. Answer (4)
Hint: Check the chirality and elements of symmetry in
product.
Solution :
P1
Optically active
P2
Optically active
P3
Optically active
P4
Optically active
P5
Optically inactive
P6
Optically inactive
P7
Optically inactive
P8
Optically inactive
25. Answer (4)
Hint :
Number of moles CaC2O
4 = Number of moles of CaCI
2
= Number of moles of organic acid
Solution :
(Oxalic acid) N1V
1 = N
2V
2 (KMnO
4)
2 × no. of moles = 80
5 0.051000
= 0.01 moles
Number of moles Oxalic acid = number of moles
of CaC2O
4 = number of moles of CaCI
2
Mass of CaCI2 = 0.01 × 111 = 1.11 g
% of CaCI2 in mixture =
1.11100 60%
1.8
26. Answer (2)
Hint :
2
2Cd H S CdS 2H
���⇀↽���
Solution :
0.5 × 10–3 moles of Cd2+, 0.5 × 10–3 moles of S–2 ions
are required to precipitate.
0.5 ×10–3 moles of S2– 0.5 × 10–3 moles H2S
1×10–3 moles of H+
Number of moles of H+ from HCl = 50 × 0.16
= 8 ×10–3 moles
Total moles of H+ = 9 × 10–3 moles
–3–29 10
H 1000 10 molar900
27. Answer (6)
Hint: Diborane has two 3c –2e
H
H
H H
H HB B
Solution : Each 3c – 2e– bond involve 3 orbital, one
from each B and H.
28. Answer (7)
Hint: Consider the changes which occurs inthermodynamic properties during a spontaneousprocess like vapourisation.
Solution : Except (f) all statements are correct
statement (f) is false because when the temperature is
smaller than equilibrium temperature, the H term
dominates and G becomes +ve for the vapourisation
process.
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
7/14
29. Answer (A, C)
Hint: According to Nernst equation cell
RTE E lnQ
nf
cell
0.059E E logQ
n
y = c + mx
so, 1
mn
x = 0.059 logQ
Solution : if n = 1
tan = 1
= 45°
E°cell
= x (magnitude wise)
for hydrogen electrode
cell
2
0.059 [H ]E E log
1 H
0.770.059 log ( 0.6 0.77 and E 0)
0.6 ∵
= 0
30. Answer (A, B, D)
Hint: AlCl3 in C
6H
6 present as dimer.
Solution : Borax loses water of crystallisation on
heating.
31. Answer (A)
Hint: Consider the factors affecting hydration energy.
Solution :
As size of cation increases the size of hydrated ions
in water decreases and thus ionic mobility increases.
Solubilities of bromides and hydroxides increases down
the group because their lattice energy change is more
than hydration energy on moving down the group.
32. Answer (A, C)
Hint: General properties of group-16 hydrides.
Solution : Correct order of M.P. is
H2S < H
2Se < H
2Te
as increase in mass of the molecule will cause
increase in intermolecular binding energy.
And bond angle of hydrides does not differ much on
moving down the group, from S to Te.
33. Answer (B, D)
Hint: Check the electron donating and withdrawing
extent of the group.
Solution :
Activating group activates the ring towards E.A.S. while
deactivating group deactivates it.
34. Answer (A, B, D)
Hint: F2 is produced from anhyd. HF.
Solution : Water should not be present during
electrolysis otherwise O2 is liberated.
35. Answer (A, C, D)
Hint: Check all the possibilities of structures which
can be obtained after the reaction.
Solution :
H , Ni2 Cl , h2 3 products
H , Ni2 Cl , h
2
3 products
36. Answer (A, B, C)
Hint: Apply concepts of gram equivalents.
Solution :
meq of KMnO4 = 25 × 0.2 = 5
meq of FeSO4 = 40 × 0.2 = 8
meq of H3ASO
3 = 30 × 0.1 × 2 = 6
meq of H2O
2 = 15 × 0.1 × 2 = 3
meq of SnCl2 = 25 × 0.1 × 2 = 5
37. Answer (D)
Hint: ∵ B can be taken as an intermediate of the
reaction.
d[B]
0dt
Solution : 1K
A B
2K
B F
1
d[A]k dt
dt
1 2
d[B]k [A] K [B]
dt
∵ 1K t
t 0[A] [A] e
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
8/14
PART - III (MATHEMATICS)
(∵ it has been assumed that all the reactions are
first order)
Now,
1K t
1 0 2
d[B]K [A] e K [B] 0
dt
1K t
02
1
[A] eK
K [B]
38. Answer (D)
Hint: We have to consider the given situation
i.e. K1 > K
2. So more of A is consumed as compared to
B.
Solution : ∵ K1 > K
2, the amount B becomes fairly
large before droping eventually to zero.
39. Answer (A, D)
Hint: As the solution is a good source of e–,
The solution will behave as a good reducing agent.
Solution : NaNH2 will be produced after a very long
time.
40. Answer (A, B)
Hint: Reaction of Na to H2O is highly exothermic due
to high polarity of H2O. So blue colour solution is not
obtained, as free electrons are absent.
Solution : Concentrated solutions are copper bronze
in colour.
Alkali metal solution in ammonia on evaporation yield
metals.
41. Answer (7)
Hint : Make cases
Solution : Here 3 vowels A, I, A can be placed in the
following five available places:
× N × G × N × N ×
OR
× N × G × G × G ×
OR
× N × G × G × N ×
Required number of ways =
3
3! 4! 4! 4!5
2! 3! 3! 2! 2!
C
= 5 4 3! 3 2! 4 3! 4 3! 4 3 2!
3! 2 1 2! 3! 3! 2! 2
= 30(4 + 4 + 6) = 420
4207
60 60
42. Answer (4)
Hint : A.M G.M.
Solution : Let , , , be the roots of given equation.
+ + + = 4 and = 1
Now A.M. of roots = 4
14 4
and G.M. of roots =
1 1
4 4( ) (1) 1
A.M. = G.M.
Which is possible only if
= = = = 1
Given equation will be of the form (x – 1)4 = 0
x4 – 4x3 + 4C2x2 – 4C
3x + 1 = 0
x4 – 4x3 + 6x2 – 4x + 1 = 0
Comparing, we get a = 6, b = – 4
a + b + 2 = 6 – 4 + 2 = 4
43. Answer (3)
Hint : Draw graph of y = x3 + 2x2 + x and y = –6
Solution : Given equation is x3 + 2x2 + x + 6 = 0
x3 + 2x2 + x = – 6
Let f(x) = x3 + 2x2 + x and g(x) = – 6
Here, f (x) = 3x2 + 4x + 1
and f (x) = 6x + 4
When f (x) = 0 3x2 + 3x + x + 1 = 0
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
9/14
3x (x + 1) + (x + 1) = 0
(x + 1) + (3x + 1) = 0
1
1,3
x
f(–1) = – 6 + 4 = – 2 < 0
and 1 1
6 4 2 03 3
f
f(x) has local maxima at x = – 1 and local minima
at x = 1
3
Also, max. f(x) = f(–1) = –1 + 2 – 1 = 0
and min. 1 1 2 1
( )3 27 9 3
f x f
1 6 9 4
27 27
( 2) 8 8 2 2 (1) 1 2 1 4
( 3) 27 18 3 12
f f
f
0
–3 –2 –1
x
y
y = g x( ) = – 6
(–3, –12)
(–2, –2)
y
x
1
3
From graph it is clear that one real ‘’ will lie between
(–3) and (–2) –3 < < –2
[ ] 3 3
44. Answer (8)
Hint : SP + SP = 2A
Solution : Locus of Z is ellipse
2,0 0, 2 ,2 4 s s A
1
2 22
ss Ae e
distance between directrices 2 4
81/ 2
A
e
45. Answer (1)
Hint : Solve Limit
Solution : From given equation
1
2
lim tan sin cos63 1
k
kx
kx kx k x
1
2
3lim tan sin
213 1
k
k x
xk x x
k k
= tan 33
2
3
3 1
x
x x
3x2 – 9x + 3 = x
3x2 – 10x + 3 = 0 = 1
46. Answer (5)
Hint : 1 2Area ( ) ( )
b
a
f x f x dx
Solution : y = x2 + x – 2, y = 2x
x2 + x – 2 = 2x x2 – x – 2 = 0
x2 – 2x + x – 2 = 0 x (x – 2) +1(x – 2) = 0
x = –1, 2
intersection points of y = x2 + x – 2 and y = 2x, are
(–1, –2) and (2, 4)
(–1, –2)
(–2, 0) xx
(2, 4)
y x = 2
y x + x – = 22
0
–1
(1, 0)
Here y = x2 + x – 2 is a parabolic curve with vertex
1 9,
2 4
which cuts x-axis at (–2, 0) and (1, 0) and
y-axis at (0, –2)
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
10/14
= Required area
2 2
2 2
1 1
(2x x x 2)dx ( x x 2)dx
23 2
1
x x2x
3 2
8 1 12 4 2
3 3 2
1 95
2 2
91 1 5
2
47. Answer (2)
Hint : f (x) < 0 at point P
Solution : At any point of maxima,
f(x) = 0 and f(x) < 0
f(x) = x2 – f2(x) –9 < 0
Let P(, )
2 – 2 – 9 < 0 and 2 – 2 = a2
a2 – 9 < 0 a (–3, 3)
48. Answer (6)
Hint : Centroid of a triangle divides the line joining
orthocentre and circumcentre in the ratio 2 : 1 internally.
Solution : Since centroid divides the line joining
orthocentre (2, 4) and circumcentre 7 5,
2 2
in the ratio
2 : 1
7 52 1 2 2 1 4
2 2, (3,3)2 1 2 1
A
2 2(3 0) (3 0) 3 2OA
A(3, 3)
y
xO
BP
L M N
45°
45°
45°
3 2
3
2AB
3
2 AP PB
3 33 ,3
2 2
B
+ = 6
49. Answer (A, B, C)
Hint : Solve integration, then limit
Solution : We have 2 2
1a
ndx
n x
2
2
1
1( )a
dxn
xn
11 1tan
11
a
x
n
nn
11tan ( )
a
n nx
n
1 1 1tan tan tan ( )2
na an
1
2 2lim lim tan
21n n
a
ndxL an
n x
, if 02 2
0, if 02
, if 02 2
a
a
a
, if 0
, if 02
0, if 0
a
a
a
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
11/14
50. Answer (A, B)
Hint : Find AB
Solution :
2 2 23 3 3
6 6 6
ax bx cx
AB a b c
ax bx cx
Now, tr(AB) = tr(C)
3ax2 + b + 6 cx = (x + 2)2 + 2x + 5x2 x R
3ax2 + 6 cx + b = 6x2 + 6x + 4
Comparing, we get a = 2, b = 4, c = 1
51. Answer (A, B, C, D)
Hint : Use properties of determinants
Solution : Given equation is
2 2
2 2
2 2
1 sin cos 2sin4
sin 1 cos 2sin4 0
sin 1 2sin4
A A
A A
A cos A
By R1 R
1 – R
2 and R
2 R
2 – R
3 we have
2 2
1 1 0
0 1 1 0
sin cos 1 2sin4A A
By C2 C
2 + C
3, we have
2 2
1 1 0
0 0 1 0
sin cos 1 2sin4 1 2sin4A A
– 0 + 0 – (–1) [cos2A + 1 + 2sin4 + sin2A ] = 0
2(1 + sin4) = 0
sin4 1 sin2
4 ( 1)2
nn
( 1) ( 1)4 8 4 8
n nn n
Which is independent of A.
When n = 0, = 0 ( 1)8 8
31,
4 8 8n
When n = –1, 2
4 8 8 8
Options (A), (B), (C), (D) are correct.
52. Answer (A, D)
Hint : Find dy
dxfrom quadratic equation.
Solution : Given differential equation is
2
( ) 0dy dy
y x y xdx dx
2( ) ( ) 4 ( )
2
x y x y y xdy
dx y
2( ) ( )
2
y x x y
y
( )
2
y x x y
y
or2 2
y x x y y x x y
y y
1ordy x
dx y
when 1dy
dx
0dx dy
x – y + c = 0 ....(1)
and when dy x
dx y
ydy + xdx = 0
2 2 2
2 2 2
x y k
x2 + y2 = k2 ....(2)
But curves passes through the point (1, 2)
1 – 2 + c = 0 c = 1
1st curve is the straight line
x – y + 1 = 0
and from (2), 1 + 4 = k2
2nd curve is x2 + y2 = 5
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
12/14
53. Answer (A, C, D)
Hint : Take dot product of both sides with
, , andu v a w
�� � �
.
Solution : Given u v w a �� � �
....(1)
Taking dot product of both sides of (1) with
, , andu v a w
�� � �
, we get
2 3
2u u v u w a u
�� � � � � �
31
2u v u w � � � �
1
2u v u w � � � �
....(2)
2 7
4u v v v w a v
�� � � � � �
71
4u v v w � � � �
7 31
4 4u v v w � � � �
....(3)
2u a a v a w a a a
� � � � � �� � �
3 7
42 4
a w � �
13 34
4 4a w � �
....(4)
and 2
�� � � � � �
u w v w w a w
31
4
� � � �
u w v w
3 11
4 4
� � � �
u w v w ....(5)
By [(2) + (3) + (5)], we get
u v u w u v v w u v v w � � � � � � � � � � � �
1 3 1 2 3 11
2 4 4 4
1
2u v v w u w � � � � � �
....(6)
By [(6) – (2)] 0v w � �
By [(6) – (3)], 1 3 2 3 1
2 4 4 4w u
� �
From (6), 1 1
04 2
u v � �
1 1 2 1 3
2 4 4 4u v
� �
54. Answer (C, D)
Hint : 2 cosAcosB = cos(A + B) + cos(A – B)
Solution : Here 0
1( ) 2cos cos( )
4f x t x t dx
0
1cos ( ) cos ( )
4t t x t t x t dt
0
0
cos 1[ ] cos(2 )
4 4
xt t x dt
0
cos 1 sin(2 )
4 4 2
x t x
cos 1sin 2 sin 0
4 8
xx x
cos 1sin sin
4 8
xx x
cos( )
4
xf x
Which is clearly continuous and
differentiable is (0, 2)
( ) sin4
f x x
( ) cos4
f x x
When ( ) 0 sin 04
f x x
sin x = 0
x = 0, , 2,
At x = 0, 2, ( ) 1 04
f x
x = 0, 2 are the points of local maxima
Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
13/14
and at x = , ( ) ( 1) 04
f x
x = is the point of local minima
At x = , ( ) sin 04
f x
atleast one c = (0, 2) such that f(c) = 0
and maximum value of ( ) cos4 4
f x
Max. ( ) at ( 0, 2 )4
f x x
55. Answer (A, C, D)
Hint : Draw diagram.
Solution :
A
B
P(3, 4)C
(0, 1)
Power of point P(3, 4) w.r.t. circle
S x2 + y2 – 2y – 3 = 0 is defined as
PA PB = S1
= 9 + 16 – 8 – 30 = 14 option (A) is correct
2 2(3 0) (4 1) 9 9 3 2PC
CA = radius of given circle 2 2(0) ( 1) ( 3) = 2
From CAP,
2 2sin
33 2
CA
CP
1 2sin
3
Angle between tangents
1 22 2sin
3
Option (B) is wrong.
Also CP is a diameter of circumcircle of PAB
Equation of circumcircle is
(x – 0) (x – 3) + (y – 1)(y – 4) = 0
x2 – 3x + y2 – 4y – y + 4 = 0
x2 + y2 – 3x – 5y + 4 = 0
Option (C) is correct.
Also, area of quadrilateral PACB
12 2 14 2 14
2
Option (D) is correct.
56. Answer (A, B, C, D)
Hint : Sun of two perfect square = 0
Solution : Given: 2a2 + 4b2 + c2 – 4ab – 2ac = 0
2 2 2 22 2 (2 ) ( 2 ) 0a a b b a c ac
(a – 2b)2 + (a – c)2 = 0
a = 2b = c ,2
ab c a
22 2
2 2 2
4cos
2 2
aa a
c a bB
ca a a
2
2
12
74
82
a
a
Option (A) is correct.
Again, a = c 2R sinA = 2R sinC, by sine rule
A = C A – C = 0
cos(A – C) = cos = 1
Option (B) is correct
Again 1
1r s a as
r s s
s a
1
2
a
a b c
21
2
a
aa a
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)
14/14
4 5 4 11
5 5 5
a
a
Option (C) is correct.
Again, By sine rule
sin sin sinA B C
a b c
sin sin sin
2
A B C
aa a
sin sin sin
11 1
2
A B C
1sin : sin : sin 1: : 1 2 :1: 2
2A B C
Solution for Q. No. 57 and 58:
Given, 2
0
( ) ( )
x
tf x x e f x t dt
2 ( )
0
( )
x
x tf x x e f x x t dt
2
0
( ) ( )
x
x tf x x e e f t dt
....(1)
Differentiating both sides of (1) w.r.t. x, we get
0
( ) 2 ( ) 1 ( )
x
x x x tf x x e e f x e e f t dt
....(2)
By [(1) + (2)
f(x) + f (x) = x2 + 2x + f(x)
Integrating both sides we get
3 2
( ) 23 2
x xf x c
3
2( )3
xf x x c ....(1)
When x = 0, f(0) = 0 + 0 + c c = f(0)
But 0
2
0
(0) (0) (0 ) 0tf e f t dt
c = f(0) = 0
Hence from (1)
32( )
3
xf x x
57. Answer (A, B, D)
Solution : 2
3( )
( 3)
g x
x x
g(x) is discontinuous at x = 0 and –3.
58. Answer (C)
Solution : 2
1 1 1( )
3( 3) 3
g x dx dx
x xx
1 1 3ln
3
xC
x x
59. Answer (A, D)
Solution : ( , , ) , ,2 2 2
a b c
a = 2, b = 2, c = 2 8 = ± 32 × 6 = ± 24
60. Answer (B)
Solution : Let P = (, , ).Now PA PB ( – a) + (– b) + r = 0
a + b = 2 + 2 + 2
PBPC b + c = 2 + 2 + 2
1 1 1
a b c
2 2 2 2 2 2
,2 2
a b and
2 2 2
2
c
abc = ± 6 × 32
(2 + 2 + 2)3 = ± 192 × 8 × (x2 + y2 + z2) = ± 1536 xyz
�����
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
1/14
1. (6)
2. (7)
3. (5)
4. (3)
5. (3)
6. (2)
7. (1)
8. (2)
9. (C, D)
10. (B, C)
11. (B, D)
12. (A, C)
13. (A, B, C)
14. (A, C)
15. (A, C)
16. (A, D)
17. (A, B, D)
18. (A, B, D)
19. (A, C)
20. (A, D)
21. (7)
22. (6)
23. (2)
24. (4)
25. (4)
26. (6)
27. (6)
28. (3)
29. (A, B, C)
30. (A, C, D)
31. (A, B, D)
32. (B, D)
33. (A, C)
34. (A)
35. (A, B, D)
36. (A, C)
37. (D)
38. (D)
39. (A, D)
40. (A, B)
41. (6)
42. (2)
43. (5)
44. (1)
45. (8)
46. (3)
47. (4)
48. (7)
49. (A, B, C, D)
50. (A, C, D)
51. (C, D)
52. (A, C, D)
53. (A, D)
54. (A, B, C, D)
55. (A, B)
56. (A, B, C)
57. (A, B, D)
58. (C)
59. (A, D)
60. (B)
ANSWERS
Test Date: 17/02/2019
PHYSICS CHEMISTRY MATHEMATICS
All India Aakash Test Series for JEE (Advanced)-2019
MOCK TEST - 1 (Paper-2) - Code-D
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
2/14
PART - I (PHYSICS)
ANSWERS & HINTS
1. Answer (6)
Hint :
Fmax
= 2 mAg
Solution :
Fmax
= 2 mAg
2. Answer (7)
Hint : Length of string will remain constant
Solution :
V0 cos 37 = V
0
4
5 V V
VV0
5
4
3. Answer (5)
Hint : hc
Emin
max
Solution :
1 hC
ev
If V increase 1 decreases
4. Answer (3)
Hint : pr v
RC C
n–
–1
Solution :
T2 P3 = C
2
3 5 2
PVP C P V C
nR
2/5pV C
3 19–
22 6–1
5
pr
R R RC
Q = n Cpr T R
Q19
2 1006
RQ
1900
3
32 100
2
Ru
u = 300 R
1900 1000– – 300
3 3
R RW Q u R
5. Answer (3)
Hint : dP
B VdV
–
Solution :
3 24, 4
3V a dV a da
–
–/
dP dPB V
dV V dV
MgdP
A
3
2
4–
3 4
a MgB
A a da
da Mg
a BA–3
6. Answer (2)
L
H
Hint : 2dpx
dx
Solution :
22
LAL ghA
Lh
g
2 2
2
7. Answer (1)
Hint : E A2 2
0
1
4 2
Solution :
x1
7cm
8 x
2
17cm
8
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
3/14
2 212.5
4 2U E A
2 2
2
12.5
4 2
TU E A
v
2 2
2
5 1 1 4 100100 4
2 4 2 100 100U E
25U E 2
15
U
8. Answer (2)
Hint : O B A BT T
1
2
Solution :
mgx = fR mg
f xR
g ga x
R R
R T R
T tg g
2 ,4 2
9. Answer (C, D)
Hint : After reflection from mirror there will be phase
change of Solution :
D 2D
y
O
Screen
d
D
y
P
Q
OP dOP d
D D2 2
d OROR d
D D3
3
y = 2d = 1 mm
also = yd
nD
1–
2 2
or y d n 5.5, n 15.5
10. Answer (B, C)
Hint : Apply superposition principle
Solution :
00
1
4
R
C
x dxV
x
0
4
R
02
P
RV
11. Answer (B, D)
Hint : ax
VI
Z
max
m
Solution :
15
I2
60
30 v
I1
I
I t1
100sin 50
2 3
I t2
100sin 50 –
2 6
I t50 2 sin 5012
Pam
= Vrm
Irm
cos
100 5 2cos15
2 2
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
4/14
12. Answer (A, C)
Hint : Apply I ��
Solution :
(2 2mR + mg2
)N
fO
0 = I
0
RmR mg mR mR2 2 2
2 2 2 22
2
4
g R
R
13. Answer (A, B, C)
Hint : Use R – C circuit analysis.
Solution :
K Q KQV
d a
1–
0 0
1
QaQ
d
t
RcQ Q e–
1 t = Rc ln3 = 4
0 .aR ln3
1
3
2 2 2
22 200
8 .2 4 .
VQ a Q a
dd d
14. Answer (A, C)
Hint : Apply energy conservation and momentum
conservation
Solution :
T
2ma0
Ring
T cos 37° = ma
For 2 m
T cos 37° + 2 ma0 = (2 m) g sin 37°
We get the result
15. Answer (A, C)
Hint : Ea
02
Solution :
( , )a a
( )p
3
1
x
y2
31 2
0 0 0
1 1
2 . 2 . 2 . 2 2 2pE j i i j
a a a
��� � � �
3 3
2 1
0 0
1 1
2 . 2 2 . 2
��� �
pE i ja a
16. Answer (A, D)
Hint :
Apply Ampere’s law
Solution :
Amplere’s law
0 enclosed. .B dl I
�� ���
�
0–
ˆ –2
��
B dxi
17. Answer (A, B, D )
18. Answer (A, B, D)
Hint : f Nmax
1 1 1
Solution for Q. Nos. 17 and Q. Nos. 18
4 kg
2 kg
1 kg f1
f2
f3
14 N
ca
14 – 7
7
= 1 m/s2
max
37 Nf
max
12 Nf
max
23 Nf
max
1a = 2 m/s2
max
2a = 1 m/s2
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
5/14
19. Answer (A, C)
20. Answer (A, D)
Hints :
Rt
LI I e–
max1–
in R - L circuit
Solution for Q. Nos. 19 and Q. Nos. 20
1
2
1 V
1
2
I
4H
3
Circuit can be reduced as
t
LI I e
R–
max1–
PART - II (CHEMISTRY)
21. Answer (7)
Hint: Consider the changes which occurs inthermodynamic properties during a spontaneousprocess like vapourisation.
Solution : Except (f) all statements are correct
statement (f) is false because when the temperature is
smaller than equilibrium temperature, the H term
dominates and G becomes +ve for the vapourisation
process.
22. Answer (6)
Hint: Diborane has two 3c –2e
H
H
H H
H HB B
Solution : Each 3c – 2e– bond involve 3 orbital, one
from each B and H.
23. Answer (2)
Hint :
2
2Cd H S CdS 2H
���⇀↽���
Solution :
0.5 × 10–3 moles of Cd2+, 0.5 × 10–3 moles of S–2 ions
are required to precipitate.
0.5 ×10–3 moles of S2– 0.5 × 10–3 moles H2S
1×10–3 moles of H+
Number of moles of H+ from HCl = 50 × 0.16
= 8 ×10–3 moles
Total moles of H+ = 9 × 10–3 moles
–3–29 10
H 1000 10 molar900
24. Answer (4)
Hint :
Number of moles CaC2O
4 = Number of moles of CaCI
2
= Number of moles of organic acid
Solution :
(Oxalic acid) N1V
1 = N
2V
2 (KMnO
4)
2 × no. of moles = 80
5 0.051000
= 0.01 moles
Number of moles Oxalic acid = number of moles
of CaC2O
4 = number of moles of CaCI
2
Mass of CaCI2 = 0.01 × 111 = 1.11 g
% of CaCI2 in mixture =
1.11100 60%
1.8
4
3R
4 2 4
4 2 3
R
t
I e
–4
31 4
1–4 3
Imax
1A
4
I = 0.25 (1 – e–t) A
also L1I1 = L
2I2
I t I I t I1 2
2 1,
3 3
tI t e–
1
11–
6 , tI t e
–
2
11–
12
Total
1 4 1 1J
2 3 16 24
U
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
6/14
25. Answer (4)
Hint: Check the chirality and elements of symmetry in
product.
Solution :
P1
Optically active
P2
Optically active
P3
Optically active
P4
Optically active
P5
Optically inactive
P6
Optically inactive
P7
Optically inactive
P8
Optically inactive
26. Answer (6)
Hint: First convert molecules tetrahedral geometry into
Fisher form.
Solution :
(A) CH3
H
OH
2
4
3
1
(R)
(1)C H2 5
(B)
HH
Cl
24
3
1
HO C H2 5
(R)
(1)
(C)
HH
Cl2
4
3
Br (S)(2)
CH3
1
(D)
HH
C H2 5
Br
4
3
2
(S)(2)
CH3
1
27. Answer (6)
Hint: To obtain aromatic character R can lose a
hydride H–, and become sp2 hybridised.
Solution :
C C
H
+
+
H
+ C(Ph)3
+ H – C(Ph)3
(R)
–
BF4
–
BF4H
a a
Number of electrons in ring R is 6.
28. Answer (3)
Hint: Kp remain constant for a reaction, even in
simultaneous equilibria.
Solution :
a 2b a
PQ(s) P(g) Q(g)
���⇀↽��� K
p = 4.8 × 10–6
2
2b a
RQ (s) R(s) 2Q(g)
���⇀↽��� K
p = 5.76 × 10–6
(2b + a)2 = 5.76 × 10–6
(2b + a) = 2.4 × 10–3
(2.4 × 10–3) × a = 4.8 × 10–6
a = 2 × 10–3 = 2 × 10–x
x = 3
29. Answer (A, B, C)
Hint: Apply concepts of gram equivalents.
Solution :
meq of KMnO4 = 25 × 0.2 = 5
meq of FeSO4 = 40 × 0.2 = 8
meq of H3ASO
3 = 30 × 0.1 × 2 = 6
meq of H2O
2 = 15 × 0.1 × 2 = 3
meq of SnCl2 = 25 × 0.1 × 2 = 5
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
7/14
30. Answer (A, C, D)
Hint: Check all the possibilities of structures which
can be obtained after the reaction.
Solution :
H , Ni2 Cl , h2 3 products
H , Ni2 Cl , h
2
3 products
31. Answer (A, B, D)
Hint: F2 is produced from anhyd. HF.
Solution : Water should not be present during
electrolysis otherwise O2 is liberated.
32. Answer (B, D)
Hint: Check the electron donating and withdrawing
extent of the group.
Solution :
Activating group activates the ring towards E.A.S. while
deactivating group deactivates it.
33. Answer (A, C)
Hint: General properties of group-16 hydrides.
Solution : Correct order of M.P. is
H2S < H
2Se < H
2Te
as increase in mass of the molecule will cause
increase in intermolecular binding energy.
And bond angle of hydrides does not differ much on
moving down the group, from S to Te.
34. Answer (A)
Hint: Consider the factors affecting hydration energy.
Solution :
As size of cation increases the size of hydrated ions
in water decreases and thus ionic mobility increases.
Solubilities of bromides and hydroxides increases down
the group because their lattice energy change is more
than hydration energy on moving down the group.
35. Answer (A, B, D)
Hint: AlCl3 in C
6H
6 present as dimer.
Solution : Borax loses water of crystallisation on
heating.
36. Answer (A, C)
Hint: According to Nernst equation cell
RTE E lnQ
nf
cell
0.059E E logQ
n
y = c + mx
so, 1
mn
x = 0.059 logQ
Solution : if n = 1
tan = 1
= 45°
E°cell
= x (magnitude wise)
for hydrogen electrode
cell
2
0.059 [H ]E E log
1 H
0.770.059 log ( 0.6 0.77 and E 0)
0.6 ∵
= 0
37. Answer (D)
Hint: ∵ B can be taken as an intermediate of the
reaction.
d[B]
0dt
Solution : 1K
A B
2K
B F
1
d[A]k dt
dt
1 2
d[B]k [A] K [B]
dt
∵ 1K t
t 0[A] [A] e
(∵ it has been assumed that all the reactions are
first order)
Now,
1K t
1 0 2
d[B]K [A] e K [B] 0
dt
1K t
02
1
[A] eK
K [B]
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
8/14
PART - III (MATHEMATICS)
41. Answer (6)
Hint : Centroid of a triangle divides the line joining
orthocentre and circumcentre in the ratio 2 : 1 internally.
Solution : Since centroid divides the line joining
orthocentre (2, 4) and circumcentre 7 5,
2 2
in the ratio
2 : 1
7 52 1 2 2 1 4
2 2, (3,3)2 1 2 1
A
2 2(3 0) (3 0) 3 2OA
A(3, 3)
y
xO
BP
L M N
45°
45°
45°
3 2
3
2AB
3
2 AP PB
3 33 ,3
2 2
B
+ = 6
42. Answer (2)
Hint : f (x) < 0 at point P
Solution : At any point of maxima,
f(x) = 0 and f(x) < 0
f(x) = x2 – f2(x) –9 < 0
Let P(, )
2 – 2 – 9 < 0 and 2 – 2 = a2
a2 – 9 < 0 a (–3, 3)
43. Answer (5)
Hint : 1 2Area ( ) ( )
b
a
f x f x dx
Solution : y = x2 + x – 2, y = 2x
x2 + x – 2 = 2x x2 – x – 2 = 0
x2 – 2x + x – 2 = 0 x (x – 2) +1(x – 2) = 0
x = –1, 2
intersection points of y = x2 + x – 2 and y = 2x, are
(–1, –2) and (2, 4)
(–1, –2)
(–2, 0) xx
(2, 4)
y x = 2
y x + x – = 22
0
–1
(1, 0)
Here y = x2 + x – 2 is a parabolic curve with vertex
38. Answer (D)
Hint: We have to consider the given situation
i.e. K1 > K
2. So more of A is consumed as compared to
B.
Solution : ∵ K1 > K
2, the amount B becomes fairly
large before droping eventually to zero.
39. Answer (A, D)
Hint: As the solution is a good source of e–,
The solution will behave as a good reducing agent.
Solution : NaNH2 will be produced after a very long
time.
40. Answer (A, B)
Hint: Reaction of Na to H2O is highly exothermic due
to high polarity of H2O. So blue colour solution is not
obtained, as free electrons are absent.
Solution : Concentrated solutions are copper bronze
in colour.
Alkali metal solution in ammonia on evaporation yield
metals.
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
9/14
1 9,
2 4
which cuts x-axis at (–2, 0) and (1, 0) and
y-axis at (0, –2)
= Required area
2 2
2 2
1 1
(2x x x 2)dx ( x x 2)dx
23 2
1
x x2x
3 2
8 1 12 4 2
3 3 2
1 95
2 2
91 1 5
2
44. Answer (1)
Hint : Solve Limit
Solution : From given equation
1
2
lim tan sin cos63 1
k
kx
kx kx k x
1
2
3lim tan sin
213 1
k
k x
xk x x
k k
= tan 33
2
3
3 1
x
x x
3x2 – 9x + 3 = x
3x2 – 10x + 3 = 0 = 1
45. Answer (8)
Hint : SP + SP = 2A
Solution : Locus of Z is ellipse
2,0 0, 2 ,2 4 s s A
1
2 22
ss Ae e
distance between directrices 2 4
81/ 2
A
e
46. Answer (3)
Hint : Draw graph of y = x3 + 2x2 + x and y = –6
Solution : Given equation is x3 + 2x2 + x + 6 = 0
x3 + 2x2 + x = – 6
Let f(x) = x3 + 2x2 + x and g(x) = – 6
Here, f (x) = 3x2 + 4x + 1
and f (x) = 6x + 4
When f (x) = 0 3x2 + 3x + x + 1 = 0
3x (x + 1) + (x + 1) = 0
(x + 1) + (3x + 1) = 0
1
1,3
x
f(–1) = – 6 + 4 = – 2 < 0
and 1 1
6 4 2 03 3
f
f(x) has local maxima at x = – 1 and local minima
at x = 1
3
Also, max. f(x) = f(–1) = –1 + 2 – 1 = 0
and min. 1 1 2 1
( )3 27 9 3
f x f
1 6 9 4
27 27
( 2) 8 8 2 2 (1) 1 2 1 4
( 3) 27 18 3 12
f f
f
0
–3 –2 –1
x
y
y = g x( ) = – 6
(–3, –12)
(–2, –2)
y
x
1
3
From graph it is clear that one real ‘’ will lie between
(–3) and (–2) –3 < < –2
[ ] 3 3
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
10/14
47. Answer (4)
Hint : A.M G.M.
Solution : Let , , , be the roots of given equation.
+ + + = 4 and = 1
Now A.M. of roots = 4
14 4
and G.M. of roots =
1 1
4 4( ) (1) 1
A.M. = G.M.
Which is possible only if
= = = = 1
Given equation will be of the form (x – 1)4 = 0
x4 – 4x3 + 4C2x2 – 4C
3x + 1 = 0
x4 – 4x3 + 6x2 – 4x + 1 = 0
Comparing, we get a = 6, b = – 4
a + b + 2 = 6 – 4 + 2 = 4
48. Answer (7)
Hint : Make cases
Solution : Here 3 vowels A, I, A can be placed in the
following five available places:
× N × G × N × N ×
OR
× N × G × G × G ×
OR
× N × G × G × N ×
Required number of ways =
3
3! 4! 4! 4!5
2! 3! 3! 2! 2!
C
= 5 4 3! 3 2! 4 3! 4 3! 4 3 2!
3! 2 1 2! 3! 3! 2! 2
= 30(4 + 4 + 6) = 420
4207
60 60
49. Answer (A, B, C, D)
Hint : Sun of two perfect square = 0
Solution : Given: 2a2 + 4b2 + c2 – 4ab – 2ac = 0
2 2 2 22 2 (2 ) ( 2 ) 0a a b b a c ac
(a – 2b)2 + (a – c)2 = 0
a = 2b = c ,2
ab c a
22 2
2 2 2
4cos
2 2
aa a
c a bB
ca a a
2
2
12
74
82
a
a
Option (A) is correct.
Again, a = c 2R sinA = 2R sinC, by sine rule
A = C A – C = 0
cos(A – C) = cos = 1
Option (B) is correct
Again 1
1r s a as
r s s
s a
1
2
a
a b c
21
2
a
aa a
4 5 4 11
5 5 5
a
a
Option (C) is correct.
Again, By sine rule
sin sin sinA B C
a b c
sin sin sin
2
A B C
aa a
sin sin sin
11 1
2
A B C
1sin : sin : sin 1: : 1 2 :1: 2
2A B C
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
11/14
50. Answer (A, C, D)
Hint : Draw diagram.
Solution :
A
B
P(3, 4)C
(0, 1)
Power of point P(3, 4) w.r.t. circle
S x2 + y2 – 2y – 3 = 0 is defined as
PA PB = S1
= 9 + 16 – 8 – 30 = 14 option (A) is correct
2 2(3 0) (4 1) 9 9 3 2PC
CA = radius of given circle 2 2(0) ( 1) ( 3) = 2
From CAP,
2 2sin
33 2
CA
CP
1 2sin
3
Angle between tangents
1 22 2sin
3
Option (B) is wrong.
Also CP is a diameter of circumcircle of PAB
Equation of circumcircle is
(x – 0) (x – 3) + (y – 1)(y – 4) = 0
x2 – 3x + y2 – 4y – y + 4 = 0
x2 + y2 – 3x – 5y + 4 = 0
Option (C) is correct.
Also, area of quadrilateral PACB
12 2 14 2 14
2
Option (D) is correct.
51. Answer (C, D)
Hint : 2 cosAcosB = cos(A + B) + cos(A – B)
Solution : Here 0
1( ) 2cos cos( )
4f x t x t dx
0
1cos ( ) cos ( )
4t t x t t x t dt
0
0
cos 1[ ] cos(2 )
4 4
xt t x dt
0
cos 1 sin(2 )
4 4 2
x t x
cos 1sin 2 sin 0
4 8
xx x
cos 1sin sin
4 8
xx x
cos( )
4
xf x
Which is clearly continuous and
differentiable is (0, 2)
( ) sin4
f x x
( ) cos4
f x x
When ( ) 0 sin 04
f x x
sin x = 0
x = 0, , 2,
At x = 0, 2, ( ) 1 04
f x
x = 0, 2 are the points of local maxima
and at x = , ( ) ( 1) 04
f x
x = is the point of local minima
At x = , ( ) sin 04
f x
atleast one c = (0, 2) such that f(c) = 0
and maximum value of ( ) cos4 4
f x
Max. ( ) at ( 0, 2 )4
f x x
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
12/14
52. Answer (A, C, D)
Hint : Take dot product of both sides with
, , andu v a w
�� � �
.
Solution : Given u v w a �� � �
....(1)
Taking dot product of both sides of (1) with
, , andu v a w
�� � �
, we get
2 3
2u u v u w a u
�� � � � � �
31
2u v u w � � � �
1
2u v u w � � � �
....(2)
2 7
4u v v v w a v
�� � � � � �
71
4u v v w � � � �
7 31
4 4u v v w � � � �
....(3)
2u a a v a w a a a
� � � � � �� � �
3 7
42 4
a w � �
13 34
4 4a w � �
....(4)
and 2
�� � � � � �
u w v w w a w
31
4
� � � �
u w v w
3 11
4 4
� � � �
u w v w ....(5)
By [(2) + (3) + (5)], we get
u v u w u v v w u v v w � � � � � � � � � � � �
1 3 1 2 3 11
2 4 4 4
1
2u v v w u w � � � � � �
....(6)
By [(6) – (2)] 0v w � �
By [(6) – (3)], 1 3 2 3 1
2 4 4 4w u
� �
From (6), 1 1
04 2
u v � �
1 1 2 1 3
2 4 4 4u v
� �
53. Answer (A, D)
Hint : Find dy
dxfrom quadratic equation.
Solution : Given differential equation is
2
( ) 0dy dy
y x y xdx dx
2( ) ( ) 4 ( )
2
x y x y y xdy
dx y
2( ) ( )
2
y x x y
y
( )
2
y x x y
y
or2 2
y x x y y x x y
y y
1ordy x
dx y
when 1dy
dx
0dx dy
x – y + c = 0 ....(1)
and when dy x
dx y
ydy + xdx = 0
2 2 2
2 2 2
x y k
x2 + y2 = k2 ....(2)
Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
13/14
But curves passes through the point (1, 2)
1 – 2 + c = 0 c = 1
1st curve is the straight line
x – y + 1 = 0
and from (2), 1 + 4 = k2
2nd curve is x2 + y2 = 5
54. Answer (A, B, C, D)
Hint : Use properties of determinants
Solution : Given equation is
2 2
2 2
2 2
1 sin cos 2sin4
sin 1 cos 2sin4 0
sin 1 2sin4
A A
A A
A cos A
By R1 R
1 – R
2 and R
2 R
2 – R
3 we have
2 2
1 1 0
0 1 1 0
sin cos 1 2sin4A A
By C2 C
2 + C
3, we have
2 2
1 1 0
0 0 1 0
sin cos 1 2sin4 1 2sin4A A
– 0 + 0 – (–1) [cos2A + 1 + 2sin4 + sin2A ] = 0
2(1 + sin4) = 0
sin4 1 sin2
4 ( 1)2
nn
( 1) ( 1)4 8 4 8
n nn n
Which is independent of A.
When n = 0, = 0 ( 1)8 8
31,
4 8 8n
When n = –1, 2
4 8 8 8
Options (A), (B), (C), (D) are correct.
55. Answer (A, B)
Hint : Find AB
Solution :
2 2 23 3 3
6 6 6
ax bx cx
AB a b c
ax bx cx
Now, tr(AB) = tr(C)
3ax2 + b + 6 cx = (x + 2)2 + 2x + 5x2 x R
3ax2 + 6 cx + b = 6x2 + 6x + 4
Comparing, we get a = 2, b = 4, c = 1
56. Answer (A, B, C)
Hint : Solve integration, then limit
Solution : We have 2 2
1a
ndx
n x
2
2
1
1( )a
dxn
xn
11 1tan
11
a
x
n
nn
11tan ( )
a
n nx
n
1 1 1tan tan tan ( )2
na an
1
2 2lim lim tan
21n n
a
ndxL an
n x
, if 02 2
0, if 02
, if 02 2
a
a
a
, if 0
, if 02
0, if 0
a
a
a
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)
14/14
�����
Solution for Q. Nos. 57 and 58:
Given, 2
0
( ) ( )
x
tf x x e f x t dt
2 ( )
0
( )
x
x tf x x e f x x t dt
2
0
( ) ( )
x
x tf x x e e f t dt
....(1)
Differentiating both sides of (1) w.r.t. x, we get
0
( ) 2 ( ) 1 ( )
x
x x x tf x x e e f x e e f t dt
....(2)
By [(1) + (2)
f(x) + f (x) = x2 + 2x + f(x)
Integrating both sides we get
3 2
( ) 23 2
x xf x c
3
2( )3
xf x x c ....(1)
When x = 0, f(0) = 0 + 0 + c c = f(0)
But 0
2
0
(0) (0) (0 ) 0tf e f t dt
c = f(0) = 0
Hence from (1)
32( )
3
xf x x
57. Answer (A, B, D)
Solution : 2
3( )
( 3)
g x
x x
g(x) is discontinuous at x = 0 and –3.
58. Answer (C)
Solution : 2
1 1 1( )
3( 3) 3
g x dx dx
x xx
1 1 3ln
3
xC
x x
59. Answer (A, D)
Solution : ( , , ) , ,2 2 2
a b c
a = 2, b = 2, c = 2 8 = ± 32 × 6 = ± 24
60. Answer (B)
Solution : Let P = (, , ).Now PA PB ( – a) + (– b) + r = 0
a + b = 2 + 2 + 2
PBPC b + c = 2 + 2 + 2
1 1 1
a b c
2 2 2 2 2 2
,2 2
a b and
2 2 2
2
c
abc = ± 6 × 32
(2 + 2 + 2)3 = ± 192 × 8 × (x2 + y2 + z2) = ± 1536 xyz