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Test - 1A (Paper - 2) (Code-G) (Answers) All India Aakash Test Series for JEE (Advanced)-2020
1/12
PHYSICS
1. (A, B, C)
2. (A, B, D)
3. (A, B)
4. (A, B, D)
5. (A, D)
6. (A, B)
7. (18)
8. (16)
9. (12)
10. (15)
11. (02)
12. (12)
13. (15)
14. (03)
15. (A)
16. (A)
17. (B)
18. (B)
CHEMISTRY
19. (A, C, D)
20. (A, B, C, D)
21. (A, B)
22. (A, D)
23. (A)
24. (D)
25. (03)
26. (15)
27. (06)
28. (20)
29. (40)
30. (43)
31. (65)
32. (98)
33. (C)
34. (D)
35. (A)
36. (B)
MATHEMATICS
37. (C, D)
38. (B, C)
39. (A, B)
40. (A, D)
41. (B, C)
42. (B, C)
43. (81)
44. (02)
45. (01)
46. (00)
47. (11)
48. (11)
49. (31)
50. (01)
51. (C)
52. (B)
53. (D)
54. (C)
Test Date : 18/11/2018
ANSWERS
TEST - 1A (Paper-2) - Code-G
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (A, B, C)
Hint : Write down the x coordinate of A3 in terms of
� and . dx
vdt
Solution :
x = 2(3l cos + 2l cos + l cos )
x = 12l cos
v = 1
12 sinsin
d dl
dt dt
⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠ �
2 3B
dv l
dt
⎛ ⎞ ⎜ ⎟⎝ ⎠
6 m/s4sin
v
2. Answer (A, B, D)
Hint : Write down the expression for tangential
acceleration and acceleration in y-direction. Use the
fact that aj = a
t
Solution :
( sin )tdv
g gdt
( sin )y
dvg g
dt
v
d
ytdvdv
dt dt
vT – v
0 = –v
0sin v
T – v
0 =
03
5
v
vT =
02
5
v
3. Answer (A, B)
Hint : Velocity of cylinder along normal relative to
connected bodies is zero.
Solution :
3
ˆ ˆ
v xi yj �
x = +10
2 2
ˆ
v v i�
v2
v2
sin 37° = x sin 37° + y cos 37°
23 4
65 5
v y
3v2
= 30 + 4y
4. Answer (A, B, D)
Hint : Draw F.B.D. and equation of motion. Write
down the constraint motion.
Solution :
T cos 37° = ma1
3mg – T = 3ma2
1 2
4
5a a⎛ ⎞ ⎜ ⎟⎝ ⎠
1
5
4T ma
3mg – T = 1
12
5ma
3mg = 1
73
20ma
1
60
73
ga
5. Answer (A, D)
Hint : Draw FBD and write equation of motion with
constrain relation
Solution :
2mg – T1 = 2ma
1
T1 – T
2 = 2ma
1
2
02
Tma
2a1 = a
0
1
4
ga
6. Answer (A, B)
Hint : Acceleration of block B with respect to A is
2
0v
�
Solution :
T – 0.05 (2 mg) = 2 ma0
T + ma0
=
2mv
�
22
T mgT mg
⎛ ⎞ ⎜ ⎟⎝ ⎠
T
ma0
5
3
mgT
03
ga
Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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7. Answer (18)
Hint : Along an incline maximum range is
2
(1 sin )
u
g
Solution :
2
10(1 sin )
u
g
2 4100 1
5u
⎛ ⎞ ⎜ ⎟⎝ ⎠
u
2 = 20 × 9
8. Answer (16)
Hint : Use the concept that
2
c
va
R
Solution :
216 m/s
dy
dt
2
2
216 m/s
x
d xa
dt
2
16
x
vR
a
9. Answer (12)
Hint : Use addition of vector, net force will be along
PO.
Solution :
2
202 1
10k OP
����
= 40
k = 2 N/m
10. Answer (15)
Hint : Direction of friction will change, hence
magnitude of acceleration
Solution :
2
16 m/s
2
Mg Mga
M
0
1
1
vt
a
2
0
12
vs
a
2
2 0
2 2
1
1
2 2
va t
a
0
2
1 2
vt
a a
0
1 1 2
1 1vT
a a a
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎣ ⎦
0
1 2
1 2
va a
a a
11. Answer (02)
Hint : Apply NSL to get acceleration of each and find
relative acceleration along y direction.
Solution :
2 mg sin37° – mg sin 53° = 3 ma
23
5
ga
2
15
ga
21 3 4
2 5 5a t h⎛ ⎞ ⎜ ⎟⎝ ⎠
210 7
15 5t h
⎛ ⎞ ⎜ ⎟⎝ ⎠
214
15
th
12. Answer (12)
Hint : Apply concept of similar triangle
Solution :
h Y
x
�
hY
x �
2
dY hv
dt x �
4hv
⎛ ⎞ ⎜ ⎟⎝ ⎠�
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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13. Answer (15)
Hint : Write down the dimension of each and
compare.
Solution :
= M1L–3
� = M0L1T0
V = M0L1T–1
d = M0L1T0
= M1L–1T–1
P = ML–1T–2
14. Answer (03)
Hint : Length of string remains constant.
Solution :
2 22( ) ( )L h y h x
0dL
dt
2 2
1
2
A
B
xvV
h x
0
3
1 4
52
4
v
h
0
3
10v
15. Answer (A)
Hint : Write down the dimension and compare.
Solution :
dvF A
dx
1F A T
1
2
vA
A
1
2f A v
b c1 3 1 1 2 2 1 1
ML MLT L LT ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a = 1
c = –2
b = –1
16. Answer (A)
Hint : Initial static friction will act and then kinetic
friction will act.
Solution :
fs = applied force
fk = N
N will decrease linearly with time.
17. Answer (B)
Hint : Write down vAB
which is constant
Solution :
At the time of minimum separation vAB
is
perpendicular to relative separation.
(4, 3)
37°53°
vA
8°
(3, –4)B
10 2 sin(8)l
0
10 2 cos(8)
5t
2 2 cos8
18. Answer (B)
Hint : Relative motion will start when s
f f
Solution :
(f1)max
= × 1 × 10
= 10
Write down the equation of motion of each.
A Ba a also
maxsf f
Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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PART - II (CHEMISTRY)
19. Answer (A, C, D)
Hint : In T.B.P. geometry, more electronegative atom
attached at axial position.
Solution :
P
B
B
A
A
A
On replacementof A by C
0 = 0
P
C
C
C
B
B
On replacementof A by C P
C
C
B
B
B
= 0
P
B
B
B
A
A 0
20. Answer (A, B, C, D)
Hint : N – N x
H
H
H
H
N – N y
F
F
F
F
y < x
as E.N atom has 'p' character so 's' character in
N–N bond and bond length decrease.
Solution : N2H
4 > N
2F
4 i.e.,
NF N
F
F
F
H
N
H
N H
H
<
1. It can be explained on the basis of Bent's rule. In
N2F
4, N – N bond has more s-character hence
bond strength increases and bond length
decreases.
While in N2H
4, N – N bond has less s-character
hence bond length increases.
2. As the positive charge increases (Zeff
) nuclear
charge increases.
3. As the negative charge increases, ionic radius
increases.
21. Answer (A, B)
Hint : In NO, HOMO is 2 = 2p px y
* * while in N2 it is
2pz.
Solution : In NO, HOMO is 2 = 2p px y
* * while in N2
it is 2pz.
22. Answer (A, D)
Hint : 2H2 + O
2
At 100ºC 2H2O
at 100ºC H2O exist as water.
Solution :
At 100ºC
2 2 2(g) (g) (g)
2H O 2H O
Before reaction 2 2 0
volume in litre
After reaction 0 1 2
volume in litre
(A) Mole of H2O formed
2PV 1 2n 8.9 10 mole
RT 0.0821 273
Mass of H2O formed = 8.9 × 10–2 × 18
= 160.2 × 10–2 g = 1.602 g
(B) O2 is left as unreacted = 1 litre
At STP mole of O2 left
= 21
4.4 10 mole22.4
Mass of O2 left = 4.4 × 10–2 × 32 = 1.408 g
(C) At 100ºC H2O also exists as vapours
Total mole present at 100ºC
2 2
H O On n left
2 2 28.9 10 4.4 10 13.3 10 moles
Volume of the vessel = 2 L
2nRT 13.3 10 0.0821 373
P 2.03 atmV 2
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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(D) Volume of O2 used for formation of H
2O = 1 L
2
2
O
1n 4.4 10 mole
22.4
23. Answer (A)
Hint : 2
aP V b RT
V
⎛ ⎞ ⎜ ⎟⎝ ⎠
Solution : If a = 0 if b = 0
Z = Pb
1RT
Z = aP
1RT
24. Answer (D)
Hint : If labelling of oleum is (100 + x)% the free SO3 (in g)
is x 80
18
Solution : Initial moles of free SO3 present in oleum
= No. of moles H2O used
12 2moles
18 3
Moles of free SO3 combined with H
2O after 4.5 g of
H2O added
3SO
4.5 1n mole
18 4
Moles of free SO3 remains
2 1 5
3 4 12 moles.
Volume of free SO3 at STP
522.4 9.33 L
12
25. Answer (03)
Hint : If labelling of H2SO
4 is (100 + x)% then mole
of SO3
x
18
mol of H2SO
4 =
x100 80
18
98
⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Solution : 109% labelled oleum means it has 9 g
H2O which reacts with free SO
3 to give H
2SO
4 i.e.
SO3 + H
2O H
2SO
4.
Mass of SO3 which will react with 8 g
H2O
80 940 g
18
Let 100 g sample of 109% labelled oleum contains
40 g SO3 and 60 + 38 = 98 g of H
2SO
4.
Moles of H2SO
4 = x = 1.0 mole
Moles of SO3 = y = 0.5 mole
x + y = 1.5 moles
x – y = 0.5 mole
x y
x y
= 03
26. Answer (15)
Hint : For photoelectric effect.
Ein = E
Thr + K.E.
Solution :
According to photoelectric effect.
E = E0 + K.E.
h = h0 + K.E.
h1
= K.E1
+ h
0.....(i)
h2
= K.E2
+ h
0.....(ii)
2h1
– 2h2 = h
0(∵ 2 K.E
1 = K.E
2)
(21
– 2) =
0
0
= 1 2 1 2
2c c 2 1c
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
0
=
8
10
3 10 2 1
2000 180010
⎡ ⎤⎢ ⎥⎣ ⎦
(∵ 1 Å = 10–10 m).
0
= 1.33 × 1015 s–1.
27. Answer (06)
Hint : Number of spectral line = n (n 1)
2
Solution :
2
n 2
13.6 ZE eV
n
Also E = En – E
0
(E = Ionisation potential)
Energy absorb by 1 atom 51
12.754
2
13.612.75 ( 13.6)
n
2
13.612.75 13.6
n
n2 = 16 n = 4
Thus, the number of spectral lines emitted during the
de-excitation will be n(n 1) 4(4 1)
62 2
Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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28. Answer (20)
Hint & Solution :
10.50 0.39Al 0.39 1
27 0.39 ⇒
15.1 0.39K 0.39 1
39 0.39 ⇒
24.96 0.78S 0.78 2
32 0.39 ⇒
49.92 3.12O 3.12 8
16 0.39 ⇒
The empirical formula of the salt is KAlS2O
8
Molecular formula of hydrated salt is KAlS2O
8 yH
2O
When
18 y100 45.6
39 27 128 64 18y
y = 12
So the value of x + y = 8 + 12 = 20
29. Answer (40)
Hint : PM = dRT
Molar mass of mixture = M
Solution :
dRT 600 1 0.08 300P
M 760 M
⇒
M = 30.4 gm/mole
mol of N2 = x, mole of O
2 = 1 – x
30.4 = 32 (1 – x) + 28 (x)
x = 0.4
30. Answer (43)
Hint :
Rate of diffusion
1
Molar mass
Solution : Now for diffusion of gaseous mixture and
pure O2.
2
2
O mix
mix O
r M
r M
(or) 2
2 2
O M Mix
M O O
V T M
V T M
mixM1 232
1 200 32
Mmix
= 43
31. Answer (65)
Hint : For van der Waal's gas
2
2
n aP (V nb) nRT
V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Solution :
Given n = 4 moles, V = 1 L, T = 37 + 273 = 310 K
a = 3.592 : b = 0.0427
using van der Waal's equation for 4 moles
2
2
n aP (V nb) nRT
V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
24 3.592P (1 4 0.0427) 4 0.0821 310
1
⎛ ⎞ ⎜ ⎟⎝ ⎠
P = 65.3 65 atm
32. Answer (98)
Hint : Mass of solution = (v × d)solution
= 1000 × 1.26
Mass of H2SO
4 = 10.15 × 98
% H2SO
4 =
10.15 98 100
1000 1.26
Solution :
% by mass = 79
Molarity = 10.15
d = 1.26 gm/ml
M =
%by mass 10 d 79 10 1.2610.15
Molar mass Molar mass
⇒
Molar Mass = 98
33. Answer (C)
Hint : Covalent character decrease as size of cation
increase.
Solution : Number of d – p = P4O
10 > SO
3 > SO
2
4 2 1
Number of unpaired electron : 1 2
2 2 2O O O
Covalent character : 2 2 2
MgCl CaCl SrCl
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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34. Answer (D)
Hint & solution :
N
SiH3
SiH3
H Si3
sp2
sp3
2NO sp
, sp2
3NO
sp sp3 3
3 3N(CH ) ,
sp d sp3 3
6 4PCl , PCl
35. Answer (A)
Hint & solution :
BF3 Non-polar, planar
N3H Polar, planar
NH3 Polar, non-planar
36. Answer (B)
Hint : Radius =
2
0
na
Z
⎛ ⎞⎜ ⎟⎝ ⎠
Separation energy =
2
0 2
ZE
n
Solution : Radius of 4th orbit of Be+3
=
2
0 0
4a 4a
4
Separation energy of electron in 2nd orbit of Li2+ ion
=
2
0 02
3 9E E
42
PART - III (MATHEMATICS)
37. Answer (C, D)
Hint : 8pq = P
Solution :
2 4 68 sin sin sin
7 7 7pq
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 3sin sin sin
7 7 7p
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1
8q
Let 2 3
cos , cos , cos7 7 7
a b c ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠,
1
8abc
2 2 4 68 1 cos 1 cos 1 cos
7 7 7p
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= 2 3
1 cos 1 cos 1 cos7 7 7
⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
= (1 – b) (1 + c) (1 + a)
= 1 + a – b + c + ac – ab – bc – abc
ac – ab – bc = 1 3 2
2cos cos 2cos cos2 7 7 7 7
⎡ ⎢⎣
2 32cos cos
7 7
⎤ ⎥⎦
= 1 4 2 3
cos cos cos cos2 7 7 7 7
⎡ ⎢⎣
5cos cos
7 7
⎤ ⎥⎦
=
2 3cos cos cos
7 7 7b a c
a – b + c + ac – ab – bc = 0
8p2
=
1 7 71
8 8 8p ⇒
38. Answer (B, C)
Hint : Write cos7 in terms of cos
Solution :
∵ cos7 = 64cos7 – 112cos5 + 56 cos3 – 7cos
cos (7) = cos (2)
7 = 2n ± 2
Taking + sign
= 2
5
n
= 2 4
0, , ,...5 5
for +ve values
Taking – sign
= 2
9
n
= 2 4
0 , , ,...9 9
for +ve values
Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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39. Answer (A, B)
Hint : 2
sinlog (cos ) 0, 1
xx
Solution :
y
5 –1
2y
2x
O
y x= sin
2
For intersection points of f(x) and g(x),
logsinx
(cos2x) = sgn(tan x)
If tan x = 0, cos2x = 1
sinx = 0, no solution
if tanx > 0, cos2x = sinx sin2x + sinx – 1 = 0
sinx = 1 1 4
2
sinx =
1 5
2
sinx = 5 1
2
Only one solution
If tanx < 0, cos2 x = cosec x, (no solution)
40. Answer (A, D)
Hint : Put z = x + iy
Solution :
Put z = x + iy; x, y R in 2z z
z
z z
2 2
2 2
2 2
22 2( )
x yz zz x iy
zz x y
⇒
z1 = 1
Put z = x + iy in 1
Re( ) 2z z z i
z z
1( ) ( ) 2
( ) ( )x x iy x iy i
x iy x iy
x = 0, 2 22
iiy i
y 4y
2 – 4y + 1 = 0
1
2y
41. Answer (B, C)
Hint : If coefficients are real, then complex roots
exists in pair which are conjugate to each other.
Solution :
x2 + bx + c = 0
roots are 1 + i and 1 – i
b = –2, c = 2
Six quadratic equations are possible
2– 2 2 0x x ...... (1)
22 – 2 0x x ...... (2)
22 – 2 0x x .......(3)
22 – 2 1 0x x .......(4)
2–2 2 0x x .......(5)
2–2 2 1 0x x .......(6)
(1) & (4) have imaginary roots. (2), (3), (5), (6) have
irrational roots.
42. Answer (B, C)
Hint : Draw graph
Solution :
–2 –1 O 1 2 3 4
y = 2
y
f x = |x – 2x |( ) – 12
2g x( ) 2 sin x
⎛ ⎞ ⎜ ⎟
⎝ ⎠
x
2 2 + 1– 2 + 1
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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43. Answer (81)
Hint : Make perfect square
Solution :
2 3sin 2cos
2a a
= 2 23 1
1 cos 2cos (cos 1)2 2
a a a
cos2x + 4sinx + 12 = 1 – sin2
x + 4 sinx + 12
= 17 – (sinx – 2)2
f(x) is maximum when cos a = –1 and sin x = –1
f(x)max
=
1ln
12–
ln8 3
= 1
–3
–4 = 81
44. Answer (02)
Hint : Case 1 : Both roots are negative
Case 2 : Exactly one root is negative
Solution :
D = (– 3)2 – 4= 2 – 9 – 6– 4= (– 5)2 – 16
= (– 9) (– 1)
D 0, (–, 1] [9, )
Case-1: If both roots are negative
D 0, –3 < > 0 (0, 1]
Case-2: If exactly one root is negative and other root
+ve or zero.
D 0, 0 (–, 0]
(–, 0] (0, 1]
(–, 1]
45. Answer (01)
Hint : Draw graphs
Solution :
y
(0, 1)
(1, 0)
(0, –1)
(–1, 0) Ox
A B = P B (A ) = { } ( (A )) = 1n P B
(1, 1)(–1, 1)
x
(–1, –1)
y
(1, –1)
46. Answer (00)
Hint : cos 0, 12x
⎛ ⎞ ⎜ ⎟⎝ ⎠
Solution :
When 1, 3, 5, ..... cos 02
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
When 0, 2, 4, ..... cos 12
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
there is no integral value of x belonging to the
domain of f(x)
47. Answer (11)
Hint : ( ) 4sin 3cos6 6
f x x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Solution :
cos sin3 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∵ and
sin cos3 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
f(x) = 4sin 3cos6 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
–5 f(x) 5
48. Answer (11)
Hint : Solve for sin x
Solution :
4(1– sin2x)sinx – 2sin2
x = 3sinx
4sinx – 4sin3x – 2sin2
x = 3sinx
sinx = 0 OR 4sin2x + 2sinx – 1 = 0
sinx = 2 4 16 1 5
8 4
y
5 –1
4y
2x
O–2
5 –1
4y –
49. Answer (31)
Hint : If z = x + iy has negative principal argument
then x > 0, y < 0 or x < 0, y < 0
Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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Solution :
z = (3 – 42 + 2) + i (2 – 18 – 175)
z has negative principal argument
3 – 42 + 2 < 0 and 2 – 18 – 175 < 0
OR 3 – 42 + 2 > 0 and 2 – 18 – 175 < 0
2 – 18 – 175 < 0
2 – 25 – 7 – 175 < 0
( – 25) + 7( – 25) < 0 (–7, 25)
50. Answer (01)
Hint : Solve for z
Solution :
z4 + z3 + 2z
2 + z + 1 = 0
z4 + z2 + z3 + z + z2 + 1 = 0
z2(z2 + 1) + z(z2 + 1) + 1(z2 + 1) = 0
(z2 + 1)(z2 + z + 1) = 0
z2 + 1 = 0 OR z2 + z + 1 = 0
z = ± i OR z = 1 1 4 1 3
2 2
i
z1
= i, z2
= –i, z3
= 1 3
2
i , z
4
= 1 3
2
i
4
4 4 4 4
1 2 3 4
1 31 1 2Re
2
iz z z z
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
1 31 1 2Re
2 2
i⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
= 1 + 1 – 1 = 1
51. Answer (C)
Hint : Separate coefficients of a, b and c
Solution :
(P)
2
01 1
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
1 1
xx
x
⇒ ⇒
(Q)
2
1 10
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1
1
xx
x
⇒ ⇒
(R)
2
01 1
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
1 1
xx
x
⇒ ⇒
(S)
2
1 10
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1
1
xx
x
⇒ ⇒
52. Answer (B)
Hint : Use properties of functions to find domain
Solution :
(P) 1 – x2 0 and 10 – x2 0 1 x2 10
Integral values of x are ± 1, ± 2, ± 3
(Q) 4[x] – [x]2 – 1 0 [x]2 – 4[x] + 1 0
([x] – 2)2 3
3 [ ] 2 3 2 3 [ ] 3 2x x ⇒
Integral values of x are 1, 2, 3
(R) x3 + 1 > 0,
(x + 1)(x2 – x + 1) > 0
x > –1
x3 + 1 1,
x 0
2x – x2 + 9 > 1
x2 – 2x
– 8 < 0
(x – 1)2 < 9
–3 < x – 1 < 3 –2 < x < 4 and x > –1
–1 < x < 4 and x 0
Integral values of x are 1, 2, 3
(S) 34{f(x)} = 36 – 2x 1 36 – 2x < 34
∵ {f(x)} [0, 1)
2x 35 .....(i)
and
2x > 2 .....(ii)
Integral values of x satisfying both (i) & (ii) are
2, 3, 4, 5
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)
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53. Answer (D)
Hint : Find general solution
Solution :
(P) sin3x – sinx = 0
sinx (sin2x – 1) = –1
cos2x = cosec x
(Not possible)
(Q) cos3x – cosx +1 = 0
cosx (cos2x – 1) = –1
sin2x = sec x
(Not possible)
(R) tanx tan2x – 1 = 0
tan2x = cot x = tan2
x⎛ ⎞⎜ ⎟
⎝ ⎠
2x = 2
n x
(2 1)
,6
nx n I
5 7 3 11, , , , ,
6 2 6 6 2 6x
But 3
,2 2
x
(S) tan2x + sin2
x – sec2x – sinx – 1 = 0
sin2x – sin x – 2 = 0
sin x = – 1, sin x = 2
When sin x = –1, sec x is not defined.
54. Answer (C)
Hint : If cos(+) = 1 then sin(+) = 0
Solution :
z1z
2 = cos( + ) + i sin( + )
Re(z1z
2) = cos( + ) = 1
+ = 2 = 2 –
z1 =
cos – i sin , z
2 = cos + i sin
� � �
Test - 1A (Paper - 2) (Code-H) (Answers) All India Aakash Test Series for JEE (Advanced)-2020
1/12
PHYSICS
1. (A, B)
2. (A, D)
3. (A, B, D)
4. (A, B)
5. (A, B, D)
6. (A, B, C)
7. (03)
8. (15)
9. (12)
10. (02)
11. (15)
12. (12)
13. (16)
14. (18)
15. (B)
16. (B)
17. (A)
18. (A)
CHEMISTRY
19. (D)
20. (A)
21. (A, D)
22. (A, B)
23. (A, B, C, D)
24. (A, C, D)
25. (98)
26. (65)
27. (43)
28. (40)
29. (20)
30. (06)
31. (15)
32. (03)
33. (B)
34. (A)
35. (D)
36. (C)
MATHEMATICS
37. (B, C)
38. (B, C)
39. (A, D)
40. (A, B)
41. (B, C)
42. (C, D)
43. (01)
44. (31)
45. (11)
46. (11)
47. (00)
48. (01)
49. (02)
50. (81)
51. (C)
52. (D)
53. (B)
54. (C)
Test Date : 18/11/2018
ANSWERS
TEST - 1A (Paper-2) - Code-H
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)
2/12
PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (A, B)
Hint : Acceleration of block B with respect to A is
2
0v
�
Solution :
T – 0.05 (2 mg) = 2 ma0
T + ma0
=
2mv
�
22
T mgT mg
⎛ ⎞ ⎜ ⎟⎝ ⎠
T
ma0
5
3
mgT
03
ga
2. Answer (A, D)
Hint : Draw FBD and write equation of motion with
constrain relation
Solution :
2mg – T1 = 2ma
1
T1 – T
2 = 2ma
1
2
02
Tma
2a1 = a
0
1
4
ga
3. Answer (A, B, D)
Hint : Draw F.B.D. and equation of motion. Write
down the constraint motion.
Solution :
T cos 37° = ma1
3mg – T = 3ma2
1 2
4
5a a⎛ ⎞ ⎜ ⎟⎝ ⎠
1
5
4T ma
3mg – T = 1
12
5ma
3mg = 1
73
20ma
1
60
73
ga
4. Answer (A, B)
Hint : Velocity of cylinder along normal relative to
connected bodies is zero.
Solution :
3
ˆ ˆ
v xi yj �
x = +10
2 2
ˆ
v v i�
v2
v2
sin 37° = x sin 37° + y cos 37°
23 4
65 5
v y
3v2
= 30 + 4y
5. Answer (A, B, D)
Hint : Write down the expression for tangential
acceleration and acceleration in y-direction. Use the
fact that aj = a
t
Solution :
( sin )tdv
g gdt
( sin )y
dvg g
dt
v
d
ytdvdv
dt dt
vT – v
0 = –v
0sin v
T – v
0 =
03
5
v
vT =
02
5
v
6. Answer (A, B, C)
Hint : Write down the x coordinate of A3 in terms of
� and . dx
vdt
Solution :
x = 2(3l cos + 2l cos + l cos )
x = 12l cos
Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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v = 1
12 sinsin
d dl
dt dt
⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠ �
2 3B
dv l
dt
⎛ ⎞ ⎜ ⎟⎝ ⎠
6 m/s4sin
v
7. Answer (03)
Hint : Length of string remains constant.
Solution :
2 22( ) ( )L h y h x
0dL
dt
2 2
1
2
A
B
xvV
h x
0
3
1 4
52
4
v
h
0
3
10v
8. Answer (15)
Hint : Write down the dimension of each and
compare.
Solution :
= M1L–3
� = M0L1T0
V = M0L1T–1
d = M0L1T0
= M1L–1T–1
P = ML–1T–2
9. Answer (12)
Hint : Apply concept of similar triangle
Solution :
h Y
x
�
hY
x �
2
dY hv
dt x �
4hv
⎛ ⎞ ⎜ ⎟⎝ ⎠�
10. Answer (02)
Hint : Apply NSL to get acceleration of each and find
relative acceleration along y direction.
Solution :
2 mg sin37° – mg sin 53° = 3 ma
23
5
ga
2
15
ga
21 3 4
2 5 5a t h⎛ ⎞ ⎜ ⎟⎝ ⎠
210 7
15 5t h
⎛ ⎞ ⎜ ⎟⎝ ⎠
214
15
th
11. Answer (15)
Hint : Direction of friction will change, hence
magnitude of acceleration
Solution :
2
16 m/s
2
Mg Mga
M
0
1
1
vt
a
2
0
12
vs
a
2
2 0
2 2
1
1
2 2
va t
a
0
2
1 2
vt
a a
0
1 1 2
1 1vT
a a a
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎣ ⎦
0
1 2
1 2
v
a a
a a
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)
4/12
12. Answer (12)
Hint : Use addition of vector, net force will be along
PO.
Solution :
2
202 1
10k OP
����
= 40
k = 2 N/m
13. Answer (16)
Hint : Use the concept that
2
c
va
R
Solution :
216 m/s
dy
dt
2
2
216 m/s
x
d xa
dt
2
16
x
vR
a
14. Answer (18)
Hint : Along an incline maximum range is
2
(1 sin )
u
g
Solution :
2
10(1 sin )
u
g
2 4100 1
5u
⎛ ⎞ ⎜ ⎟⎝ ⎠
u2 = 20 × 9
15. Answer (B)
Hint : Relative motion will start when s
f f
Solution :
(f1)max
= × 1 × 10
= 10
Write down the equation of motion of each.
A Ba a also
maxsf f
16. Answer (B)
Hint : Write down vAB
which is constant
Solution :
At the time of minimum separation vAB
is
perpendicular to relative separation.
(4, 3)
37°53°
vA
8°
(3, –4)B
10 2 sin(8)l
0
10 2 cos(8)
5t
2 2 cos8
17. Answer (A)
Hint : Initial static friction will act and then kinetic
friction will act.
Solution :
fs = applied force
fk = N
N will decrease linearly with time.
18. Answer (A)
Hint : Write down the dimension and compare.
Solution :
dvF A
dx
1F AT
1
2
vA
A
1
2f A v
b c1 3 1 1 2 2 1 1
ML MLT L LT ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a = 1
c = –2
b = –1
Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
5/12
PART - II (CHEMISTRY)
19. Answer (D)
Hint : If labelling of oleum is (100 + x)% the free SO3 (in g)
is x 80
18
Solution : Initial moles of free SO3 present in oleum
= No. of moles H2O used
12 2moles
18 3
Moles of free SO3 combined with H
2O after 4.5 g of
H2O added
3SO
4.5 1n mole
18 4
Moles of free SO3 remains
2 1 5
3 4 12 moles.
Volume of free SO3 at STP
522.4 9.33 L
12
20. Answer (A)
Hint : 2
aP V b RT
V
⎛ ⎞ ⎜ ⎟⎝ ⎠
Solution : If a = 0 if b = 0
Z = Pb
1RT
Z = aP
1RT
21. Answer (A, D)
Hint : 2H2 + O
2
At 100ºC 2H2O
at 100ºC H2O exist as water.
Solution :
At 100ºC
2 2 2(g) (g) (g)
2H O 2H O
Before reaction 2 2 0
volume in litre
After reaction 0 1 2
volume in litre
(A) Mole of H2O formed
2PV 1 2n 8.9 10 mole
RT 0.0821 273
Mass of H2O formed = 8.9 × 10–2 × 18
= 160.2 × 10–2 g = 1.602 g
(B) O2 is left as unreacted = 1 litre
At STP mole of O2 left
= 21
4.4 10 mole22.4
Mass of O2 left = 4.4 × 10–2 × 32 = 1.408 g
(C) At 100ºC H2O also exists as vapours
Total mole present at 100ºC
2 2
H O On n left
2 2 28.9 10 4.4 10 13.3 10 moles
Volume of the vessel = 2 L
2nRT 13.3 10 0.0821 373
P 2.03 atmV 2
(D) Volume of O2 used for formation of H
2O = 1 L
2
2
O
1n 4.4 10 mole
22.4
22. Answer (A, B)
Hint : In NO, HOMO is 2 = 2p px y
* * while in N2 it is
2pz.
Solution : In NO, HOMO is 2 = 2p px y
* * while in N2
it is 2pz.
23. Answer (A, B, C, D)
Hint : N – N x
H
H
H
H
N – N y
F
F
F
F
y < x
as E.N atom has 'p' character so 's' character in
N–N bond and bond length decrease.
Solution : N2H
4 > N
2F
4 i.e.,
NF N
F
F
F
H
N
H
N H
H
<
1. It can be explained on the basis of Bent's rule. In
N2F
4, N – N bond has more s-character hence
bond strength increases and bond length
decreases.
While in N2H
4, N – N bond has less s-character
hence bond length increases.
2. As the positive charge increases (Zeff
) nuclear
charge increases.
3. As the negative charge increases, ionic radius
increases.
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)
6/12
24. Answer (A, C, D)
Hint : In T.B.P. geometry, more electronegative atom
attached at axial position.
Solution :
P
B
B
A
A
A
On replacementof A by C
0 = 0
P
C
C
C
B
B
On replacementof A by C P
C
C
B
B
B
= 0
P
B
B
B
A
A 0
25. Answer (98)
Hint : Mass of solution = (v × d)solution
= 1000 × 1.26
Mass of H2SO
4 = 10.15 × 98
% H2SO
4 =
10.15 98 100
1000 1.26
Solution :
% by mass = 79
Molarity = 10.15
d = 1.26 gm/ml
M =
%by mass 10 d 79 10 1.2610.15
Molar mass Molar mass
⇒
Molar Mass = 98
26. Answer (65)
Hint : For van der Waal's gas
2
2
n aP (V nb) nRT
V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
Solution :
Given n = 4 moles, V = 1 L, T = 37 + 273 = 310 K
a = 3.592 : b = 0.0427
using van der Waal's equation for 4 moles
2
2
n aP (V nb) nRT
V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
24 3.592P (1 4 0.0427) 4 0.0821 310
1
⎛ ⎞ ⎜ ⎟⎝ ⎠
P = 65.3 65 atm
27. Answer (43)
Hint :
Rate of diffusion
1
Molar mass
Solution : Now for diffusion of gaseous mixture and
pure O2.
2
2
O mix
mix O
r M
r M
(or) 2
2 2
O M Mix
M O O
V T M
V T M
mixM1 232
1 200 32
Mmix
= 43
28. Answer (40)
Hint : PM = dRT
Molar mass of mixture = M
Solution :
dRT 600 1 0.08 300P
M 760 M
⇒
M = 30.4 gm/mole
mol of N2 = x, mole of O
2 = 1 – x
30.4 = 32 (1 – x) + 28 (x)
x = 0.4
29. Answer (20)
Hint & Solution :
10.50 0.39Al 0.39 1
27 0.39 ⇒
Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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15.1 0.39K 0.39 1
39 0.39 ⇒
24.96 0.78S 0.78 2
32 0.39 ⇒
49.92 3.12O 3.12 8
16 0.39 ⇒
The empirical formula of the salt is KAlS2O
8
Molecular formula of hydrated salt is KAlS2O
8 yH
2O
When
18 y100 45.6
39 27 128 64 18y
y = 12
So the value of x + y = 8 + 12 = 20
30. Answer (06)
Hint : Number of spectral line = n (n 1)
2
Solution :
2
n 2
13.6 ZE eV
n
Also E = En – E
0
(E = Ionisation potential)
Energy absorb by 1 atom 51
12.754
2
13.612.75 ( 13.6)
n
2
13.612.75 13.6
n
n2 = 16 n = 4
Thus, the number of spectral lines emitted during the
de-excitation will be n(n 1) 4(4 1)
62 2
31. Answer (15)
Hint : For photoelectric effect.
Ein = E
Thr + K.E.
Solution :
According to photoelectric effect.
E = E0 + K.E.
h = h0 + K.E.
h1
= K.E1
+ h
0.....(i)
h2
= K.E2
+ h
0.....(ii)
2h1
– 2h2 = h
0(∵ 2 K.E
1 = K.E
2)
(21
– 2) =
0
0
= 1 2 1 2
2c c 2 1c
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
0
=
8
10
3 10 2 1
2000 180010
⎡ ⎤⎢ ⎥⎣ ⎦
(∵ 1 Å = 10–10 m).
0
= 1.33 × 1015 s–1.
32. Answer (03)
Hint : If labelling of H2SO
4 is (100 + x)% then mole
of SO3
x
18
mol of H2SO
4 =
x100 80
18
98
⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Solution : 109% labelled oleum means it has 9 g
H2O which reacts with free SO
3 to give H
2SO
4 i.e.
SO3 + H
2O H
2SO
4.
Mass of SO3 which will react with 8 g
H2O
80 940 g
18
Let 100 g sample of 109% labelled oleum contains
40 g SO3 and 60 + 38 = 98 g of H
2SO
4.
Moles of H2SO
4 = x = 1.0 mole
Moles of SO3 = y = 0.5 mole
x + y = 1.5 moles
x – y = 0.5 mole
x y
x y
= 03
33. Answer (B)
Hint : Radius =
2
0
na
Z
⎛ ⎞⎜ ⎟⎝ ⎠
Separation energy =
2
0 2
ZE
n
Solution : Radius of 4th orbit of Be+3
=
2
0 0
4a 4a
4
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)
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PART - III (MATHEMATICS)
37. Answer (B, C)
Hint : Draw graph
Solution :
–2 –1 O 1 2 3 4
y = 2
y
f x = |x – 2x |( ) – 12
2g x( ) 2 sin x
⎛ ⎞ ⎜ ⎟
⎝ ⎠
x
2 2 + 1– 2 + 1
38. Answer (B, C)
Hint : If coefficients are real, then complex roots
exists in pair which are conjugate to each other.
Solution :
x2 + bx + c = 0
roots are 1 + i and 1 – i
b = –2, c = 2
Separation energy of electron in 2nd orbit of Li2+ ion
=
2
0 02
3 9E E
42
34. Answer (A)
Hint & solution :
BF3 Non-polar, planar
N3H Polar, planar
NH3 Polar, non-planar
35. Answer (D)
Hint & solution :
N
SiH3
SiH3
H Si3
sp2
sp3
2NO sp
, sp2
3NO
sp sp3 3
3 3N(CH ) ,
sp d sp3 3
6 4PCl , PCl
36. Answer (C)
Hint : Covalent character decrease as size of cation
increase.
Solution : Number of d – p = P4O
10 > SO
3 > SO
2
4 2 1
Number of unpaired electron : 1 2
2 2 2O O O
Covalent character : 2 2 2
MgCl CaCl SrCl
Six quadratic equations are possible
2– 2 2 0x x ...... (1)
22 – 2 0x x ...... (2)
22 – 2 0x x .......(3)
22 – 2 1 0x x .......(4)
2–2 2 0x x .......(5)
2–2 2 1 0x x .......(6)
(1) & (4) have imaginary roots. (2), (3), (5), (6) have
irrational roots.
39. Answer (A, D)
Hint : Put z = x + iy
Solution :
Put z = x + iy; x, y R in 2z z
z
z z
2 2
2 2
2 2
22 2( )
x yz zz x iy
zz x y
⇒
z1 = 1
Put z = x + iy in 1
Re( ) 2z z z i
z z
1( ) ( ) 2
( ) ( )x x iy x iy i
x iy x iy
Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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x = 0, 2 22
iiy i
y 4y2 – 4y + 1 = 0
1
2y
40. Answer (A, B)
Hint : 2
sinlog (cos ) 0, 1
xx
Solution :
y
5 –1
2y
2x
O
y x= sin
2
For intersection points of f(x) and g(x),
logsinx
(cos2x) = sgn(tan x)
If tan x = 0, cos2x = 1
sinx = 0, no solution
if tanx > 0, cos2x = sinx sin2x + sinx – 1 = 0
sinx = 1 1 4
2
sinx =
1 5
2
sinx = 5 1
2
Only one solution
If tanx < 0, cos2 x = cosec x, (no solution)
41. Answer (B, C)
Hint : Write cos7 in terms of cos
Solution :
∵ cos7 = 64cos7 – 112cos5 + 56 cos3 – 7cos
cos (7) = cos (2)
7 = 2n ± 2
Taking + sign
= 2
5
n
= 2 4
0, , ,...5 5
for +ve values
Taking – sign
= 2
9
n
= 2 4
0 , , ,...9 9
for +ve values
42. Answer (C, D)
Hint : 8pq = P
Solution :
2 4 68 sin sin sin
7 7 7pq
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 3sin sin sin
7 7 7p
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1
8q
Let 2 3
cos , cos , cos7 7 7
a b c ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠,
1
8abc
2 2 4 68 1 cos 1 cos 1 cos
7 7 7p
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= 2 3
1 cos 1 cos 1 cos7 7 7
⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
= (1 – b) (1 + c) (1 + a)
= 1 + a – b + c + ac – ab – bc – abc
ac – ab – bc = 1 3 2
2cos cos 2cos cos2 7 7 7 7
⎡ ⎢⎣
2 32cos cos
7 7
⎤ ⎥⎦
= 1 4 2 3
cos cos cos cos2 7 7 7 7
⎡ ⎢⎣
5cos cos
7 7
⎤ ⎥⎦
=
2 3cos cos cos
7 7 7b a c
a – b + c + ac – ab – bc = 0
8p2
=
1 7 71
8 8 8p ⇒
43. Answer (01)
Hint : Solve for z
Solution :
z4 + z3 + 2z2 + z + 1 = 0
All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)
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z4 + z2 + z3 + z + z2 + 1 = 0
z2(z2 + 1) + z(z2 + 1) + 1(z2 + 1) = 0
(z2 + 1)(z2 + z + 1) = 0
z2 + 1 = 0 OR z2 + z + 1 = 0
z = ± i OR z = 1 1 4 1 3
2 2
i
z1
= i, z2
= –i, z3
= 1 3
2
i , z
4
= 1 3
2
i
4
4 4 4 4
1 2 3 4
1 31 1 2Re
2
iz z z z
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
1 31 1 2Re
2 2
i⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
= 1 + 1 – 1 = 1
44. Answer (31)
Hint : If z = x + iy has negative principal argument
then x > 0, y < 0 or x < 0, y < 0
Solution :
z = (3 – 42 + 2) + i (2 – 18 – 175)
z has negative principal argument
3 – 42 + 2 < 0 and 2 – 18 – 175 < 0
OR 3 – 42 + 2 > 0 and 2 – 18 – 175 < 0
2 – 18 – 175 < 0
2 – 25 – 7 – 175 < 0
( – 25) + 7( – 25) < 0 (–7, 25)
45. Answer (11)
Hint : Solve for sin x
Solution :
4(1– sin2x)sinx – 2sin2x = 3sinx
4sinx – 4sin3x – 2sin2x = 3sinx
sinx = 0 OR 4sin2x + 2sinx – 1 = 0
sinx = 2 4 16 1 5
8 4
y
5 –1
4y
2x
O–2
5 –1
4y –
46. Answer (11)
Hint : ( ) 4sin 3cos6 6
f x x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Solution :
cos sin3 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∵ and
sin cos3 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
f(x) = 4sin 3cos6 6
x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
–5 f(x) 5
47. Answer (00)
Hint : cos 0, 12x
⎛ ⎞ ⎜ ⎟⎝ ⎠
Solution :
When 1, 3, 5, ..... cos 02
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
When 0, 2, 4, ..... cos 12
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
there is no integral value of x belonging to the
domain of f(x)
48. Answer (01)
Hint : Draw graphs
Solution :
y
(0, 1)
(1, 0)
(0, –1)
(–1, 0) Ox
A B = P B (A ) = { } ( (A )) = 1n P B
(1, 1)(–1, 1)
x
(–1, –1)
y
(1, –1)
49. Answer (02)
Hint : Case 1 : Both roots are negative
Case 2 : Exactly one root is negative
Solution :
D = (– 3)2 – 4= 2 – 9 – 6– 4= (– 5)2 – 16
= (– 9) (– 1)
D 0, (–, 1] [9, )
Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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Case-1: If both roots are negative
D 0, –3 < > 0 (0, 1]
Case-2: If exactly one root is negative and other root
+ve or zero.
D 0, 0 (–, 0]
(–, 0] (0, 1]
(–, 1]
50. Answer (81)
Hint : Make perfect square
Solution :
2 3sin 2cos
2a a
= 2 23 1
1 cos 2cos (cos 1)2 2
a a a
cos2x + 4sinx + 12 = 1 – sin2x + 4 sinx + 12
= 17 – (sinx – 2)2
f(x) is maximum when cos a = –1 and sin x = –1
f(x)max
=
1ln
12–
ln8 3
= 1
–3
–4 = 81
51. Answer (C)
Hint : If cos(+) = 1 then sin(+) = 0
Solution :
z1z
2 = cos( + ) + i sin( + )
Re(z1z
2) = cos( + ) = 1
+ = 2 = 2 – z
1 =
cos – i sin , z
2 = cos + i sin
52. Answer (D)
Hint : Find general solution
Solution :
(P) sin3x – sinx = 0
sinx (sin2x – 1) = –1
cos2x = cosec x
(Not possible)
(Q) cos3x – cosx +1 = 0
cosx (cos2x – 1) = –1
sin2x = sec x
(Not possible)
(R) tanx tan2x – 1 = 0
tan2x = cot x = tan2
x⎛ ⎞⎜ ⎟
⎝ ⎠
2x = 2
n x
(2 1)
,6
nx n I
5 7 3 11, , , , ,
6 2 6 6 2 6x
But 3
,2 2
x
(S) tan2x + sin2x – sec2x – sinx – 1 = 0
sin2x – sin x – 2 = 0
sin x = – 1, sin x = 2
When sin x = –1, sec x is not defined.
53. Answer (B)
Hint : Use properties of functions to find domain
Solution :
(P) 1 – x2 0 and 10 – x2 0 1 x2 10
Integral values of x are ± 1, ± 2, ± 3
(Q) 4[x] – [x]2 – 1 0 [x]2 – 4[x] + 1 0
([x] – 2)2 3
3 [ ] 2 3 2 3 [ ] 3 2x x ⇒
Integral values of x are 1, 2, 3
(R) x3 + 1 > 0,
(x + 1)(x2 – x + 1) > 0
x > –1
x3 + 1 1,
x 0
2x – x2 + 9 > 1
x2 – 2x – 8 < 0
(x – 1)2 < 9
–3 < x – 1 < 3 –2 < x < 4 and x > –1
–1 < x < 4 and x 0
Integral values of x are 1, 2, 3
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� � �
(S) 34{f(x)} = 36 – 2x 1 36 – 2x < 34
∵ {f(x)} [0, 1)
2x 35 .....(i)
and
2x > 2 .....(ii)
Integral values of x satisfying both (i) & (ii) are
2, 3, 4, 5
54. Answer (C)
Hint : Separate coefficients of a, b and c
Solution :
(P)
2
01 1
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
1 1
xx
x
⇒ ⇒
(Q)
2
1 10
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1
1
xx
x
⇒ ⇒
(R)
2
01 1
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
1 1
xx
x
⇒ ⇒
(S)
2
1 10
x xa b c
x x
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1
1
xx
x
⇒ ⇒