Solutions AIATS JEE (Adv) - 2018 Test-1 Paper-2 (Code-C & D) (16!07!2017) XII Studying
Solutions AIATS JEE (Adv) - 2018 Test-1 Paper-2 (Code-C & D) (16!07!2017) XII Studying
Solutions AIATS JEE (Adv) - 2018 Test-1 Paper-2 (Code-C & D) (16!07!2017) XII Studying
ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (A) 21. (A) 41. (C)
2. (C) 22. (B) 42. (D)
3. (D) 23. (A) 43. (C)
4. (A, B, C) 24. (A, B, C) 44. (A, C)
5. (A, B, D) 25. (B, D) 45. (A, B, C, D)
6. (A, C, D) 26. (B, D) 46. (B, D)
7. (A, B, C, D) 27. (A, B, C, D) 47. (B, C)
8. (A, B, C, D) 28. (A, C, D) 48. (A, B)
9. (C) 29. (A, D) 49. (A, B, C, D)
10. (C) 30. (B) 50. (B)
11. (C) 31. (B) 51. (A)
12. (D) 32. (D) 52. (B)
13. (B) 33. (D) 53. (A)
14. (B) 34. (C) 54. (C)
15. (A) 35. (A) 55. (D)
16. A(R) 36. A (R) 56. A(P, Q)
B(S, T) B (S, T) B(T)
C(S, T) C (P, T) C(P, Q)
D(R) D (Q, T) D(P, Q)
17. A(S, T) 37. A (P, R) 57. A(P, R, S)
B(R) B (Q, S) B(P, R, S)
C(P) C (P, T) C(T)
D(Q) D (Q, S) D(Q)
18. (12) 38. (20) 58. (99)
19. (30) 39. (99) 59. (49)
20. (16) 40. (22) 60. (00)
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-C) (Answers & Hints)
Gain in PE of capacitor =
1
2
C V F2 – VI2 5
5
0.4
0.2 7.5 750 cm
7. Answer (A, B, C, D)
1
= (1 F) 1 [42 – 22] = 6 J 8. Answer (A, B, C, D)
2
We have A1 = 4r2
Net heat produced = 8 – 6 = 2 J
A2 = 4(2r)2 = 4A1
3. Answer (D)
4. Answer (A, B, C) 9
A3 4(3r )2 9 A1 A2
– 4
– +
+
1 = 2 = 3 through same solid angle
Repulsive force on q1 is less than attractive force.
1 Q Q Q Q
5. Answer (A, B, D)
1 = ∫ EdA ∫ dA
4 0 r 2 4 0 4 0 4 0
Field due to q1 at other capacitor
⎛ 1 1 Q Q
q1 1 ⎞ 2q 1l
E ⎜⎜ 2 – 2 ⎟
⎟ (r l ) Flux passing through A2 within area A1= 4 4 = 16
4 0 ⎝r (r l ) ⎠ 4 0 r 3 0 0
d /2
1 1⎡ ⎤
d
6. Answer (A, C, D) dx dx
⎢∫ ∫ ⎥
C A ⎢⎣ 0 K 0 x d /2 K 0 (d – x ) ⎦⎥
5 5
Resistances of wires in Ratio : =4:1
r 2 4 r 2
A
C
4 ⎡ 2K d ⎤
Potential difference across 1st wire 2 1.6 V 2ln ⎢ 0 ⎥
5 ⎣ 2K 0 ⎦
1
Potential difference across 2nd wire 2 0.4 V 10. Answer (C)
5
11. Answer (C)
Hence cell of emf 1 volt and cell 1.5 volt will be balanced
in the first wire but the cell 1.8 V, 2 V will be balanced 12. Answer (D)
on second wire. 13. Answer (B)
Length required to balance first cell 14. Answer (B)
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Test - 1 (Paper - II) (Code-C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
15. Answer (A) 17. Answer A(S, T); B(R); C(P); D(Q)
v0 t = ; = 1; a = 1; b = 2
v R 2v 0 – t (2t0 t 6t0)
2t0
2 2.1 2
R
v R –v 0 (6t0 t 7t0) ⎛ b ⎞ 3 ln 2 3ln 2
3t ln ⎜ ⎟
⎝a⎠
v0
vR t – 8v 0 (7t0 t 8t0)
t0 2
1
v a b
Current is a function of time is R
R c
v0 t = ; 1= 1; 2 = 2 ; a = 1
0 t t0; i t
Rt0
b=2
dq v 0 c=4
t
dt Rt0
1 2 .1.2 2
R
⎛ b c⎞ ⎛ 2 4 ⎞ 3ln2
q
v0 t t ⎜ 2 ln 1 ln ⎟ ⎜ 2ln 1ln ⎟
∫ dq a b⎠ 1 2⎠
Rt0 ∫0
t dt ⎝ ⎝
0
v0 2 b
a
q t (parabolic curve) 1
2Rt0
In the interval l
v0 l 1
I R 0.5
R ab .1.2 2
q t
v0
∫
v 0t 0
dq
R t∫
dt
0
2R
1
a
v t v
q 0 0 0 (t – t0 ) linear variation b
2R R
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-C) (Answers & Hints)
19. Answer (30)
⎡ 1 1⎤
R –
4 ⎢⎣ a b ⎥⎦ h h
0b 0b
Hint: C1 16 , C2 8
1 ⎡1 1 ⎤ 1 d d
= – 0.04
4 ⎣⎢1 2 ⎦⎥ 8
0 bh 0 bh
C3 , C4
4d 2d
a 1 1 1 1 1
b C C1 C 2 C 3 C 4
0 bh
l C
30d
20. Answer (16)
1 = 1; 2 = 2; l = , a = 1, b = 2
1 2l 1.2.
R 2 2 2
R
( 1a 2 (b – a )) (1.1 2.(4 – 1))
2d
2 2
Fe = N sin
(1 6) 7
mg = Ncos
= 0.3 Fe = mg tan
18. Answer (12)
1 q2 d
mg
5 4 0 4 d 2 R 2
–d2
Hint : Rab 1st case =
11
160 d 3 mg
5 q
Rab 2nd case =
12 R2 d 2
PART - II (CHEMISTRY)
21. Answer (A) 25. Answer (B, D)
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Test - 1 (Paper - II) (Code-C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
0.08 10 18 –14
4
M 360 ⎡OH– ⎤ K w 10 1.0
⎣ ⎦ [H ] 10 –14
M = 25
Ksp = [M2+][OH–]2
35. Answer (A)
K sp 1.0 10 –20
1 w 18 [M2 ] = 10–20
[OH]2 (1)2
3 54 100
2
w = 100 M(aq) 2e – M(s)
92
U238 90Th234 + 2He4 -emission
= 0.38 – 0.03(20)
36
Kr87 37Rb87 + –1e0 –-emission 0.38 – 0.6 = – 0.22
79
Au194 + –1e0 78Pt194 K-capture – 22 × 10–2 = – a × 10–2
6
C14 7N14 + –1e0 –-emission a = 22
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-C) (Answers & Hints)
1 > f(x)
2
f(x + 4) = sin ( x 4) cos4 ( x 4)
8 8 f(x) (–, 1)
f(0) = 0
1 2x
Now, sin–1 sin2 = sin
f(0) = 0 1 x2
42. Answer (D)
sin–1sin2 = 2 if
1 4 4
(P) domain is (–, 0) ...(i)
|x| x x = tan, – 1 x 1
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Test - 1 (Paper - II) (Code-C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
44. Answer (A, C)
1 2x
2 = sin ,–1x1
1 x2 1
Replacing x by and solving we get
x
1 2x
2tan–1x = sin , –1 x 1
1 x2 1 ⎛ 6 ⎞
f(x) = ⎜ 8 x 10 ⎟
28 ⎝ x ⎠
1 tan2 1 x2
(Q) cos2 =
1 tan2 1 x2 1 ⎛ 6 ⎞
f(x) = ⎜ 8 2⎟0
28 ⎝ x ⎠
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-C) (Answers & Hints)
49. Answer (A, B, C, D)
⎡ 1 1⎤ ⎡ 1 1⎤
2
⎡ 1 2⎤
Now A = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ (f(x)2 = 4 – x2
⎣0 1⎦ ⎣0 1⎦ ⎣ 0 1⎦
x2 + (f(x))2 = 4
⎡1 3⎤ ⎡ 1 10 ⎤ x = 2cos
A3 = ⎢
0 1 ⎥ A10 = ⎢ ⎥
⎣ ⎦ ⎣0 1 ⎦ and f(x) = 2sin
QQT = (PAPT)(PAPT)T
= ⎡⎣2 13, 2 13 ⎤⎦
T T T
= (PAP )(PA P )
50. Answer (B)
= PA(PTP)ATPT 51. Answer (A)
I ⎛x⎞
f ( x ) f ⎜ ⎟ = 26x3
⎝3⎠
47. Answer (B, C)
x
x + 2y = 3 Replace x by
3
7x – y = 6 3
⎛x⎞ ⎛ x ⎞ ⎛x⎞
2x – y = [k] f ⎜ ⎟ f ⎜ 2 ⎟ 26 ⎜ ⎟
⎝3⎠ ⎝3 ⎠ ⎝3⎠
To exist the solution of the system of equations Again replace
f⎜ 2⎟f⎜ 3 ⎟ 26 ⎜ 2 ⎟
48. Answer (A, B) ⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠
..................................
tan–1x = sin–1(x + 3k)
3
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞
f ⎜ n ⎟ f ⎜ n 1 ⎟ 26 ⎜ n ⎟
⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠
⎛ x ⎞ ⎛ 1 ⎞
f ( x ) f ⎜ n 1 ⎟ = 26 x 3 ⎜ 1 3 ... ⎟
⎝3 ⎠ ⎝ 3 ⎠
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Test - 1 (Paper - II) (Code-C) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
52. Answer (B) |AB| = 10 × 12{8 – 7} – 13 × {72 – 60} + 14 × {63 – 60}
= 120 – 13 × 12 + 42
⎛ 1 22k 1 ⎞ ⎛ 2k 1 ⎞
∑ cot 1 ⎜⎝ 2k 1 ⎟⎠ = ∑ tan1 ⎜⎜ 1 2k 2k 1 ⎟⎟ = 162 – 156
k 1 k 1 ⎝ ⎠
=6
⎛ ⎛ 2 2 k k 1 ⎞⎞ |adj(AB)| = |AB|(n–1) = (6)2 = 36
= ∑ ⎜⎜ tan1 ⎜⎜ 1 2k 2k 1 ⎟⎟ ⎟⎟
k 1⎝ ⎝ ⎠⎠ |adj(adj(A))| = |A|(n–1)2 = |A|4
Now |A| = 2 × (2 – 2) – 1(1 – 2) + 3(1 – 2)
= ∑ tan1 2k tan1 2k 1
= 0 – 1(–1) + 3(–1)
k 1
=1 – 3 = – 2
=
2 4 4 |adj(AB)| + |(adj(adj(A))| = 36 + 16 = 52
⎛ 2r 1 ⎞ 56. Answer A(P, Q); B(T); C(P, Q); D(P, Q)
and ∑ tan1 ⎜⎜ 1 22r 1 ⎟⎟ 4
r 1 ⎝ ⎠
5x 51 x
(A) f(x) + f(1 – x) =
53. Answer (A) 5x 5 51 x 5
⎛ 2r ⎞
∑ tan1 ⎜ 1 (1 r 2 r 4 ⎟ 5x
5
r 1 ⎝ ⎠ = x
5 5 5 5x
2 2
1 ⎛ ( r r 1) ( r r 1) ⎞ =1
= ∑ tan ⎜
⎜ 1 (1 r r 2 )(1 r r 2 ⎟⎟
r 1 ⎝ ⎠ 128
⎛ r ⎞
∑ f ⎜⎝ 129 ⎟⎠ = 64
∑ tan1(r 2 r 1) tan1(r 2 r 1)
r 1
=
r 1
1
(P) sin–1(sinx) =
= 10
4
54. Answer (C) By using graph we get total no. of solution = 64
55. Answer (D) (B) Put x = y = 1
Solution for Q. Nos. 54 and 55 f(2) = f(1) + f(1) – 1 – 1
⎡ 4 1 3 ⎤ =0
⎢ 0 ⎥⎥
3A – 2B = ⎢ 1 0 put x = 2, y = 1 f(3) = f(2) + f(1) – 2 – 1
⎣⎢ 1 1 1 ⎥⎦
= 0 + 1 – 3 = –2
|3A – 2B| = 0 Put x = 3, y = 1 f(4) = f(3) + f(1) – 3 – 1
(–1){–1 + (3 – )} = 0 = 2 f(4) = –2 + 1 – 4 = – 5
AB = ⎢⎢ 9 12 12 ⎥⎥ 5
⎢⎣ 5 7 8 ⎥⎦ 80 ∑ f (r ) = 80 – 15 = 65
r 1
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-C) (Answers & Hints)
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Test - 1 (Paper - II) (Code-D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (D) 21. (A) 41. (C)
2. (C) 22. (B) 42. (D)
3. (A) 23. (A) 43. (C)
4. (C) 24. (A, D) 44. (A, B, C, D)
5. (A, B, C, D) 25. (A, C, D) 45. (A, B)
6. (A, B, C, D) 26. (A, B, C, D) 46. (B, C)
7. (A, C, D) 27. (B, D) 47. (B, D)
8. (A, B, D) 28. (B, D) 48. (A, B, C, D)
9. (A, B, C) 29. (A, B, C) 49. (A, C)
10. (B) 30. (C) 50. (C)
11. (A) 31. (A) 51. (D)
12. (D) 32. (D) 52. (B)
13. (B) 33. (D) 53. (A)
14. (C) 34. (B) 54. (B)
15. (C) 35. (B) 55. (A)
16. A(S, T) 36. A (P, R) 56. A(P, R, S)
B(R) B (Q, S) B(P, R, S)
C(P) C (P, T) C(T)
D(Q) D (Q, S) D(Q)
17. A(R) 37. A (R) 57. A(P, Q)
B(S, T) B (S, T) B(T)
C(S, T) C (P, T) C(P, Q)
D(R) D (Q, T) D(P, Q)
18. (16) 38. (22) 58. (00)
19. (30) 39. (99) 59. (49)
20. (12) 40. (20) 60. (99)
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-D) (Answers & Hints)
A 5
C 5 0.2 7.5 750 cm
⎡ 2K d ⎤ 0.4
2ln ⎢ 0 ⎥
⎣ 2K 0 ⎦ 8. Answer (A, B, D)
Field due to q1 at other capacitor
5. Answer (A, B, C, D)
We have A1 = 4r2 q1 ⎛ 1 1 ⎞ 2q 1l
E ⎜⎜ 2 – 2 ⎟
⎟ (r l )
4 0 ⎝r (r l ) ⎠ 4 0 r 3
A2 = 4(2r)2 = 4A1
F = q2[E(r) – E(r + l)]
2 9
A3 4(3r ) 9 A1 A2
4 q 1q 2 l ⎡ 1 1 ⎤ 3q 1q 2 l
2
= 4 ⎢ 3 – ⎥
1 = 2 = 3 through same solid angle 0 ⎣r (r l ) 3 ⎦ 2 0 r 4
1 Q Q Q Q 9. Answer (A, B, C)
1 = ∫ EdA ∫ dA 4 4 4
4 0 r 2 0 0 0 – +
– +
1 Q Q Repulsive force on q1 is less than attractive force.
Flux passing through A2 within area A1= 4 4 = 16 10. Answer (B)
0 0
11. Answer (A)
1 4Q Q
Flux passing through A3 within area A2 = 9 4 = 9 P.D across resistance as a function of time
0 0
v0
6. Answer (A, B, C, D) vR t (0 t t0)
t0
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Test - 1 (Paper - II) (Code-D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
v0 1
v R 2v 0 – t (2t0 t 6t0)
2t0
a b
v R –v 0 (6t0 t 7t0) 120°
v0
vR t – 8v 0 (7t0 t 8t0) t = ; = 1; a = 1; b = 2
t0
2 2.1 2
v R
Current is a function of time is R ⎛ b ⎞ 3 ln 2 3ln 2
3t ln ⎜ ⎟
R ⎝a⎠
v0
0 t t0; i t 2
Rt0
1
a b
dq v 0
t c
dt Rt0
t = ; 1= 1; 2 = 2 ; a = 1
q t
v0
∫ dq Rt0 ∫ t dt b=2
0 0
c=4
v0 2 1 2
q t (parabolic curve) .1.2 2
2Rt0 R
⎛ b c⎞ ⎛ 2 4 ⎞ 3ln2
t ⎜ 2 ln 1 ln ⎟ ⎜ 2ln 1ln ⎟
⎝ a b⎠ ⎝ 1 2⎠
In the interval
q t
l = , = 1, a = 1, b = 2
v
∫ dq 0 ∫ dt
Rt
v 0t 0 l 1
2R
0
R 0.5
ab .1.2 2
v 0t0 v 0
q (t – t0 ) linear variation
2R R
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-D) (Answers & Hints)
Fe = N sin
⎡ 1 1⎤
R –
4 ⎢⎣ a b ⎥⎦
mg = Ncos
Fe = mg tan
1 ⎡1 1 ⎤ 1
= – 0.04 1 q2 d
4 ⎢⎣1 2 ⎥⎦ 8 mg
4 0 4 d 2 R 2
–d2
a 160 d 3 mg
b q
R2 d 2
2 2 1 1 1 1 1
(1 6) 7 C C1 C 2 C 3 C 4
= 0.3
0 bh
17. Answer A(R); B(S, T); C(S, T); D(R) C
30d
18. Answer (16) 20. Answer (12)
5
Hint: Rab 1st case =
R 11
2d 5
Rab 2nd case =
12
PART - II (CHEMISTRY)
21. Answer (A) 28. Answer (B, D)
Fact For reversible reaction
22. Answer (B)
dx
23. Answer (A) K1[P][Q] – K 2 [R][S]
dt
24. Answer (A, D)
As2S3 is negatively charged and Fe(OH)3 is positively dx
Reaction ceases when 0
charged so mutual coagulation occurs. dt
25. Answer (A, C, D) 29. Answer (A, B, C)
i n
º
- Ered of any electrode is relative w.r.t. reference
n = number of ions after dissociation
electrode.
26. Answer (A, B, C, D)
Fact º
- Ecell and Ecell cannot depend on mass of the system
27. Answer (B, D)
- The solid nature of oxidizing and reducing agents
º
Ecell must be positive for a spontaneous process. prevent direct contact.
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Test - 1 (Paper - II) (Code-D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
30. Answer (C) Ksp = [M2+][OH–]2
po ps n2
K sp 1.0 10 –20
ps n1 [M2 ] = 10–20
[OH]2 (1)2
po – ps loss in weight of solvent
ps loss in weight of solution
2
M(aq) 2e – M(s)
0.08 10 18
4 M 360
º 0.06 ⎡ 1 ⎤
M = 25 Ered Ered – log ⎢ ⎥
2 ⎣10 –20 ⎦
31. Answer (A)
= 0.38 – 0.03(20)
1 w 18
3 54 100 0.38 – 0.6 = – 0.22
Au 194
+ –1e 78Pt
0 194
K-capture M=
79 4
6
C 7N + –1e
14 14 0
–-emission
37. Answer A(R); B(S, T); C(P, T); D(Q, T) M 99
38. Answer (22)
40. Answer (20)
pH = 14
The number of milli moles of electrolyte required per
[H+] = 10–14 litre of colloidal solution is called flocculation value
–14
⎡OH– ⎤ K w 10
0.73 gm HCl = 0.02 moles
1.0
⎣ ⎦ [H ] 10 –14
= 20 millimoles 1000 ml
2 tan 2x 2x
1
sin2 =
1 tan 2
1 x2 2 = sin ,–1x1
1 x2
1 2x 2x
Now, sin–1 sin2 = sin 2tan–1x = sin
1
, –1 x 1
1 x2 1 x2
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-D) (Answers & Hints)
(Q) 3x = 3 – 3f(x)
1 tan2 1 x2
(Q) cos2 = log3x = log(3 – 3f(x))
1 tan2 1 x2
x = log3((3 – 3f(x))
2
1 1 x
cos–1cos2 = cos 3 – 3f(x) > 0
1 x2
1 > f(x)
cos–1 cos2 = 2 if 0 2 <
f(x) (–, 1)
2
1 x
2 = cos1 , 0 (R) 1 | sin x | | cos x | 2
1 x 2 2
2
x2 ⎡ 2, 1⎤
1 1 | sin x | | cos x | ⎣ ⎦
2tan x = cos
–1
,x0
1 x2
x
2 tan
2x (S) sin1 ⎛⎜ ⎞⎟ domain is [–10, 10]
(R) tan2 = 2 2 10
⎝ ⎠
1 tan 1 x
⎛ x ⎞
2x cos1 ⎜ ⎟ domain is [–20, 20]
tan–1 tan2 = tan1 ⎝ 20 ⎠
1 x2
tan–1tan2 = 2 if 2 100 x 2 domain is [–10, 10]
2 2
1
domain is (–25, 25)
4 4
625 x 2
2x
– 1 < x < 1 2tan–1x = tan1 if |x| < 1 domain of f(x) is [–10, 10]
1 x2
43. Answer (C)
(S) Let sin–1x = , 1
2 2 (P) Period of 5x + 5 – [5x + 5] is and period of
5
sin3 = 3sin – 4sin3 = 3x – 4x3
sin–1sin3 = sin–1(3x – 4x3) x
tan is 8.
8
sin–1sin3 = 3 if 3 Period of f(x) = 8
2 2
1 2 x x
x
1 (Q) f(x) = sin cos4
and 8 8
6 6 2 2
1 2
3sin–1x = sin–1(3x – 4x3), x
1 f(x + 4) = sin ( x 4) cos4 ( x 4)
2 2 8 8
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Test - 1 (Paper - II) (Code-D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
46. Answer (B, C)
2 x ⎛ 2 x ⎞ 2 x
= cos ⎜ 1 sin ⎟ sin
8 ⎝ 8 ⎠ 8 x + 2y = 3
4 x x 7x – y = 6
= cos sin2
8 8
2x – y = [k]
= f(x)
To exist the solution of the system of equations
16
[k] = 1 k [1, 2)
(R) ∑ sin1 xi = 8 x1 = x2 = x3 = ... x16 = 1
i 1
47. Answer (B, D)
(S) f(0) = 0
⎡ 3 1 ⎤
f(0) = 0 ⎢ ⎥
⎡ 1 1⎤ 2 2 ⎥
A= ⎢ ⎥ ,P= ⎢
f(0) = 0 ⎣ 0 1⎦ ⎢ 1 3⎥
⎢ ⎥
44. Answer (A, B, C, D) ⎣ 2 2 ⎦
(f(x)2 = 4 – x2 Q = PAPT
= ⎣⎡2 13, 2 13 ⎦⎤ ⎡1 3⎤ ⎡ 1 10 ⎤
A3 = ⎢ ⎥ A10 = ⎢0 1 ⎥
⎣ 0 1⎦ ⎣ ⎦
45. Answer (A, B)
R = A10
det(R) = 1
QQT = (PAPT)(PAPT)T
Number of solution of the above equation = 1
= (PAPT)(PATPT)
m=1
a+b+c=3 = PAATPT
is 3 + 3–1C3–1 = 5C2 = 10 I
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All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-D) (Answers & Hints)
48. Answer (A, B, C, D)
⎡10 13 14 ⎤
AB = ⎢⎢ 9 12 12 ⎥⎥
AT = – A
BT = – B ⎢⎣ 5 7 8 ⎥⎦
CT = C
(A) (A3B2 – B2A3)T = (A3B2)T – (B2A3)T = (B2)T(A3)T |AB| = 10 × 12{8 – 7} – 13 × {72 – 60} + 14 × {63 – 60}
= 0 – 1(–1) + 3(–1)
= (C3)T(B5)T – (B5)T(C3)T
=1 – 3 = – 2
= –C3B5 + B5C3 = B5C3 – C3B5
|adj(AB)| + |(adj(adj(A))| = 36 + 16 = 52
(D) (A3B4C5 – C5B4A3)T = (A3B4C5)T – (C5B4A3)T
= –C5B4A3 + A3B4C5 52. Answer (B)
⎡2 1 3 ⎤ ⎡1 2 4⎤
= ∑ tan1(r 2 r 1) tan1(r 2 r 1)
r 1
⎢2 3 3 ⎥
A = ⎢⎢ 1 2 2 ⎥⎥ , B = ⎢ ⎥
⎢⎣ 1 1 1⎥⎦ ⎢⎣2 2 1⎥⎦ =
4
8/10
Test - 1 (Paper - II) (Code-D) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018
54. Answer (B) f(x – 4) + f(x + 4) = f(x) ...(4)
55. Answer (A) replace x by x + 4
Solution for Q. Nos. 54 and 55 f(x) + f(x + 8) = f(x + 4) ...(5)
Adding (4) & (5) we get
⎛x⎞
f ( x ) f ⎜ ⎟ = 26x3 f(x – 4) + f(x + 8) = 0 f(x) = f(x + 24) ...(6)
⎝3⎠
period = 24
x
Replace x by f(x) = f(x + 24)
3
x
⎛x⎞ ⎛ x ⎞ ⎛x⎞
3 (C) 1 1
f ⎜ ⎟ f ⎜ 2 ⎟ 26 ⎜ ⎟ 24
3
⎝ ⎠ 3
⎝ ⎠ ⎝3⎠
x [–24, 24]
Again replace
57. Answer A(P, Q); B(T); C(P, Q); D(P, Q)
3
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞
f⎜ 2⎟f⎜ 3 ⎟ 26 ⎜ 2 ⎟ 5x 51 x
3
⎝ ⎠ ⎝3 ⎠ ⎝3 ⎠ (A) f(x) + f(1 – x) =
5x 5 51 x 5
..................................
5x 5
3 =
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ x
5 5 5 5x
f⎜ n ⎟ f ⎜ n 1 ⎟ 26 ⎜ n ⎟
⎝3 ⎠ ⎝ 3 ⎠ ⎝3 ⎠
=1
⎛ x ⎞ ⎛ 1 ⎞
f ( x ) f ⎜ n 1 ⎟ = 26 x 3 ⎜ 1 3 ... ⎟ 128
⎛ r ⎞
⎝3 ⎠ ⎝ 3 ⎠ ∑ f ⎜⎝ 129 ⎟⎠ = 64
r 1
f(x) = 27x + 1 3
And f(x) = 28 gives exactly one positive real solution. f(2) = f(1) + f(1) – 1 – 1
9/10
All India Aakash Test Series for JEE (Advanced)-2018 Test - 1 (Paper - II) (Code-D) (Answers & Hints)
= tan[tan–1a10 – tan–1a1]
Hence no solution.
59. Answer (49) ⎛ a1 ⎞ ⎞ a10 a1
1 ⎛ a
= tan ⎜ tan ⎜ 10 ⎟⎟ = 1 a a
4 –1 2 –1 2
ax + sin (x – 2x + 2) + cos (x – 2x + 2) + x + 3 = 0 ⎝ ⎝ 1 a1a10 ⎠ ⎠ 1 10
x=1
a1 9d a1 9d
=
1 a1a10 1 a1a10
a 1 3 = 0 a = 4
2 2
3
⎛ x ⎞ 9
⎟ = x 3 1 b 50 x
2
sin ⎜ 2 11 27 121
2 3 = =
⎝ ⎠ 2 25 52 11 1421
1
11 11
x= 3 b = 50 2 3 = 53 2
1421 1421 27 11
a + b = 49 S = 99
3 3 1421
10/10