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Test - 3A (Paper - 1) (Code-E) (Answers)
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A (Paper-1) - Code-E Test Date : 24/02/2019
ANSWERS PHYSICS
CHEMISTRY
1.
(A)
21.
(B)
41.
(C)
2.
(C)
22.
(D)
42.
(A)
3.
(B)
23.
(D)
43.
(B)
4.
(C)
24.
(A)
44.
(C)
5.
(A)
25.
(B)
45.
(B)
6.
(A)
26.
(B)
46.
(C)
7.
(A, C)
27.
(A, C, D)
47.
(A, B, C)
8.
(A, B, C)
28.
(B, C, D)
48.
(A, C)
9.
(B, C)
29.
(B, C, D)
49.
(B)
10.
(A, C)
30.
(A)
50.
(B)
11.
(B, C)
31.
(B)
51.
(C)
12.
(A, D)
32.
(A, C)
52.
(B)
13.
(A, B)
33.
(A, B, C)
53.
(C)
14.
(A, B)
34.
(A, B, C)
54.
(A)
15.
(B, C)
35.
(B, C)
55.
(B)
16.
A (R)
36.
56.
A (Q, S)
17.
A (P, T)
MATHEMATICS
B (Q, T)
B (P, Q, R, S, T)
B (Q, R, S, T)
C (Q, T)
C (P)
C (R, S, T)
D (P, S)
D (P, Q, R, S)
D (R, S, T)
A (Q, T)
37.
A (Q, R, T)
57.
A (R, T)
B (R)
B (P, S)
B (Q, R, T)
C (P)
C (Q, R, T)
C (Q, R, T)
D (Q, S, T)
D (Q, R, T)
D (Q, R, S, T)
18.
(03)
38.
(07)
58.
(04)
19.
(03)
39.
(06)
59.
(05)
20.
(09)
40.
(04)
60.
(02)
1/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
HINTS & SOLUTIONS PART - I (PHYSICS) 1.
Answer (A)
3.
Answer (B) Hint:
I Hint: T = 2 Mgd
Based on Doppler effect observer will observe the frequency of wave which is emitted by source.
Solution:
Solution:
M 2 M 2 I = 12 16
0 (v ) = (v v cos ) s
vs
cot v sin vs
7M 2 I= 48
vs = v cos T=
2
7M
2
48 Mg 4
2.
4.
Solution:
Hint: Add the total no. of mole by
dn
Pdv RT
Solution: T T0 T = T0 L x L
PA dx = dn RT n
PA dx dn R T T 0 L
TL PV n = (T T )R ln T L 0 0
2/10
Answer (C)
Calculate the effective phase difference at point P.
Answer (C)
T0
4 3 , cos = 5 5
Hint:
7 3g
using :
0
sin2
sin =
2 7 = 2 3g =
2 2 3 cos 3 6 6 cos
2 6 3
3 cos
1 3 3
8 cos 9 = cos–1 (8/9) Y tan D
17 Y 8 D
P
S1 S2
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
5.
12. Answer (A, D)
Answer (A) Hint:
Hint and Solution of Q. Nos. 10, 11 to 12
Temperature of each body will change based on their heat capacity.
Hint:
Solution: ms
dT1 kA (T2 T1 ) dt
2ms
t 6.
dT2 kA (T2 T1 ) dt
2ms 4 n 3kA 3
Answer (A)
Process for system will be adiabatic. More over work done by external force will be used to raise the internal energy. Solution: fext dx = (P1 – P2) dV = 2nCvdT 1 1 nRT dV 2nCv dT V0 V V0 V ( 1)
V
0 V2 0
T V dV ln 2 V T0
Hint:
13. Answer (A, B)
Heat flow through conductor is constant.
14. Answer (A, B)
Solution:
15. Answer (B, C)
Req
(b r ) 4k br
Hint and Solution of Q. Nos. 10, 11 to 12 Hint: Based on equation of travelling waves
dT b(T1 T2 ) (b r )r dr
Solution:
7.
Answer (A, C)
P = B
8.
Answer (A, B, C)
9.
Answer (B, C) Hint and Solution of Q. Nos. 7, 8 to 9 Hint: Draw the corresponding PV diagram and calculate corresponding quantities. Solution: P = kV
RT kV V RT = kV 2 2nV = nT 2
dV dT V T
2
dV 1 VdT T
10. Answer (A, C) 11. Answer (B, C)
dy dx
y vp = v s x P vp = v s B0 v s ap = B t 0
v s P0 = B x vs 0 0
ap
v s2 p0 B0 x 0
16. Answer A(R), B(Q, T), C(Q, T), D(P, S) Hint: Heat flow across section will be constant. Temperature gradient is inversely proportional to area. Solution: In all case heat current flow will be same. Use electrical analogy for heat circuit to compute temperature and gradient.
3/10
All India Aakash Test Series for JEE (Advanced)-2020 17. Answer A(Q, T), B(R), C(P), D(Q, S, T)
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) Solution:
Hint: Distance will be maximum when velocity of each is same. Solution:
31 4 2 1 =
This has to be solved using
1 = 600 Hz
Phasor diagram
A
3A
2Asin =
19. Answer (03) Hint: T = constant
= 3
Solution :
B
= 2 or 2 – 2 2 4 or 3 3
=
2 3
d 1 dT 3 T 20. Answer (09) Hint:
When they are at same position.
x = A sin 2
A0 2
=
A
B
2
Solution :
N 2
18. Answer (03)
N=8
Hint:
9 nodes
0 = 2
PART - II (CHEMISTRY) Solution:
21. Answer (B) Hint: R R
C = CH2
Reductive ozonolysis
R
C =O & H – C – H
Solution: CH 2
Non-aromatic
O
R
H B
Antiaromatic
CH 2
B CH 2
CH 2
O3 6HCHO Zn–H2 O
H
Aromatic CH 2
CH 2
22. Answer (D) Hint: Aromatic is non-aromatic compound as it is non-planar molecule.
4/10
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020 Solution:
23. Answer (D) Hint: Basic strength lone pair concentration.
CH = CH 2
Solution: (+I) Et
(+I) Et
N
NH 2
CH – CH 3
NH 2
>
>
CH – CH 3
HCl
Ph (–R) >
Cl
NH
Ph (–R)
NO2 (–I, –R)
Cl–
CH3
hydride ion shift
CH 3
Cl–1
ring expansion
24. Answer (A) Hint:
Cl Na/ dry ether
CH2 – (CH 2)4 – CH 2 Br
28. Answer (B, C, D)
Br
Hint: 1, 2 & 1, 4 addition of HCl
Solution: CH4
Solution:
Cl2/h
D CH3 – CH = CH 2
CH3 – Cl A alc KOH
NBS
Br – CH2 – CH = CH2 E
(Et) 2CuLi
CH 3 – CH – CH3 C Cl
Na, dry ether
Na/ dry ether
CH 3 – CH2 – CH3 B
CH3 – CH = CH – CH = CH 2
Cl2/h
CH3 – CH = CH – CH – CH 3
CH2 = CH – CH2 – CH 2 F CH = CH 2
CH 2 –(CH 2)4 – CH2 – Br G Br
Hint: Properties of canonical structure.
CH3 – CH = CH – CH – CH3 Cl– CH3 – CH – CH = CH – CH3 Cl
–
CH3 – CH – CH = CH – CH3 Cl 1, 4 addition product
29. Answer (B, C, D) Hint: Electrophilic addition reaction Solution:
Solution: One canonical structure will not explain all the properties of molecule. 26. Answer (B) Hint: H
Ring opening
CH 2 – CH = CH 2
Solution:
H
carbonium ion
rearrangment
CH2 – CH = CH 2
H
Cl (1-2-addition product)
HBr/ Proxide
25. Answer (B)
O
H
H
–H
H2 O
CH2 – CH = CH2
HBr & HCl both follow Markownikoff with carbonium ion rearrangement
Br2
Br
30. Answer (A) Hint: Br2/high temperature will do allylic substitution Solution: O–H
O
O–H
CH2 = CH – CH 2 – OH
H
27. Answer (A, C, D) Hint: Carbonium ion rearrangement.
5/10
All India Aakash Test Series for JEE (Advanced)-2020 OH
OH
OH
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 34. Answer (A, B, C) Hint:
Ring
Expansion
Deprotonation (–H +)
Electrophilic aromatic substitution Solution: (P)
CH2
NBS
OH
E
O More activating
CH2 O (major)
Br
(Q)
E
31. Answer (B) Hint: bonds which are not the part of aromatic ring, can undergo addition reaction. Solution:
35. Answer (B, C) Hint: Friedel Craft alkylation reaction
OH
OH
OH
alc KOH
Br 2
Br
Me
CH3 &
Br Br (Q)
Solution:
(R)
CH3
has greater electron Me
Me
(S) CH3
density than
32. Answer (A, C) Hint:
36. Answer A(P, T); B(P, Q, R, S, T); C(P), D(P, Q, R, S)
HO
Hint: Alkene decolourise Br2/CCl4.
is aromatic compound.
Solution: Solution:
+ H 2/Ni CH3 – CH 2 – CH3
OH
Compound R is
It is aromatic cyclic and stable Has acidic H-atom
,
CH2 &
, CH
33. Answer (A, B, C) Hint:
are unsaturated compound can give bromine water test
Electrophilic aromatic substitution
CH3 – C CH + CH 3MgBr
CH4 + CH3 – C C Mg Br
Solution: O
O – O – C – CH 3
(activating
(deactivating i.e. – R) –Cl deactivating –O–CH3 more activating –NO2 deactivating
6/10
+
R)
&
– C – CH 3
Can react with Na as it has acidic hydrogen atom
CH3 – C CH H
Hydrogen is acidic due to aromaticity
aromatic
–
+
–
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
37. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T)
39. Answer (06)
Hint:
Hint: Decarboxylation reaction
Reaction intermediate singlet carbene is sp 2 hybridised.
Solution: O
Solution:
O COOH
R1
COOH COOH
HOOC
C
COONa
NaOH NaOOC
R2
COONa COONa (salt) electrolysis
Singlet carbene hybridisation is sp2. and have a vacant 2p orbital hence can behave like an electrophile.
O
38. Answer (07) + 4 CO 2 (m = 4)
Hint: Kjeldahl method Solution:
Tautomerise
m.e of H2SO4 used for NH3
OH
= (25 × 2) – 30 × 0.5 × 2 = 50 – 30 = 20 No. of m.e of NH3 = 20 % mass of nitrogen in organic compound =
=
Hint: RMgX RH
(20)(14) 100 (1000)(4)
= 20
(DBE = 4)
2m d 6 2 40. Answer (04)
Solution:
1.4 4
C6H5 SO3H,
C 6H5OH
NO 2
2 14 7 4
OH ,
i.e 7
OH
NH2
PART - III (MATHEMATICS) Solution:
41. Answer (C) Hint: z z z – z 2
|x| + |y| = 1
|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2) | – z3|2 + |z1 – z2|2 = 2(2)
Solution:
zz z–z 2
|x| + |y| = 1
Locus of z is a square and z i z – i 2 Locus of z is a straight line z = –i satisfies the given equations
|z1 – z2|2 = 3 | z1 z2 | 3 Similarly | z2 z3 |2 | z3 z1 | 3 Thus, whose vertices are z1, z2 and z3 is equilateral triangle and its side is of length
3
3 4
3
42. Answer (A) Hint: Area =
3 2 a 4
Therefore, area of triangle
2
3 3 4
7/10
All India Aakash Test Series for JEE (Advanced)-2020 43. Answer (B)
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 47. Answer (A, B, C)
Hint : Number of exhaustive cases = 74
Hint:
Solution:
2 – 4 0 48. Answer (A, C)
Total no. of ways = (7)4 7
Hint:
4
Now no. of favourable case C2 (2 2) 21 14 Now probability of required event
21 14 (7)4
6 49
49. Answer (B) Hint: z –
44. Answer (C) Hint: greatest value of series when x1 = x2 = .....xn Solution: The value of the sum is greatest when x1 = x2 = ... xn
2 – 4 > 0
and required value is n sin n n
45. Answer (B)
Solution for Q. Nos. 47 to 49 z2 + z + 1 = 0 z
2 4 2
Case 1 –2 < < 2 2 – 4 < 0
Hint: sinx = cosecx and tany = coty
i 4 2 2
Solution:
z
sin x cosecx 2 and tany + coty 2
y 2 1 x2 2 x2 + y2 = 1
So, sinx = 1, tany = 1 , y 2 4
x
Case 2
y tan 2 1 which satisfies t2 + 2t – 1 = 0 2 46. Answer (C)
Put
n
sin t
1 lim 2 t 0 t (1 t 2 )
log l lim – sin t log(t 2 (1 t 2 )) t 0
2log(t ) log(1 t 2 ) t 0 cosect
lim
2 sin2 t =0l=1 t 0 t cos t
lim
8/10
2 – 4 > 0 2 4 2 4 x ,y 0 2 2
2 4 2 4 z1 ,0 , z2 ,0 2 2
t
1 4 = lim t t 0 1 2 1 t
>2
z
Solution:
1
x
sin t
One root lies inside the unit circle and other will lie outside the unit circle. Case 3 When is very large
2 4 z z 2
2 4 1 2 2 2 4 50. Answer (B) z
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 51. Answer (C)
All India Aakash Test Series for JEE (Advanced)-2020 Now | z |2
52. Answer (B) Hint: Use projection formula
cos( A B )
4 5
sin( A B )
3 3 , tan( A B ) 5 4
AB 1 tan 2 3
Hint:
(A)
z z i
2 1 | z |2
z = unimodular |z|
Radius of the circumcircle =
1 3 5 × hypotenuse 2 2
53. Answer (C)
z z 2i 2i |z| |z|
z unimodular complex no. |z|
1 6 3 = 9 sq. units 2
9 3 3 5 s 363 5 2 2
and lies on perpendicular bisector of 2i and –2i
z 1 z | z | a real number |z| Im (z) = 0
(B) = ei z
1 2 1 e 2i 2cos ei
(C) 2b = 8 a2 = 7
2be = 6
54. Answer (A) e=
55. Answer (B) 2 2 (1 )2
3 4
Equation of ellipse
x2 y 2 1 7 16
(D) Iz – 2 – 3i| = 1
Solution: | z |2
zz 1 | z |2
Solution:
So triangle ABC is right angle triangle
2
1 | z |2
56. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
ab C 1 cot ab 2 3
Hint: z
2 1 | z |2
2i ( z z ) (1 )
C 90
r
zz
zz
2i zz 1
( A B) 2 tan 2 3 AB 4 1 tan2 2
Area of ABC
1 | z |2
2 ( z z ) (1 )
AB is an acute angle ∵ 2
| z |2
Solution:
1 1 1 1 | z |2 1 1 1
( x 2)2 ( y 3)2 1
2 2 2
(1 )
2 2
(1 )
1
x 2 cos , y 3 sin x [1, 3], y [2, 4]
9/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 59. Answer (05)
57. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T) Hint: x – 1 x
Hint:
Solution: (A) lim(sin x cos x )cosecx e
z1 C is
sin x cos x 1 x 0 sin x lim
10
x 0
3 z2 C is 10
=e 4 n 10 (B) L lim n 5 Le
L L
lim n
n
(1 form)
Solution:
1 4 (10) n 5 n 5
z1 C is
3 z2 C is 10
1 ln(10) 5 e
3 3 z1 z2 cos cos + i sin sin 10 10 10 10
1 5 10
10
L5 2 5
10 2 5 10 2 5 1 z1 z2 i 2 4
xx 1 (C) xlim 0 2 L= 2
| z1 z2 | 2
5 x x x 3
(D) L lim
60. Answer (02) Hint:
x x x 3 1 3 3
|a – d|, |a|, |a + d| be three distances Solution:
5 x 5 x 5 x 1 x 3 x 3 x 3 lim
x
3 5 2
As perpendicular distances are in A.P. Let |a – d|, |a|, |a + d| be these distances
5 x 5 x 3 3
so the co-ordinates of point P are (±(a – d), ± a, ± (a + d)
58. Answer (04) Solution:
According to question
sin sin 2 sin3 0 sin2 (1 + 2cos) = 0 ..(1)
2a2 + d2 – 2ad = 5
...(i)
2a2 + 2d2 = 10
...(ii)
2
2
cos cos 2 cos 3 0 cos2 (1 + 2cos) = 0 ..(2)
2a + d + 2ad = 13
From (1) and (2)
we get
1 2 Number of values of [–2, 2] is 4
a = ±2, d = ±1
Solving (i), (ii) and (iii)
cos –
so points P (±1, ±2, ±3) Now = 1, = 2, = 3
10/10
...(iii)
Test - 3A (Paper - 1) (Code-F) (Answers)
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A (Paper-1) - Code-F Test Date : 24/02/2019
ANSWERS PHYSICS
CHEMISTRY
1.
(A)
21.
(B)
41.
(C)
2.
(A)
22.
(B)
42.
(B)
3.
(C)
23.
(A)
43.
(C)
4.
(B)
24.
(D)
44.
(B)
5.
(C)
25.
(D)
45.
(A)
6.
(A)
26.
(B)
46.
(C)
7.
(A, C)
27.
(A, C, D)
47.
(A, B, C)
8.
(A, B, C)
28.
(C, D)
48.
(A, C)
9.
(B, C)
29.
(B, C, D)
49.
(B)
10.
(A, C)
30.
(A)
50.
(B)
11.
(B, C)
31.
(B)
51.
(C)
12.
(A, D)
32.
(A, C)
52.
(B)
13.
(A, B)
33.
(A, B, C)
53.
(C)
14.
(A, B)
34.
(A, B, C)
54.
(A)
15.
(B, C)
35.
(B, C)
55.
(B)
16.
A (Q, T)
36.
56.
A (R, T)
17.
A (Q, R, T)
MATHEMATICS
B (R)
B (P, S)
B (Q, R, T)
C (P)
C (Q, R, T)
C (Q, R, T)
D (Q, S, T)
D (Q, R, T)
D (Q, R, S, T)
A (R)
37.
A (P, T)
57.
A (Q, S)
B (Q, T)
B (P, Q, R, S, T)
B (Q, R, S, T)
C (Q, T)
C (P)
C (R, S, T)
D (P, S)
D (P, Q, R, S)
D (R, S, T)
18.
(09)
38.
(04)
58.
(02)
19.
(03)
39.
(06)
59.
(05)
20.
(03)
40.
(07)
60.
(04)
1/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
HINTS & SOLUTIONS PART - I (PHYSICS) 1.
Answer (A)
4.
Hint:
Hint:
Heat flow through conductor is constant.
Based on Doppler effect observer will observe the frequency of wave which is emitted by source.
Solution:
Req
2.
Solution:
(b r ) 4k br
0 (v ) = (v v cos ) s
dT b(T1 T2 ) (b r )r dr Answer (A) Hint:
cot v sin vs
Temperature of each body will change based on their heat capacity.
vs = v cos
Solution:
0
dT kA ms 1 (T2 T1 ) dt dT kA 2ms 2 (T2 T1 ) dt
sin2
sin = 5.
2ms ⎛ 4 ⎞ t n⎜ ⎟ 3kA ⎝3⎠ 3.
Answer (B)
Answer (C) Hint: Add the total no. of mole by
Answer (C) Hint:
using :
Calculate the effective phase difference at point P.
2 ⎤ ⎡ 2 ⎢ 3 cos 3 ⎥ 6 ⎣ ⎦ 6 cos
2 6 3
3 cos
1 3 3
⎛ T T0 T = T0 ⎜ L ⎝ L
P
Y tan D Y 17 D 8
2/10
⎞ ⎟x ⎠
PA dx = dn RT
S1 S2
T
n
PA 0 dx ∫ dn R T∫ T 0 L
⎛8⎞ cos ⎜ ⎟ ⎝9⎠ =
Pdv
∫ dn ∫ RT
Solution:
Solution:
cos–1
4 3 , cos = 5 5
⎛ TL ⎞ PV n = (T T )R ln ⎜ T ⎟ L 0 ⎝ 0⎠
(8/9) 6.
Answer (A) Hint: T = 2
I Mgd
vs
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
Solution:
Solution:
⎛ M 2 M 2 ⎞ I = ⎜⎜ 12 16 ⎟⎟ ⎝ ⎠
fext dx = (P1 – P2) dV = 2nCvdT
I=
7M 2 48
⎡ 1 1 ⎤ nRT ⎢ ⎥ dV 2nCv dT V V V 0 V ⎦ ⎣ 0 ( 1)∫
V
0
T=
=
2
7M 2 48 Mg 4
2 7 2 3g
=
7 3g
7.
Answer (A, C)
8.
Answer (A, B, C)
9.
Answer (B, C) Hint and Solution of Q. Nos. 7 to 9 Hint: Draw the corresponding PV diagram and calculate corresponding quantities. Solution: P = kV
RT kV V RT = kV 2 2nV = nT
2 2
dV dT V T dV 1 VdT T
V02
⎛T ⎞ V dV ln ⎜ ⎟ 2 V ⎝ T0 ⎠
13. Answer (A, B) 14. Answer (A, B) 15. Answer (B, C) Hint and Solution of Q. Nos. 13 to 15 Hint: Based on equation of travelling waves Solution: P = B
dy dx
⎛ y ⎞ vp = v s ⎜ ⎟ ⎝ x ⎠ ⎛P ⎞ vp = v s ⎜ ⎟ ⎝ B0 ⎠ v s ⎛ ⎞ ap = B ⎜ t ⎟ ⎠ 0 ⎝ v s P0 = B x vs 0 0
ap
v s2 p0 B0 x 0
16. Answer A(Q, T), B(R), C(P), D(Q, S, T) Hint: Distance will be maximum when velocity of each is same. Solution:
10. Answer (A, C)
This has to be solved using
11. Answer (B, C)
Phasor diagram
12. Answer (A, D)
2Asin =
Hint and Solution of Q. Nos. 10 to 12 Hint: Process for system will be adiabatic. More over work done by external force will be used to raise the internal energy.
=
3A
3
= 2 or 2 – 2 =
2 4 or 3 3
A B
3/10
All India Aakash Test Series for JEE (Advanced)-2020 When they are at same position.
N=8
A
⎛ ⎞ x = A sin ⎜ ⎟ 2 ⎝ ⎠
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
B
9 nodes 19. Answer (03)
A0 = 2
Hint: T = constant
17. Answer A(R), B(Q, T), C(Q, T), D(P, S)
Solution :
Hint:
d 1 dT 3 T
Heat flow across section will be constant. Temperature gradient is inversely proportional to area.
20. Answer (03)
Solution: In all case heat current flow will be same. Use electrical analogy for heat circuit to compute temperature and gradient.
0 = 2 Solution:
18. Answer (09)
31 4 2
Hint:
Hint:
2
1 =
Solution :
N 2
2 3
1 = 600 Hz
PART - II (CHEMISTRY) 21. Answer (B) Na/ dry ether
CH2 – (CH2)4 – CH2
Hint: H
Ring opening
CH 2 – CH = CH 2
Br
Br
Solution:
Solution:
H
carbonium ion
rearrangment
CH2 – CH = CH 2
H
O
H
–H
H2 O
CH2 – CH = CH2
CH2 = CH – CH 2 – OH
CH4
Cl2/h
D CH3 – CH = CH 2
CH3 – Cl A alc KOH
NBS
Br – CH2 – CH = CH2 E
(Et) 2CuLi
CH 3 – CH2 – CH3 B
CH 3 – CH – CH3 C Cl
Na, dry ether
Cl2/h
CH2 = CH – CH2 – CH 2 F CH = CH 2
22. Answer (B) Hint: Properties of canonical structure. Solution: One canonical structure will not explain all the properties of molecule. 23. Answer (A) Hint:
4/10
Na/ dry ether
HBr/ Proxide
CH2 –(CH2)4 – CH2 – Br G Br
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
24. Answer (D)
27. Answer (A, C, D)
Hint: Basic strength lone pair concentration.
Hint: Carbonium ion rearrangement.
Solution:
Solution:
(+I) Et
(+I) Et
N
CH = CH 2 NH 2
NH 2
>
>
> NO2 (–I, –R)
CH – CH 3
HCl
Ph (–R)
CH – CH 3
NH
Cl
Ph (–R)
CH3
hydride ion shift
CH 3
Cl–1
25. Answer (D) Hint: is non-aromatic compound as it is non-planar molecule.
Cl–
ring expansion
Cl
28. Answer (C, D)
Solution:
Hint: 1, 2 & 1, 4 addition of HCl Solution:
Non-aromatic
CH3 – CH = CH – CH = CH 2
H
H
CH3 – CH = CH – CH – CH 3
B
Cl (1-2-addition product)
Antiaromatic B
CH3 – CH = CH – CH – CH3 Cl– CH3 – CH – CH = CH – CH3 Cl
–
CH3 – CH – CH = CH – CH3
H
Cl 1, 4 addition product
29. Answer (B, C, D)
Aromatic
Hint: Electrophilic addition reaction Solution: HBr & HCl both follow Markownikoff with carbonium ion rearrangement
Aromatic
26. Answer (B) Hint: R R
C = CH2
Br2
Br
30. Answer (A) Reductive ozonolysis
O
R R
C =O & H – C – H
Br2/high temperature will do allylic substitution Solution:
Solution: CH 2
Hint:
CH 2
O–H
O CH 2 CH 2 CH 2
O3 6HCHO Zn–H2 O
H
O–H
CH 2
5/10
All India Aakash Test Series for JEE (Advanced)-2020 OH
OH
Ring
Expansion
OH
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) 34. Answer (A, B, C) Hint:
Deprotonation (–H +)
Electrophilic aromatic substitution Solution:
(P) NBS
CH2
OH
E
O More activating
CH2 O
Br
(major)
(Q)
31. Answer (B) Hint: bonds which are not the part of aromatic ring, can undergo addition reaction. Solution:
E
35. Answer (B, C) Hint:
OH
OH
OH
alc KOH
Br 2
Br
Friedel Craft alkylation reaction Solution: Me
CH3
Br Br (Q)
(R)
&
(S)
CH3
32. Answer (A, C) Hint:
density than
HO
is aromatic compound.
has greater electron Me
Me CH3
36. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T) Hint:
Solution: OH
Compound R is
Reaction intermediate singlet carbene is sp 2 hybridised.
It is aromatic cyclic and stable Has acidic H-atom
Solution: R1
33. Answer (A, B, C)
C
Hint: R2
Electrophilic aromatic substitution Solution: O
O – O – C – CH 3
(activating
+
R)
&
– C – CH 3
Singlet carbene hybridisation is sp2. and have a vacant 2p orbital hence can behave like an electrophile. 37. Answer A(P, T); B(P, Q, R, S, T); C(P), D(P, Q, R, S)
(deactivating i.e. – R) –Cl deactivating
Hint: Alkene decolourise Br2/CCl4.
–O–CH3 more activating
Solution:
–NO2 deactivating
6/10
+ H 2/Ni CH3 – CH 2 – CH3
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020 O
O COOH
,
CH2 &
,
COOH COOH
HOOC
CH
+
COONa COONa electrolysis
CH4 + CH3 – C C Mg Br –
NaOOC
(salt)
are unsaturated compound can give bromine water test CH3 – C CH + CH 3MgBr
COONa
NaOH
–
O + 4 CO 2 (m = 4)
Can react with Na as it has acidic hydrogen atom
Tautomerise
OH
CH3 – C CH (DBE = 4)
Hydrogen is acidic due to aromaticity
H
2m d 6 2 40. Answer (07) Hint: Kjeldahl method Solution: m.e of H2SO4 used for NH3 = (25 × 2) – 30 × 0.5 × 2 = 50 – 30 = 20 No. of m.e of NH3 = 20 % mass of nitrogen in organic compound
aromatic
38. Answer (04) Hint: RMgX RH Solution: C6H5 SO3H, NO 2
C 6H5OH OH
=
, OH
(20)(14) 100 (1000)(4)
1.4 4 2 14 7 = 4 i.e 7
NH2
= 20
39. Answer (06) Hint: Decarboxylation reaction Solution:
PART - III (MATHEMATICS) 42. Answer (B) Hint: sinx = cosecx and tany = coty Solution:
41. Answer (C) Solution: Put
1 n
t
⎛ 1 ⎞ ⎜ 4 ⎟ = lim ⎜ t ⎟ t 0 ⎜⎜ 1 1 ⎟⎟ ⎝ t2 ⎠
sin x cosecx 2 and tany + coty 2
So, sinx = 1, tany = 1
sin t
⎛ ⎞ 1 lim ⎜ 2 ⎟ t 0 t (1 t 2 ) ⎝ ⎠
log l lim – sin t log(t 2 (1 t 2 )) t 0
2log(t ) log(1 t 2 ) t 0 cosect
lim
2 sin2 t =0l=1 t 0 t cos t
lim
sin t
, y 2 4 y ⎛ ⎞ tan ⎜ ⎟ 2 1 which satisfies t2 + 2t – 1 = 0 ⎝2⎠ 43. Answer (C) Hint: greatest value of series when x1 = x2 = .....xn Solution: The value of the sum is greatest when x1 = x2 = ...
x
xn
⎛⎞ and required value is n sin ⎜ ⎟ n ⎝n⎠
7/10
All India Aakash Test Series for JEE (Advanced)-2020 44. Answer (B)
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) 49. Answer (B)
4
Hint : Number of exhaustive cases = 7
Hint:
Solution:
z –
Total no. of ways = (7)4
Solution for Q. Nos. 47 to 49
Now no. of favourable case 7 C2 (24 2) 21 14 Now probability of required event
21 14 (7)
4
6 49
45. Answer (A) Hint: Area =
3 2 a 4
⇒ y 2 1 x2 2 x2 + y2 = 1 Case 2 >2
Similarly | z2 z3 |2 | z3 z1 | 3
2 – 4 > 0
Thus, whose vertices are z1, z2 and z3 is equilateral triangle and its side is of length Therefore, area of triangle
3 4
3
3
2
3 3 4
|x| + |y| = 1
|x| + |y| = 1
Locus of z is a square and z i z – i 2 Locus of z is a straight line z = –i satisfies the given equations
47. Answer (A, B, C)
–4>0 2
8/10
2 4 2 4 ⇒x ,y 0 2 2
⎛ 2 4 ⎞ ⎛ 2 4 ⎞ z1 ⎜ ,0 ⎟ , z2 ⎜ ,0 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠
Case 3 When is very large
Solution: zz z–z 2
z
One root lies inside the unit circle and other will lie outside the unit circle.
46. Answer (C) Hint: z z z – z 2
i 4 2 2
x
|z1 – z2|2 = 3 ⇒ | z1 z2 | 3
Hint:
Case 1
z
| – z3|2 + |z1 – z2|2 = 2(2)
48. Answer (A, C)
2 4 2
2 – 4 < 0
|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)
2 – 4 0
z
–2 < < 2
Solution:
Hint:
z2 + z + 1 = 0
⎛ 2 4 ⎞ ⎟ , z z⎜ ⎜ ⎟ 2 ⎝ ⎠
2 4 1 2 2 2 4 50. Answer (B) z
51. Answer (C) 52. Answer (B) Hint: Use projection formula Solution: cos( A B )
4 5
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
3 3 ⇒ sin( A B ) , tan( A B ) 5 4
⇒
⇒
2i ( z z ) (1 )
zz 1 | z |2
z z i
⎛ AB⎞ 1 tan ⎜ ⎟ ⎝ 2 ⎠ 3
2 1 | z |2
56. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T) Hint: x – 1 x Solution:
ab C 1 cot ab 2 3
lim
(A) lim(sin x cos x )cosecx e x 0
sin x cos x 1 sin x
x 0
=e
⇒ C 90
So triangle ABC is right angle triangle Area of ABC
r
2i zz 1
⎡ ( A B) ⎤ 2 tan ⎢ ⎥ ⎣ 2 ⎦ 3 ⎛ AB⎞ 4 1 tan2 ⎜ ⎟ ⎝ 2 ⎠
⎧ AB is an acute angle ⎨∵ 2 ⎩
⇒
zz 2 1 | z |2
1 6 3 = 9 sq. units 2
9 3 3 5 s ⎛363 5 ⎞ 2 ⎜⎜ ⎟⎟ 2 ⎝ ⎠
⎛ 4 n 10 ⎞ (B) L lim ⎜ n ⎝ 5 ⎟⎠ Le
lim n
1
L e5
n
(1 form)
1 ⎛ ⎞ n 5 4 (10) ⎜ ⎟n ⎜ ⎟ 5 ⎜⎝ ⎟⎠
ln(10) 1
1 3 5 Radius of the circumcircle = × hypotenuse 2 2 53. Answer (C)
L 10 5
L5 2 5
54. Answer (A)
xx 1 (C) xlim 0
55. Answer (B)
2 L= 2
2
Hint: z
2 2 (1 )2
(D) L lim
x
Solution: | z |2
Now | z |2
2 2 2
(1 )
2 2
(1 )
1
1 1 1 ⇒ 1 | z |2 1 1 1 | z |2
⎛ x ⎞ ⎡x⎤ x ⎜⎝ 3 1⎟⎠ ⎢ 3 ⎥ 3 ⎣ ⎦ 5 ⎛ x ⎞ 5 ⎡x⎤ 5 x 1 x ⎜⎝ 3 ⎟⎠ x ⎢⎣ 3 ⎥⎦ x 3 lim
x
5 ⎡x⎤ 5 x ⎢⎣ 3 ⎥⎦ 3
57. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
1 | z |2
⇒ 2 ( z z ) (1 )
5 ⎡x⎤ x ⎢⎣ 3 ⎥⎦
zz 1 | z |2
Hint:
z = unimodular |z|
9/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
Solution: (A)
Solving (i), (ii) and (iii)
z z 2i 2i |z| |z|
we get
z unimodular complex no. |z|
so points P (±1, ±2, ±3)
a = ±2, d = ±1 Now = 1, = 2, = 3
and lies on perpendicular bisector of 2i and –2i
59. Answer (05)
z 1 ⇒ z | z | ⇒ a real number |z| Im (z) = 0 (B) = z
Hint:
z1 C is
ei 1 2 1 e 2i 2cos ei
⎛ 3 ⎞ z2 C is ⎜ ⎟ ⎝ 10 ⎠ Solution:
(C) 2b = 8 a2 = 7
2be = 6
z1 C is
3 e= 4 Equation of ellipse
10
10
⎛ 3 ⎞ z2 C is ⎜ ⎟ ⎝ 10 ⎠
x2 y 2 1 7 16
(D) |z – 2 – 3i| = 1
3 ⎞ 3 ⎞ ⎛ ⎛ cos ⎟ + i ⎜ sin sin ⎟ z1 z2 ⎜ cos ⎝ ⎝ 10 10 ⎠ 10 10 ⎠
( x 2)2 ( y 3)2 1
x 2 cos , y 3 sin
⎛ 10 2 5 10 2 5 ⎞ 1 z1 z2 ⎜ ⎟ i ⎜⎝ ⎟⎠ 2 4
x [1, 3], y [2, 4]
58. Answer (02) Hint:
| z1 z2 | 2
|a – d|, |a|, |a + d| be three distances
3 5 2
60. Answer (04)
Solution: As perpendicular distances are in A.P.
Solution:
Let |a – d|, |a|, |a + d| be these distances
sin sin 2 sin 3 0 sin2 (1 + 2cos) = 0 ..(1)
so the co-ordinates of point P are
cos cos 2 cos 3 0 cos2 (1 + 2cos) = 0 ..(2)
(±(a – d), ± a, ± (a + d) According to question
From (1) and (2)
2a2 + d2 – 2ad = 5
...(i)
2a2 + 2d2 = 10
...(ii)
cos –
2a2 + d2 + 2ad = 13
...(iii)
Number of values of [–2, 2] is 4
10/10
1 2
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A (Paper-1) - Code-E Test Date : 24/02/2019
ANSWERS PHYSICS
CHEMISTRY
1.
(A)
21.
(B)
41.
(C)
2.
(C)
22.
(D)
42.
(A)
3.
(B)
23.
(D)
43.
(B)
4.
(C)
24.
(A)
44.
(C)
5.
(A)
25.
(B)
45.
(B)
6.
(A)
26.
(B)
46.
(C)
7.
(A, C)
27.
(A, C, D)
47.
(A, B, C)
8.
(A, B, C)
28.
(B, C, D)
48.
(A, C)
9.
(B, C)
29.
(B, C, D)
49.
(B)
10.
(A, C)
30.
(A)
50.
(B)
11.
(B, C)
31.
(B)
51.
(C)
12.
(A, D)
32.
(A, C)
52.
(B)
13.
(A, B)
33.
(A, B, C)
53.
(C)
14.
(A, B)
34.
(A, B, C)
54.
(A)
15.
(B, C)
35.
(B, C)
55.
(B)
16.
A (R)
36.
56.
A (Q, S)
17.
A (P, T)
MATHEMATICS
B (Q, T)
B (P, Q, R, S, T)
B (Q, R, S, T)
C (Q, T)
C (P)
C (R, S, T)
D (P, S)
D (P, Q, R, S)
D (R, S, T)
A (Q, T)
37.
A (Q, R, T)
57.
A (R, T)
B (R)
B (P, S)
B (Q, R, T)
C (P)
C (Q, R, T)
C (Q, R, T)
D (Q, S, T)
D (Q, R, T)
D (Q, R, S, T)
18.
(03)
38.
(07)
58.
(04)
19.
(03)
39.
(06)
59.
(05)
20.
(09)
40.
(04)
60.
(02)
1/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
HINTS & SOLUTIONS PART - I (PHYSICS) 1.
Answer (A)
3.
Answer (B) Hint:
I Hint: T = 2 Mgd
Based on Doppler effect observer will observe the frequency of wave which is emitted by source.
Solution:
Solution:
M 2 M 2 I = 12 16
0 (v ) = (v v cos ) s
vs
cot v sin vs
7M 2 I= 48
vs = v cos T=
2
7M
2
48 Mg 4
2.
4.
Solution:
Hint: Add the total no. of mole by
dn
Pdv RT
Solution: T T0 T = T0 L x L
PA dx = dn RT n
PA dx dn R T T 0 L
TL PV n = (T T )R ln T L 0 0
2/10
Answer (C)
Calculate the effective phase difference at point P.
Answer (C)
T0
4 3 , cos = 5 5
Hint:
7 3g
using :
0
sin2
sin =
2 7 = 2 3g =
2 2 3 cos 3 6 6 cos
2 6 3
3 cos
1 3 3
8 cos 9 = cos–1 (8/9) Y tan D
17 Y 8 D
P
S1 S2
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
5.
12. Answer (A, D)
Answer (A) Hint:
Hint and Solution of Q. Nos. 10, 11 to 12
Temperature of each body will change based on their heat capacity.
Hint:
Solution: ms
dT1 kA (T2 T1 ) dt
2ms
t 6.
dT2 kA (T2 T1 ) dt
2ms 4 n 3kA 3
Answer (A)
Process for system will be adiabatic. More over work done by external force will be used to raise the internal energy. Solution: fext dx = (P1 – P2) dV = 2nCvdT 1 1 nRT dV 2nCv dT V0 V V0 V ( 1)
V
0 V2 0
T V dV ln 2 V T0
Hint:
13. Answer (A, B)
Heat flow through conductor is constant.
14. Answer (A, B)
Solution:
15. Answer (B, C)
Req
(b r ) 4k br
Hint and Solution of Q. Nos. 10, 11 to 12 Hint: Based on equation of travelling waves
dT b(T1 T2 ) (b r )r dr
Solution:
7.
Answer (A, C)
P = B
8.
Answer (A, B, C)
9.
Answer (B, C) Hint and Solution of Q. Nos. 7, 8 to 9 Hint: Draw the corresponding PV diagram and calculate corresponding quantities. Solution: P = kV
RT kV V RT = kV 2 2nV = nT 2
dV dT V T
2
dV 1 VdT T
10. Answer (A, C) 11. Answer (B, C)
dy dx
y vp = v s x P vp = v s B0 v s ap = B t 0
v s P0 = B x vs 0 0
ap
v s2 p0 B0 x 0
16. Answer A(R), B(Q, T), C(Q, T), D(P, S) Hint: Heat flow across section will be constant. Temperature gradient is inversely proportional to area. Solution: In all case heat current flow will be same. Use electrical analogy for heat circuit to compute temperature and gradient.
3/10
All India Aakash Test Series for JEE (Advanced)-2020 17. Answer A(Q, T), B(R), C(P), D(Q, S, T)
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) Solution:
Hint: Distance will be maximum when velocity of each is same. Solution:
31 4 2 1 =
This has to be solved using
1 = 600 Hz
Phasor diagram
A
3A
2Asin =
19. Answer (03) Hint: T = constant
= 3
Solution :
B
= 2 or 2 – 2 2 4 or 3 3
=
2 3
d 1 dT 3 T 20. Answer (09) Hint:
When they are at same position.
x = A sin 2
A0 2
=
A
B
2
Solution :
N 2
18. Answer (03)
N=8
Hint:
9 nodes
0 = 2
PART - II (CHEMISTRY) Solution:
21. Answer (B) Hint: R R
C = CH2
Reductive ozonolysis
R
C =O & H – C – H
Solution: CH 2
Non-aromatic
O
R
H B
Antiaromatic
CH 2
B CH 2
CH 2
O3 6HCHO Zn–H2 O
H
Aromatic CH 2
CH 2
22. Answer (D) Hint: Aromatic is non-aromatic compound as it is non-planar molecule.
4/10
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020 Solution:
23. Answer (D) Hint: Basic strength lone pair concentration.
CH = CH 2
Solution: (+I) Et
(+I) Et
N
NH 2
CH – CH 3
NH 2
>
>
CH – CH 3
HCl
Ph (–R) >
Cl
NH
Ph (–R)
NO2 (–I, –R)
Cl–
CH3
hydride ion shift
CH 3
Cl–1
ring expansion
24. Answer (A) Hint:
Cl Na/ dry ether
CH2 – (CH 2)4 – CH 2 Br
28. Answer (B, C, D)
Br
Hint: 1, 2 & 1, 4 addition of HCl
Solution: CH4
Solution:
Cl2/h
D CH3 – CH = CH 2
CH3 – Cl A alc KOH
NBS
Br – CH2 – CH = CH2 E
(Et) 2CuLi
CH 3 – CH – CH3 C Cl
Na, dry ether
Na/ dry ether
CH 3 – CH2 – CH3 B
CH3 – CH = CH – CH = CH 2
Cl2/h
CH3 – CH = CH – CH – CH 3
CH2 = CH – CH2 – CH 2 F CH = CH 2
CH 2 –(CH 2)4 – CH2 – Br G Br
Hint: Properties of canonical structure.
CH3 – CH = CH – CH – CH3 Cl– CH3 – CH – CH = CH – CH3 Cl
–
CH3 – CH – CH = CH – CH3 Cl 1, 4 addition product
29. Answer (B, C, D) Hint: Electrophilic addition reaction Solution:
Solution: One canonical structure will not explain all the properties of molecule. 26. Answer (B) Hint: H
Ring opening
CH 2 – CH = CH 2
Solution:
H
carbonium ion
rearrangment
CH2 – CH = CH 2
H
Cl (1-2-addition product)
HBr/ Proxide
25. Answer (B)
O
H
H
–H
H2 O
CH2 – CH = CH2
HBr & HCl both follow Markownikoff with carbonium ion rearrangement
Br2
Br
30. Answer (A) Hint: Br2/high temperature will do allylic substitution Solution: O–H
O
O–H
CH2 = CH – CH 2 – OH
H
27. Answer (A, C, D) Hint: Carbonium ion rearrangement.
5/10
All India Aakash Test Series for JEE (Advanced)-2020 OH
OH
OH
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 34. Answer (A, B, C) Hint:
Ring
Expansion
Deprotonation (–H +)
Electrophilic aromatic substitution Solution: (P)
CH2
NBS
OH
E
O More activating
CH2 O (major)
Br
(Q)
E
31. Answer (B) Hint: bonds which are not the part of aromatic ring, can undergo addition reaction. Solution:
35. Answer (B, C) Hint: Friedel Craft alkylation reaction
OH
OH
OH
alc KOH
Br 2
Br
Me
CH3 &
Br Br (Q)
Solution:
(R)
CH3
has greater electron Me
Me
(S) CH3
density than
32. Answer (A, C) Hint:
36. Answer A(P, T); B(P, Q, R, S, T); C(P), D(P, Q, R, S)
HO
Hint: Alkene decolourise Br2/CCl4.
is aromatic compound.
Solution: Solution:
+ H 2/Ni CH3 – CH 2 – CH3
OH
Compound R is
It is aromatic cyclic and stable Has acidic H-atom
,
CH2 &
, CH
33. Answer (A, B, C) Hint:
are unsaturated compound can give bromine water test
Electrophilic aromatic substitution
CH3 – C CH + CH 3MgBr
CH4 + CH3 – C C Mg Br
Solution: O
O – O – C – CH 3
(activating
(deactivating i.e. – R) –Cl deactivating –O–CH3 more activating –NO2 deactivating
6/10
+
R)
&
– C – CH 3
Can react with Na as it has acidic hydrogen atom
CH3 – C CH H
Hydrogen is acidic due to aromaticity
aromatic
–
+
–
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
37. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T)
39. Answer (06)
Hint:
Hint: Decarboxylation reaction
Reaction intermediate singlet carbene is sp 2 hybridised.
Solution: O
Solution:
O COOH
R1
COOH COOH
HOOC
C
COONa
NaOH NaOOC
R2
COONa COONa (salt) electrolysis
Singlet carbene hybridisation is sp2. and have a vacant 2p orbital hence can behave like an electrophile.
O
38. Answer (07) + 4 CO 2 (m = 4)
Hint: Kjeldahl method Solution:
Tautomerise
m.e of H2SO4 used for NH3
OH
= (25 × 2) – 30 × 0.5 × 2 = 50 – 30 = 20 No. of m.e of NH3 = 20 % mass of nitrogen in organic compound =
=
Hint: RMgX RH
(20)(14) 100 (1000)(4)
= 20
(DBE = 4)
2m d 6 2 40. Answer (04)
Solution:
1.4 4
C6H5 SO3H,
C 6H5OH
NO 2
2 14 7 4
OH ,
i.e 7
OH
NH2
PART - III (MATHEMATICS) Solution:
41. Answer (C) Hint: z z z – z 2
|x| + |y| = 1
|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2) | – z3|2 + |z1 – z2|2 = 2(2)
Solution:
zz z–z 2
|x| + |y| = 1
Locus of z is a square and z i z – i 2 Locus of z is a straight line z = –i satisfies the given equations
|z1 – z2|2 = 3 | z1 z2 | 3 Similarly | z2 z3 |2 | z3 z1 | 3 Thus, whose vertices are z1, z2 and z3 is equilateral triangle and its side is of length
3
3 4
3
42. Answer (A) Hint: Area =
3 2 a 4
Therefore, area of triangle
2
3 3 4
7/10
All India Aakash Test Series for JEE (Advanced)-2020 43. Answer (B)
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 47. Answer (A, B, C)
Hint : Number of exhaustive cases = 74
Hint:
Solution:
2 – 4 0 48. Answer (A, C)
Total no. of ways = (7)4 7
Hint:
4
Now no. of favourable case C2 (2 2) 21 14 Now probability of required event
21 14 (7)4
6 49
49. Answer (B) Hint: z –
44. Answer (C) Hint: greatest value of series when x1 = x2 = .....xn Solution: The value of the sum is greatest when x1 = x2 = ... xn
2 – 4 > 0
and required value is n sin n n
45. Answer (B)
Solution for Q. Nos. 47 to 49 z2 + z + 1 = 0 z
2 4 2
Case 1 –2 < < 2 2 – 4 < 0
Hint: sinx = cosecx and tany = coty
i 4 2 2
Solution:
z
sin x cosecx 2 and tany + coty 2
y 2 1 x2 2 x2 + y2 = 1
So, sinx = 1, tany = 1 , y 2 4
x
Case 2
y tan 2 1 which satisfies t2 + 2t – 1 = 0 2 46. Answer (C)
Put
n
sin t
1 lim 2 t 0 t (1 t 2 )
log l lim – sin t log(t 2 (1 t 2 )) t 0
2log(t ) log(1 t 2 ) t 0 cosect
lim
2 sin2 t =0l=1 t 0 t cos t
lim
8/10
2 – 4 > 0 2 4 2 4 x ,y 0 2 2
2 4 2 4 z1 ,0 , z2 ,0 2 2
t
1 4 = lim t t 0 1 2 1 t
>2
z
Solution:
1
x
sin t
One root lies inside the unit circle and other will lie outside the unit circle. Case 3 When is very large
2 4 z z 2
2 4 1 2 2 2 4 50. Answer (B) z
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 51. Answer (C)
All India Aakash Test Series for JEE (Advanced)-2020 Now | z |2
52. Answer (B) Hint: Use projection formula
cos( A B )
4 5
sin( A B )
3 3 , tan( A B ) 5 4
AB 1 tan 2 3
Hint:
(A)
z z i
2 1 | z |2
z = unimodular |z|
Radius of the circumcircle =
1 3 5 × hypotenuse 2 2
53. Answer (C)
z z 2i 2i |z| |z|
z unimodular complex no. |z|
1 6 3 = 9 sq. units 2
9 3 3 5 s 363 5 2 2
and lies on perpendicular bisector of 2i and –2i
z 1 z | z | a real number |z| Im (z) = 0
(B) = ei z
1 2 1 e 2i 2cos ei
(C) 2b = 8 a2 = 7
2be = 6
54. Answer (A) e=
55. Answer (B) 2 2 (1 )2
3 4
Equation of ellipse
x2 y 2 1 7 16
(D) Iz – 2 – 3i| = 1
Solution: | z |2
zz 1 | z |2
Solution:
So triangle ABC is right angle triangle
2
1 | z |2
56. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
ab C 1 cot ab 2 3
Hint: z
2 1 | z |2
2i ( z z ) (1 )
C 90
r
zz
zz
2i zz 1
( A B) 2 tan 2 3 AB 4 1 tan2 2
Area of ABC
1 | z |2
2 ( z z ) (1 )
AB is an acute angle ∵ 2
| z |2
Solution:
1 1 1 1 | z |2 1 1 1
( x 2)2 ( y 3)2 1
2 2 2
(1 )
2 2
(1 )
1
x 2 cos , y 3 sin x [1, 3], y [2, 4]
9/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) 59. Answer (05)
57. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T) Hint: x – 1 x
Hint:
Solution: (A) lim(sin x cos x )cosecx e
z1 C is
sin x cos x 1 x 0 sin x lim
10
x 0
3 z2 C is 10
=e 4 n 10 (B) L lim n 5 Le
L L
lim n
n
(1 form)
Solution:
1 4 (10) n 5 n 5
z1 C is
3 z2 C is 10
1 ln(10) 5 e
3 3 z1 z2 cos cos + i sin sin 10 10 10 10
1 5 10
10
L5 2 5
10 2 5 10 2 5 1 z1 z2 i 2 4
xx 1 (C) xlim 0 2 L= 2
| z1 z2 | 2
5 x x x 3
(D) L lim
60. Answer (02) Hint:
x x x 3 1 3 3
|a – d|, |a|, |a + d| be three distances Solution:
5 x 5 x 5 x 1 x 3 x 3 x 3 lim
x
3 5 2
As perpendicular distances are in A.P. Let |a – d|, |a|, |a + d| be these distances
5 x 5 x 3 3
so the co-ordinates of point P are (±(a – d), ± a, ± (a + d)
58. Answer (04) Solution:
According to question
sin sin 2 sin3 0 sin2 (1 + 2cos) = 0 ..(1)
2a2 + d2 – 2ad = 5
...(i)
2a2 + 2d2 = 10
...(ii)
2
2
cos cos 2 cos 3 0 cos2 (1 + 2cos) = 0 ..(2)
2a + d + 2ad = 13
From (1) and (2)
we get
1 2 Number of values of [–2, 2] is 4
a = ±2, d = ±1
Solving (i), (ii) and (iii)
cos –
so points P (±1, ±2, ±3) Now = 1, = 2, = 3
10/10
...(iii)
Test - 3A (Paper - 1) (Code-F) (Answers)
All India Aakash Test Series for JEE (Advanced)-2020
All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A (Paper-1) - Code-F Test Date : 24/02/2019
ANSWERS PHYSICS
CHEMISTRY
1.
(A)
21.
(B)
41.
(C)
2.
(A)
22.
(B)
42.
(B)
3.
(C)
23.
(A)
43.
(C)
4.
(B)
24.
(D)
44.
(B)
5.
(C)
25.
(D)
45.
(A)
6.
(A)
26.
(B)
46.
(C)
7.
(A, C)
27.
(A, C, D)
47.
(A, B, C)
8.
(A, B, C)
28.
(C, D)
48.
(A, C)
9.
(B, C)
29.
(B, C, D)
49.
(B)
10.
(A, C)
30.
(A)
50.
(B)
11.
(B, C)
31.
(B)
51.
(C)
12.
(A, D)
32.
(A, C)
52.
(B)
13.
(A, B)
33.
(A, B, C)
53.
(C)
14.
(A, B)
34.
(A, B, C)
54.
(A)
15.
(B, C)
35.
(B, C)
55.
(B)
16.
A (Q, T)
36.
56.
A (R, T)
17.
A (Q, R, T)
MATHEMATICS
B (R)
B (P, S)
B (Q, R, T)
C (P)
C (Q, R, T)
C (Q, R, T)
D (Q, S, T)
D (Q, R, T)
D (Q, R, S, T)
A (R)
37.
A (P, T)
57.
A (Q, S)
B (Q, T)
B (P, Q, R, S, T)
B (Q, R, S, T)
C (Q, T)
C (P)
C (R, S, T)
D (P, S)
D (P, Q, R, S)
D (R, S, T)
18.
(09)
38.
(04)
58.
(02)
19.
(03)
39.
(06)
59.
(05)
20.
(03)
40.
(07)
60.
(04)
1/10
All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
HINTS & SOLUTIONS PART - I (PHYSICS) 1.
Answer (A)
4.
Hint:
Hint:
Heat flow through conductor is constant.
Based on Doppler effect observer will observe the frequency of wave which is emitted by source.
Solution:
Req
2.
Solution:
(b r ) 4k br
0 (v ) = (v v cos ) s
dT b(T1 T2 ) (b r )r dr Answer (A) Hint:
cot v sin vs
Temperature of each body will change based on their heat capacity.
vs = v cos
Solution:
0
dT kA ms 1 (T2 T1 ) dt dT kA 2ms 2 (T2 T1 ) dt
sin2
sin = 5.
2ms ⎛ 4 ⎞ t n⎜ ⎟ 3kA ⎝3⎠ 3.
Answer (B)
Answer (C) Hint: Add the total no. of mole by
Answer (C) Hint:
using :
Calculate the effective phase difference at point P.
2 ⎤ ⎡ 2 ⎢ 3 cos 3 ⎥ 6 ⎣ ⎦ 6 cos
2 6 3
3 cos
1 3 3
⎛ T T0 T = T0 ⎜ L ⎝ L
P
Y tan D Y 17 D 8
2/10
⎞ ⎟x ⎠
PA dx = dn RT
S1 S2
T
n
PA 0 dx ∫ dn R T∫ T 0 L
⎛8⎞ cos ⎜ ⎟ ⎝9⎠ =
Pdv
∫ dn ∫ RT
Solution:
Solution:
cos–1
4 3 , cos = 5 5
⎛ TL ⎞ PV n = (T T )R ln ⎜ T ⎟ L 0 ⎝ 0⎠
(8/9) 6.
Answer (A) Hint: T = 2
I Mgd
vs
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
Solution:
Solution:
⎛ M 2 M 2 ⎞ I = ⎜⎜ 12 16 ⎟⎟ ⎝ ⎠
fext dx = (P1 – P2) dV = 2nCvdT
I=
7M 2 48
⎡ 1 1 ⎤ nRT ⎢ ⎥ dV 2nCv dT V V V 0 V ⎦ ⎣ 0 ( 1)∫
V
0
T=
=
2
7M 2 48 Mg 4
2 7 2 3g
=
7 3g
7.
Answer (A, C)
8.
Answer (A, B, C)
9.
Answer (B, C) Hint and Solution of Q. Nos. 7 to 9 Hint: Draw the corresponding PV diagram and calculate corresponding quantities. Solution: P = kV
RT kV V RT = kV 2 2nV = nT
2 2
dV dT V T dV 1 VdT T
V02
⎛T ⎞ V dV ln ⎜ ⎟ 2 V ⎝ T0 ⎠
13. Answer (A, B) 14. Answer (A, B) 15. Answer (B, C) Hint and Solution of Q. Nos. 13 to 15 Hint: Based on equation of travelling waves Solution: P = B
dy dx
⎛ y ⎞ vp = v s ⎜ ⎟ ⎝ x ⎠ ⎛P ⎞ vp = v s ⎜ ⎟ ⎝ B0 ⎠ v s ⎛ ⎞ ap = B ⎜ t ⎟ ⎠ 0 ⎝ v s P0 = B x vs 0 0
ap
v s2 p0 B0 x 0
16. Answer A(Q, T), B(R), C(P), D(Q, S, T) Hint: Distance will be maximum when velocity of each is same. Solution:
10. Answer (A, C)
This has to be solved using
11. Answer (B, C)
Phasor diagram
12. Answer (A, D)
2Asin =
Hint and Solution of Q. Nos. 10 to 12 Hint: Process for system will be adiabatic. More over work done by external force will be used to raise the internal energy.
=
3A
3
= 2 or 2 – 2 =
2 4 or 3 3
A B
3/10
All India Aakash Test Series for JEE (Advanced)-2020 When they are at same position.
N=8
A
⎛ ⎞ x = A sin ⎜ ⎟ 2 ⎝ ⎠
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
B
9 nodes 19. Answer (03)
A0 = 2
Hint: T = constant
17. Answer A(R), B(Q, T), C(Q, T), D(P, S)
Solution :
Hint:
d 1 dT 3 T
Heat flow across section will be constant. Temperature gradient is inversely proportional to area.
20. Answer (03)
Solution: In all case heat current flow will be same. Use electrical analogy for heat circuit to compute temperature and gradient.
0 = 2 Solution:
18. Answer (09)
31 4 2
Hint:
Hint:
2
1 =
Solution :
N 2
2 3
1 = 600 Hz
PART - II (CHEMISTRY) 21. Answer (B) Na/ dry ether
CH2 – (CH2)4 – CH2
Hint: H
Ring opening
CH 2 – CH = CH 2
Br
Br
Solution:
Solution:
H
carbonium ion
rearrangment
CH2 – CH = CH 2
H
O
H
–H
H2 O
CH2 – CH = CH2
CH2 = CH – CH 2 – OH
CH4
Cl2/h
D CH3 – CH = CH 2
CH3 – Cl A alc KOH
NBS
Br – CH2 – CH = CH2 E
(Et) 2CuLi
CH 3 – CH2 – CH3 B
CH 3 – CH – CH3 C Cl
Na, dry ether
Cl2/h
CH2 = CH – CH2 – CH 2 F CH = CH 2
22. Answer (B) Hint: Properties of canonical structure. Solution: One canonical structure will not explain all the properties of molecule. 23. Answer (A) Hint:
4/10
Na/ dry ether
HBr/ Proxide
CH2 –(CH2)4 – CH2 – Br G Br
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
24. Answer (D)
27. Answer (A, C, D)
Hint: Basic strength lone pair concentration.
Hint: Carbonium ion rearrangement.
Solution:
Solution:
(+I) Et
(+I) Et
N
CH = CH 2 NH 2
NH 2
>
>
> NO2 (–I, –R)
CH – CH 3
HCl
Ph (–R)
CH – CH 3
NH
Cl
Ph (–R)
CH3
hydride ion shift
CH 3
Cl–1
25. Answer (D) Hint: is non-aromatic compound as it is non-planar molecule.
Cl–
ring expansion
Cl
28. Answer (C, D)
Solution:
Hint: 1, 2 & 1, 4 addition of HCl Solution:
Non-aromatic
CH3 – CH = CH – CH = CH 2
H
H
CH3 – CH = CH – CH – CH 3
B
Cl (1-2-addition product)
Antiaromatic B
CH3 – CH = CH – CH – CH3 Cl– CH3 – CH – CH = CH – CH3 Cl
–
CH3 – CH – CH = CH – CH3
H
Cl 1, 4 addition product
29. Answer (B, C, D)
Aromatic
Hint: Electrophilic addition reaction Solution: HBr & HCl both follow Markownikoff with carbonium ion rearrangement
Aromatic
26. Answer (B) Hint: R R
C = CH2
Br2
Br
30. Answer (A) Reductive ozonolysis
O
R R
C =O & H – C – H
Br2/high temperature will do allylic substitution Solution:
Solution: CH 2
Hint:
CH 2
O–H
O CH 2 CH 2 CH 2
O3 6HCHO Zn–H2 O
H
O–H
CH 2
5/10
All India Aakash Test Series for JEE (Advanced)-2020 OH
OH
Ring
Expansion
OH
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) 34. Answer (A, B, C) Hint:
Deprotonation (–H +)
Electrophilic aromatic substitution Solution:
(P) NBS
CH2
OH
E
O More activating
CH2 O
Br
(major)
(Q)
31. Answer (B) Hint: bonds which are not the part of aromatic ring, can undergo addition reaction. Solution:
E
35. Answer (B, C) Hint:
OH
OH
OH
alc KOH
Br 2
Br
Friedel Craft alkylation reaction Solution: Me
CH3
Br Br (Q)
(R)
&
(S)
CH3
32. Answer (A, C) Hint:
density than
HO
is aromatic compound.
has greater electron Me
Me CH3
36. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T) Hint:
Solution: OH
Compound R is
Reaction intermediate singlet carbene is sp 2 hybridised.
It is aromatic cyclic and stable Has acidic H-atom
Solution: R1
33. Answer (A, B, C)
C
Hint: R2
Electrophilic aromatic substitution Solution: O
O – O – C – CH 3
(activating
+
R)
&
– C – CH 3
Singlet carbene hybridisation is sp2. and have a vacant 2p orbital hence can behave like an electrophile. 37. Answer A(P, T); B(P, Q, R, S, T); C(P), D(P, Q, R, S)
(deactivating i.e. – R) –Cl deactivating
Hint: Alkene decolourise Br2/CCl4.
–O–CH3 more activating
Solution:
–NO2 deactivating
6/10
+ H 2/Ni CH3 – CH 2 – CH3
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020 O
O COOH
,
CH2 &
,
COOH COOH
HOOC
CH
+
COONa COONa electrolysis
CH4 + CH3 – C C Mg Br –
NaOOC
(salt)
are unsaturated compound can give bromine water test CH3 – C CH + CH 3MgBr
COONa
NaOH
–
O + 4 CO 2 (m = 4)
Can react with Na as it has acidic hydrogen atom
Tautomerise
OH
CH3 – C CH (DBE = 4)
Hydrogen is acidic due to aromaticity
H
2m d 6 2 40. Answer (07) Hint: Kjeldahl method Solution: m.e of H2SO4 used for NH3 = (25 × 2) – 30 × 0.5 × 2 = 50 – 30 = 20 No. of m.e of NH3 = 20 % mass of nitrogen in organic compound
aromatic
38. Answer (04) Hint: RMgX RH Solution: C6H5 SO3H, NO 2
C 6H5OH OH
=
, OH
(20)(14) 100 (1000)(4)
1.4 4 2 14 7 = 4 i.e 7
NH2
= 20
39. Answer (06) Hint: Decarboxylation reaction Solution:
PART - III (MATHEMATICS) 42. Answer (B) Hint: sinx = cosecx and tany = coty Solution:
41. Answer (C) Solution: Put
1 n
t
⎛ 1 ⎞ ⎜ 4 ⎟ = lim ⎜ t ⎟ t 0 ⎜⎜ 1 1 ⎟⎟ ⎝ t2 ⎠
sin x cosecx 2 and tany + coty 2
So, sinx = 1, tany = 1
sin t
⎛ ⎞ 1 lim ⎜ 2 ⎟ t 0 t (1 t 2 ) ⎝ ⎠
log l lim – sin t log(t 2 (1 t 2 )) t 0
2log(t ) log(1 t 2 ) t 0 cosect
lim
2 sin2 t =0l=1 t 0 t cos t
lim
sin t
, y 2 4 y ⎛ ⎞ tan ⎜ ⎟ 2 1 which satisfies t2 + 2t – 1 = 0 ⎝2⎠ 43. Answer (C) Hint: greatest value of series when x1 = x2 = .....xn Solution: The value of the sum is greatest when x1 = x2 = ...
x
xn
⎛⎞ and required value is n sin ⎜ ⎟ n ⎝n⎠
7/10
All India Aakash Test Series for JEE (Advanced)-2020 44. Answer (B)
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) 49. Answer (B)
4
Hint : Number of exhaustive cases = 7
Hint:
Solution:
z –
Total no. of ways = (7)4
Solution for Q. Nos. 47 to 49
Now no. of favourable case 7 C2 (24 2) 21 14 Now probability of required event
21 14 (7)
4
6 49
45. Answer (A) Hint: Area =
3 2 a 4
⇒ y 2 1 x2 2 x2 + y2 = 1 Case 2 >2
Similarly | z2 z3 |2 | z3 z1 | 3
2 – 4 > 0
Thus, whose vertices are z1, z2 and z3 is equilateral triangle and its side is of length Therefore, area of triangle
3 4
3
3
2
3 3 4
|x| + |y| = 1
|x| + |y| = 1
Locus of z is a square and z i z – i 2 Locus of z is a straight line z = –i satisfies the given equations
47. Answer (A, B, C)
–4>0 2
8/10
2 4 2 4 ⇒x ,y 0 2 2
⎛ 2 4 ⎞ ⎛ 2 4 ⎞ z1 ⎜ ,0 ⎟ , z2 ⎜ ,0 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠
Case 3 When is very large
Solution: zz z–z 2
z
One root lies inside the unit circle and other will lie outside the unit circle.
46. Answer (C) Hint: z z z – z 2
i 4 2 2
x
|z1 – z2|2 = 3 ⇒ | z1 z2 | 3
Hint:
Case 1
z
| – z3|2 + |z1 – z2|2 = 2(2)
48. Answer (A, C)
2 4 2
2 – 4 < 0
|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)
2 – 4 0
z
–2 < < 2
Solution:
Hint:
z2 + z + 1 = 0
⎛ 2 4 ⎞ ⎟ , z z⎜ ⎜ ⎟ 2 ⎝ ⎠
2 4 1 2 2 2 4 50. Answer (B) z
51. Answer (C) 52. Answer (B) Hint: Use projection formula Solution: cos( A B )
4 5
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
All India Aakash Test Series for JEE (Advanced)-2020
3 3 ⇒ sin( A B ) , tan( A B ) 5 4
⇒
⇒
2i ( z z ) (1 )
zz 1 | z |2
z z i
⎛ AB⎞ 1 tan ⎜ ⎟ ⎝ 2 ⎠ 3
2 1 | z |2
56. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T) Hint: x – 1 x Solution:
ab C 1 cot ab 2 3
lim
(A) lim(sin x cos x )cosecx e x 0
sin x cos x 1 sin x
x 0
=e
⇒ C 90
So triangle ABC is right angle triangle Area of ABC
r
2i zz 1
⎡ ( A B) ⎤ 2 tan ⎢ ⎥ ⎣ 2 ⎦ 3 ⎛ AB⎞ 4 1 tan2 ⎜ ⎟ ⎝ 2 ⎠
⎧ AB is an acute angle ⎨∵ 2 ⎩
⇒
zz 2 1 | z |2
1 6 3 = 9 sq. units 2
9 3 3 5 s ⎛363 5 ⎞ 2 ⎜⎜ ⎟⎟ 2 ⎝ ⎠
⎛ 4 n 10 ⎞ (B) L lim ⎜ n ⎝ 5 ⎟⎠ Le
lim n
1
L e5
n
(1 form)
1 ⎛ ⎞ n 5 4 (10) ⎜ ⎟n ⎜ ⎟ 5 ⎜⎝ ⎟⎠
ln(10) 1
1 3 5 Radius of the circumcircle = × hypotenuse 2 2 53. Answer (C)
L 10 5
L5 2 5
54. Answer (A)
xx 1 (C) xlim 0
55. Answer (B)
2 L= 2
2
Hint: z
2 2 (1 )2
(D) L lim
x
Solution: | z |2
Now | z |2
2 2 2
(1 )
2 2
(1 )
1
1 1 1 ⇒ 1 | z |2 1 1 1 | z |2
⎛ x ⎞ ⎡x⎤ x ⎜⎝ 3 1⎟⎠ ⎢ 3 ⎥ 3 ⎣ ⎦ 5 ⎛ x ⎞ 5 ⎡x⎤ 5 x 1 x ⎜⎝ 3 ⎟⎠ x ⎢⎣ 3 ⎥⎦ x 3 lim
x
5 ⎡x⎤ 5 x ⎢⎣ 3 ⎥⎦ 3
57. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
1 | z |2
⇒ 2 ( z z ) (1 )
5 ⎡x⎤ x ⎢⎣ 3 ⎥⎦
zz 1 | z |2
Hint:
z = unimodular |z|
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All India Aakash Test Series for JEE (Advanced)-2020
Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
Solution: (A)
Solving (i), (ii) and (iii)
z z 2i 2i |z| |z|
we get
z unimodular complex no. |z|
so points P (±1, ±2, ±3)
a = ±2, d = ±1 Now = 1, = 2, = 3
and lies on perpendicular bisector of 2i and –2i
59. Answer (05)
z 1 ⇒ z | z | ⇒ a real number |z| Im (z) = 0 (B) = z
Hint:
z1 C is
ei 1 2 1 e 2i 2cos ei
⎛ 3 ⎞ z2 C is ⎜ ⎟ ⎝ 10 ⎠ Solution:
(C) 2b = 8 a2 = 7
2be = 6
z1 C is
3 e= 4 Equation of ellipse
10
10
⎛ 3 ⎞ z2 C is ⎜ ⎟ ⎝ 10 ⎠
x2 y 2 1 7 16
(D) |z – 2 – 3i| = 1
3 ⎞ 3 ⎞ ⎛ ⎛ cos ⎟ + i ⎜ sin sin ⎟ z1 z2 ⎜ cos ⎝ ⎝ 10 10 ⎠ 10 10 ⎠
( x 2)2 ( y 3)2 1
x 2 cos , y 3 sin
⎛ 10 2 5 10 2 5 ⎞ 1 z1 z2 ⎜ ⎟ i ⎜⎝ ⎟⎠ 2 4
x [1, 3], y [2, 4]
58. Answer (02) Hint:
| z1 z2 | 2
|a – d|, |a|, |a + d| be three distances
3 5 2
60. Answer (04)
Solution: As perpendicular distances are in A.P.
Solution:
Let |a – d|, |a|, |a + d| be these distances
sin sin 2 sin 3 0 sin2 (1 + 2cos) = 0 ..(1)
so the co-ordinates of point P are
cos cos 2 cos 3 0 cos2 (1 + 2cos) = 0 ..(2)
(±(a – d), ± a, ± (a + d) According to question
From (1) and (2)
2a2 + d2 – 2ad = 5
...(i)
2a2 + 2d2 = 10
...(ii)
cos –
2a2 + d2 + 2ad = 13
...(iii)
Number of values of [–2, 2] is 4
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