test - 11 (code-a) (answers) all india aakash test series
TRANSCRIPT
Test - 11 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2018
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1. (2)
2. (3)
3. (2)
4. (4)
5. (2)
6. (2)
7. (2)
8. (4)
9. (3)
10. (4)
11. (1)
12. (2)
13. (4)
14. (4)
15. (2)
16. (3)
17. (1)
18. (2)
19. (1)
20. (4)
21. (1)
22. (1)
23. (1)
24. (4)
25. (4)
26. (3)
27. (3)
28. (4)
29. (2)
30. (2)
PHYSICS CHEMISTRY MATHEMATICS
31. (1)
32. (2)
33. (2)
34. (3)
35. (3)
36. (1)
37. (3)
38. (1)
39. (1)
40. (3)
41. (1)
42. (2)
43. (1)
44. (3)
45. (2)
46. (3)
47. (1)
48. (2)
49. (2)
50. (2)
51. (3)
52. (2)
53. (4)
54. (4)
55. (4)
56. (2)
57. (4)
58. (1)
59. (3)
60. (1)
61. (1)
62. (2)
63. (2)
64. (4)
65. (3)
66. (2)
67. (1)
68. (3)
69. (4)
70. (3)
71. (2)
72. (2)
73. (2)
74. (3)
75. (1)
76. (3)
77. (2)
78. (1)
79. (2)
80. (4)
81. (3)
82. (2)
83. (2)
84. (4)
85. (1)
86. (3)
87. (4)
88. (1)
89. (4)
90. (4)
Test Date : 29/03/2018
ANSWERS
TEST - 11 (Code-A)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-A) (Hints)
2/8
1. l1M = l
2M
1 M
1 =
1
2
lM
l
l1M
2 = l
2M M
2 =
2
1
lM
l
Now with this balance, the salesman gives (M1 + M
2)
instead of M + M = 2M.
Loss of the salesman = M1 + M
2 – 2M
or Loss = 1 2
2 1
– 2
Ml MlM
l l
=
2 2
1 2 1 2
1 2
– 2l l l lM
l l
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
=
2
1 2
1 2
( – )M l l
l l
⎡ ⎤⎢ ⎥⎣ ⎦
2. P = 1 mm, N = 50
LC = P
N =
1mm
50 = 0.02 mm
If 44th division on the circular scale conicides with
the reference line, then the zero of the scale should
lie above the reference line, thus the instrument has
a negative zero error.
e = –(50 – 44) × 0.02 = –0.12 mm
Main scale reading is 3 mm
Circular scale reading is 26 × 0.02 = 0.52 mm
Observed reading R0
= 3 mm + 0.52 mm
= 3.52 mm
True reading Rt
= R0 – e
= 3.52 – (–0.12) = 3.64 mm
3. We know that q = CV = (20 × 240) mc = 4.8 mc
Now 100C
C
= 1
10020
= 5%
100V
V
= 6
100240
= 2.5%
100 100 100 q C V
q C V = 5 + 2.5 = 7.5%
The error in q is 7.5
4.8100
q⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0.36 mC
Thus q = (4.8 ± 0.36) mC
4. For minimum separation, v1 = v
2
A B
a1 = 2 m/s
2
u1 = 4 m/s
a2 = 4 m/s
2
u2 = 2 m/s
10 mu
1 + a
1t = u
2 + a
2t
4 + 2t = 2 + 4t
2 = 2t
t = 1 s
S1 =
2
1 1
1
2u t a t =
14 1 2 1
2 = 5
PART - A (PHYSICS)
S2 =
2
2 2
1
2u t a t =
12 1 4 1
2 = 4 m
S1 – S
2 = (5 – 4) = 1m
Minimum separation = (10 –1) = 9 m
5.21
2mv = as2 mv2 = 2as2
2 22
n
mv asF
R R
2
2 2asv
m
2a
v s
m
2dv a
vdt m
= at
F = 2 2
t nF F =
22
2 2(2 )
asas
R
⎛ ⎞ ⎜ ⎟⎝ ⎠
8.
g
v0
v
0,v v gt
�� �
2
0
1
2
� ��
r v t gt
L r p �
� �
= 2
0 0
1
2m v t gt v gt
⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
� �� �
2
0cos ˆ(– )2
mv g tL k
�
2
0cos ( )
2
� mv g tL =
2
0 0
2
cos
2
mv g v
g
=
3
0cos
2
mv
g
9. l
x
x0
mg – 2T0cos = 0
mg – 2T0cos = ma
mg – 2T0
0
2
x x
l
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
= ma
mg – 2T0
022 xx
l l
⎛ ⎞⎜ ⎟⎝ ⎠
= ma
0
0 0
22– 2 – 2
xxmg T T
l l = ma
Acceleration = 0
4–
Tx
ml
Test - 11 (Code-A) (Hints) All India Aakash Test Series for JEE (Main)-2018
3/8
= 0
4 2T
ml T
204T
ml
l = 0
2
4
T
m
10. PC – P
D = ax ...(i)
PA – P
C= –gh = –103 × 10 × 6
= –60 × 103 N/m2
PA – P
D= (P
A – P
C) + (P
C – P
D)
= –(60 × 103) + 4 × 103 = –5.6 × 104 N/m2
12. XL = L = 110
XC =
1
C = 145
R = 20
Z = 2 2
LX R = 50 5 145
1.350 5
CX
Z
13. For R1
Frequency received by R1 from wall
1 0–
v
v u
⎛ ⎞ ⎜ ⎟⎝ ⎠
and frequency received by R1, when source moves
away from R1
2 0
v
v u
⎛ ⎞ ⎜ ⎟⎝ ⎠
SR1
R2
14. C = 0A
d
C0
= 0
1 2
1 2
1 2
– ( )
A
t td t t
k k
⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0
–3 3 3(2) 3(3)
A
d d d dd
⎛ ⎞ ⎜ ⎟⎝ ⎠
15. K = eV = 21
2mv
v = 2eV
m
So, v|| =
2coseV
m
T = 2mBe
Pitch P = V||T =
2 2cos
m eV
Be m
= 2
22 cos
mV
eB
16. r2 = 90° – 60° = 30°
30° = A
A
B C
i2
60°
i1 = 90°
r2
2
2
sin
sin
i
r ,
2sin
2sin30
i
sini2 =
12
2 i
2 = 45°
= i2 – r
2 = 45° – 30° = 15°
18. E0 = B
0c = 20 × 10–9 × 3 × 108 = 6 V/m
19. For forward bias; p-side must be a higher potential
than n-side.
20. VA – 5 + 15 + 5 = V
B
VA – V
B = –15 V
VB – V
A = 15 V
22. Activity of S1 =
1
2 activity of S
2
1N
1 =
1
2(
2N
2) or
1 2
2 12
N
N
1 1
2 2
2T N
T N
1/2
ln2T⎛ ⎞⎜ ⎟⎝ ⎠
Given N1 = 2N
2,
1
2
T
T = 4
23. KE = –U = Ui – U
f = 0 – U
f
24.1 1 1
v u f
2 2
–1 1– 0dv du
v u
2
2
vdv du
u
22
2–
v f
u fu
⎛ ⎞ ⎜ ⎟⎝ ⎠
Image size =
2
–
fl
u f
⎛ ⎞⎜ ⎟⎝ ⎠
25. = D
d
d is halved and D is doubled
fringe width will become four times.
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-A) (Hints)
4/8
26.
3
4
5
27. For an ideal gas PV = nRT
PV = nRT
V nR nR V
nRTT P T
V
1V
V T T
or = 1
T
28. Decrease in surface energy = Heat required in
vaporization
T ds = L dm
T × 2 × 4r dr = L × 4r2 dr ×
2T
rL
29.1
1
4TP
r , 2
2
4TP
r
r1 < r
2
P1 > P
2
air flows from end (1) to end (2) and volume of
bubble at end-1 will decrease.
30.
2r
F Ic
8
22 250.7 0.18 N
7 3 10
PART - B (CHEMISTRY)
31. N2 + 3H
2 2NH
3
1 mole nitrogen is combining with 6 atoms of hydrogen.
2
2
xy
6
2 moles of NH3 are formed by 6 moles of atoms of
hydrogen
1
1
2xy
6
Thus 1 2 1 2
1 2
2x x 2x – xy – y –
6 6 6
⎛ ⎞ ⎜ ⎟⎝ ⎠
32.
O
+ Ph MgBrH O
3
+
CH3 H
OH Ph
CH3 H
Ph OH
+
(Geometrical isomer)
34. (1) Cr = [Ar] 4s13d5
(2) Wavelength of -rays is of the order of 10–11 m
(3) In hydrogen, the energy depends only on the
principal quantum number.
35. From PV = nRT or PV = W
RTM
2 2N O
W W
36. HC CH—CH —C—OH2
C OO C H2 5
O
1234
4-ethoxy carbonyl but 3-ene 1-oic acid.
37.
2 – 2sp 2
2 2–sp 3 3
K (SrF ) [Sr ] [F ]
K (SrCO ) [Sr ] [CO ]
–10 – 2
–10 2–
3
7.9 10 [F ]
7.0 10 [CO ]
[F–] = 3.674 × 10–2 3.7 × 10–2 M
38. 3 3 2
0100 0
2080 20
CH COOH NaOH CH COONa H O
a
[salt]pH pK log
[acid]
20pH 4.7447 log
80 = 4.14
41. Conductivity of solution of Co2[Fe(CN)
6]
k = 2.06 × 10–6 – 4.1 × 10–7
= 1.65 × 10–6 ohm–1 cm–1
Now, 2 4–2 6 6
m Co [Fe(CN) ] m (Co ) m [Fe(CN) ]2
= 6165 cm2 mol–1
Now, m
= k 1000
M
616 =
–6
1.65 10 1000
S
S = 2.7 × 10–6 (solubility)
Ksp
= 4s3 = 4 (2.7 × 10–6)3 = 7.87 × 10–17 M3
45. – 2–2I2 –2 4
(Red ppt) (Colourless)
Hg 2I HgI Hg I
2 2 –4
From cobalt thiocyonate (Blue ppt)
Hg Co 4SCN Hg Co(SCN)
46. t 0
4
9
t for [CoCl
4]2– =
418000
9 cm–1 = 8000 cm–1
Test - 11 (Code-A) (Hints) All India Aakash Test Series for JEE (Main)-2018
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PART - C (MATHEMATICS)
61. [sinx] 0 0 sinx 1
Clearly sinx = 0, 1 does not satisfy the equation
when 0 < sinx < 1, then [sinx] = 0
LHS = 1
For solution, RHS = [2 cosx] = 1
1 2 cosx < 2 1
2 cosx < 1
0, 2 , 23 3
x ⎛ ⎤ ⎛ ⎤ ⎜ ⎜⎥ ⎥
⎝ ⎦ ⎝ ⎦(∵ 0 < sinx < 1)
Integral x = 1, 7
62.
2
2 –[(–6) – 4 5 2]
5 – 6 2
4 5
x x
5x2 – 6x + 2 1
5
– < log1/5
(5x2
– 6x + 2) 1
–1 2
1/5– tan log (5 – 6 2)2 4
x x
63. At any x, distance (vertical) between curves is
difference of values of y.
48. 2 2 3 2 2 2(X)
6Cl 2Ba(OH) Ba(ClO ) 5BaCl 6H O
3 2 2 4 3 4(Y)
Ba(ClO ) H SO 2HClO BaSO
3 2 2 2
12HClO 2ClO H O O
2
52. SN1 and S
N2 both substitutions are not possible on
bridgehead carbon.
53.
CH OH2
H CO3
CH OH2
>
CH OH2
I
>
CH OH2
O N2
>
55. Anomers have different configuration on C-1 carbon
while epimer have different configuration on any other
carbon atom besides C-1.
56. Mg(s)
+ Cl2(g)
MgCl2(s)
; H1 = –642
Mg(s)
+ 1
2Cl
2(g) MgCl
(s); H
2 = –125
2MgCl disproportionation MgCl
2 + Mg; H = ?
H = H1 – 2H
2 = (–642) – (2 × –125) = –392 kJ mol–1
57. CH3
H
H H
One hydrogen atomis antiperiplanar
(I)
I
CH3
H
HCH
3
No hydrogen is antiperiplanar
(II)
I
CH3
H
H
Two hydrogen atomsare antiperiplanar
(III)
I
H
Hence order of ease of E2 will be III > I > II.
58.
N|H
+ Br—BrBr
2
N|H
H
Br+
N|H
Br
Attack at 3 position is favoured for indole.
60. g eq. of KMnO4 = g eq. of Fe2+
NV
1000 =
W
E
M n V W
1000 E
0.02 5 12 W
1000 56
W = 0.067 g
∵ Here n-factor of KMnO4 is 5 and n-factor of Fe2+ is 1.
Now, according to the stoichiometry of reaction
4 × 56 g of Fe2+ = 2 × 33 g of NH2OH
0.067 g Fe2+ = 2 33 0.067
4 56
= 0.0198 g NH
2OH
Now, 0.0198 g NH2OH is present in 50 mL solution
1000 mL of solution will contain NH2OH
= 0.0198 1000
50
= 0.396 g of NH
2OH
Now strength = Weight of solute
1000Volume of solution
= 0.396
100010
= 39.6 gL–1
2 + sinx – cos2x = 2 + sinx – (1 – 2sin2x)
= 2sin2x + sinx + 1 = 2 1
2 sin sin 12
x x⎡ ⎤ ⎢ ⎥⎣ ⎦
=
21 1
2 sin 1–4 8
x⎛ ⎞ ⎜ ⎟⎝ ⎠
=
21 7
2 sin4 8
x⎛ ⎞ ⎜ ⎟⎝ ⎠
Its minimum value = 7
8
64. Number of permutation in which no 2 ‘‘C’’ are together
=
L L O E T
2 can be arrangedC =
6
2
5!
2!C = 900
Number of permutation in which 2 ‘‘L’’ are together
but no 2 ‘‘C’’ are together
=
LL O E T
2 can be arrangedC
= 4! × 5C2 = 240
Required number of permutations = 900 – 240 = 660
65. Number of elements till end of 20 rows
= 1 + 2 + 3 + ... + 20 = 20 21
2
= 210
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-A) (Hints)
6/8
Sum of elements of 21st row
= 211 + 212 + ... 21 terms
= 21
2 [2 × 211 + 20 × 1] = 4641
66. z3 = 8i
1
32
28 ; 0, 1, 2i k
z e K
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
3 62. ; 0, 1, 2
ki
z e K
⎛ ⎞⎜ ⎟⎝ ⎠
35
26 62. ; 2 ; 2.
i ii
z e e e
⎛ ⎞⎜ ⎟⎝ ⎠
3 ; – 3 ; – 2z i i i
(0, 2)
– 3, 1 3, 1
Required area = 1
2 3 32 = 3 3 sq. units.
67. For real and distinct roots; D > 0
D = b2 – 4a > 0 b2 > 4a
For a = 1, b2 > 4 b = 3, 4
For a = 2, b2 > 8 b = 3, 4
For a = 3, b2 > 12 b = 4
For a = 4, b2 > 16 b is nothing
5 equations can be possible.
68. Coefficient of x10 in (1 – x2)10 = 10C5(1)5 (–1)5 = –10C
5
Term independent of x in
102
–x
x
⎛ ⎞⎜ ⎟⎝ ⎠
=
5
10 5
5
–2
C xx
⎛ ⎞⎜ ⎟⎝ ⎠
= –10C5 25
Required ratio =
10
5
10 5
5
– 1
32– .2
C
C
69. A =
1 3
2 –2
⎡ ⎤⎢ ⎥⎣ ⎦
; adj(A) = –2 –2 –2 –3
–3 1 –2 1
T
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
(adj(A))2 =
–2 –3 –2 –3 10 3
–2 1 –2 1 2 7
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
70. |x| + y + sin(z) = 3, |x| + 3y + 2sin(z) = 6,
|x| + y + 3sin(z) = 1
1 1 1
1 3 2 4
1 1 3
D , 1
3 1 1
6 3 2 8
1 1 3
D
2
1 3 1
1 6 2 8
1 1 3
D , 3
1 1 3
1 3 6 – 4
1 1 1
D
|x| = 1 8
24
D
D x = 2, –2
y = 8
4 = 2 and sin(z) =
–4
4
= –1
∵ x, y, z [–, 2]
x = 2, –2; y = 2; z = – 3
,2 2
71. ∵ AB = BA
1 1 1 1
0 1 0 1
a b a b
c d c d
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a a b a c b d
c c d c d
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
a = a + c
a + b = b + d c = 0; a = d
c = c
c + d = d
72.1
( ) ; 11–
f x xx
1 1–( ( ))
1 (1– ) – 11–
1–
xf f x
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 1– –1
; 0–
x xx
x x
1( ( ( )))
– 1 – ( – 1)1–
xf f f x x
x x x
x
⎛ ⎞⎜ ⎟⎝ ⎠
For f(f(f(x))) to be real; f(f(x)), f(x) must be real.
Therefore, x 0, 1.
73. 1400 = 7 × 200, 1000 = 5 × 200
Common divisor of both will be all possible divisors
of 200.
200 = 23 × 52
Number of such divisors = (3 + 1) × (2 + 1) = 12
74. Probability of getting ‘‘15’’ = 1
20
Probability of getting token with value less than or
equal to 15 =
15
1
20
1
3
4
C
C
Required probability =
1 3
4
1
1 3 27
20 4 320C
⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
75. In ABC, tan60° = 150
y y =
150m
3
60° 45°B
CD
A
150 m
y(m) x(m)
Test - 11 (Code-A) (Hints) All India Aakash Test Series for JEE (Main)-2018
7/8
In ABD, tan 45° = 150
x y x + y = 150
x = 150
150 –3
m
Speed =
1150 1–
3
2
⎛ ⎞⎜ ⎟⎝ ⎠
m/min
= 150 ( 3 – 1) 60
2 13
m/hr. = 4500( 3 –1)
3
= 4500 ( 3 –1) ( 3 1)
3
3 3 ( 3 1)
= 4500 2 3
3 ( 3 1)
= 3000 3
( 3 1) m/hr.
76.
22
2–
10 10
i i
x
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
∑ ∑
2
24 – 5
10
ix
∑ 2
290i
x ∑22
2–
20 20
i iy
y y⎛ ⎞ ⎜ ⎟
⎝ ⎠
∑ ∑
2680
iy ∑
2 2
2( )
– (5)30
i ix y
∑ =
970 25–
30 1 =
22
3
77. ~ (~ ) ( ) ( (~ )
(Tautology)
p q q p q p q p q p q
T T F T F T
T F T F T T
F T F T T T
F F T T T T
78. (0, 0)
( , )h k ( , )a bline l
xcos + ysin = p, acos + bsin = p
– 0 –cos
– 0 sin
k
h
⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = –1 k cos = h sin ...(i)
Also, h cos + k sin = p = a cos + b sin (h – a) cos + (k – b) sin = 0 ...(ii)
Combining (i), (ii) to eliminate cos, sinWe get, h(h – a) + k (k – b) = 0
h2 + k2 – ah – bk = 0
79. Required circle is passing through intersection of
point circle (x – 1)2 + (y – 2)2 = 0 and line
4x + 3y – 10 = 0.
It will be,
(x – 1)2 + (y – 2)2 + (4x + 3y – 10) = 0
x2 + y2 – 2x – 4y + 1 + 4 + (4x + 3y – 10) = 0
x2 + y2 + 2(2 – 1)x + (3 – 4)y + (5 – 10) = 0
Its radius = 5
2
2 3 – 4(2 – 1) – (5 – 10 ) 25
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
4(42 – 4 + 1) + 92 + 16 – 24 – 20 + 40 = 100
252 = 100 = ±2
Circles are : x2 + y2 + 6x + 2y – 15 = 0 and
x2 + y2 – 10x – 10y + 25 = 0
80. Let equation of circle be
(x – r)2 + (y – 4)2 = r2
and solving with
(y – 4)2 = 4(x + 2)
we get 2 2 2r 81. f(x) is continuous when 5x = x2 + 6
i.e., x2 – 5x + 6 = 0 x = 2, 3
82. Let L =
11 { }{ }
0
(1 { })lim
x
x
x
x
e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
logL =
1
{ }
0
1 (1 { })lim log
{ }
x
x
x
x e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
LHL = –
1
{ }
0
1 (1 { })lim log
{ }
x
x
x
x e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 0
1 log(1 (1 ))lim – log
(1 ) (1 )x
xe
x x
⎡ ⎤⎢ ⎥ ⎣ ⎦
= log2 – loge = 2
loge
⎛ ⎞⎜ ⎟⎝ ⎠
For L; LHL = 2
e
RHL =
1
0
1lim log(1 ) – logx
x
x e
x
⎡ ⎤⎢ ⎥⎣ ⎦
= 0
1 log(1 )lim –1x
x
x x
⎡ ⎤⎢ ⎥⎣ ⎦
=
2 3
0
– ... –1 2 3
limx
x xx x
x x
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
= –1
2
For L; RHL = e–1/2 = 1
e
LHL RHL Limit does not exist.
83. 2 2 2( –3 4 – 4) ( – 5)x x x can be interpreted
as distance of (5, –4) from (x, y); where
2–3 4 –y x x
y2 = –3 + 4x – x2 x2 + y2 – 4x + 3 = 0
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-A) (Hints)
8/8
A
O
(2, 0)B
C
(5, –4)
Required value = (CA)2 = (CO + OA)2
= 22 2(5 – 2) 4 1 = (5 + 1)2
= 36
84. Clearly g(x) is inverse of f(x).
y = f(x) =
–
–
2
x xe e
2–1
2
x
x
ey
e
e2x – 2y ex – 1 = 0
2
2 4 4
2
xy y
e
= 2
1y y
g(x) = 2log 1x x
Now, 22
1 2( ) 1
2 11
xg x
xx x
⎡ ⎤ ⎢ ⎥⎣ ⎦
= 2
10,
1
x R
x
Also, g(x) is always real.
Tangent can’t be horizontal or vertical and since
g(x) > 0 therefore, it has no extremum.
85. I = 20
–20
sin sinx x dx
∫ =
20
–20
sin(– ) sin(– )x x dx
∫
2I = 20
–20
sin sin – sinx x x dx
∫
I =
20
0
sin sin – sinx x x dx
∫
I =
20
0
– sinx dx
∫ = –40
86.Y
C(4, 1)
B
(1, 2)A(0, 1)
O(0, 0)
X
2y
x
1y x
4
xy
Required region’s area = Area of region OABCO
= 1 4
0 1
21 – . – .
4 4
x xx dx dx
x ∫ ∫
=
1 42 2
0
1
2 2.– –
13 8 8
2
x x x x xx
⎛ ⎞⎜ ⎟⎝ ⎠
= 11
3 sq. unit
87.3( 2 )
dyx y y
dx
x dy + 2y3dy = y dx
2y dy = 2
–y dx xdy xd
yy
⎛ ⎞ ⎜ ⎟⎝ ⎠
d(y2) =
xd
y
⎛ ⎞⎜ ⎟⎝ ⎠
An integrating,
C + y2 = x
y x = y (C + y2)
88. Plane through the given line can be taken as
(2x – y + 3z + 1) + (x + y + z + 3) = 0
(2 + )x + ( – 1)y + ( + 3)z + (3 + 1) = 0
∵ It is parallel to 1 2 3
x y z
1.(2 + ) + 2( – 1) + 3( + 3) = 0
–3
2
Plane will be x – 5y + 3z = 7
89. Being coplanar :
1 1
1 1 1 0
1 –1
m
m
m
1[m + (m + 1)] – 1[m – (m + 1)] + m[–1 – 1] = 0
2m + 1 + 1 – 2m = 0 2 = 0 (Not possible)
90. Let, –1
–3cos
5
⎛ ⎞ ⎜ ⎟⎝ ⎠
–3
cos5
; also 2
2sin 1– cos =
91–
25 =
4
5
sin(2) = 2sin.cos = 4 (–3)
25 5
= –24
25
� � �
Test - 11 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2018
1/8
1. (2)
2. (2)
3. (4)
4. (3)
5. (3)
6. (4)
7. (4)
8. (1)
9. (1)
10. (1)
11. (4)
12. (1)
13. (2)
14. (1)
15. (3)
16. (2)
17. (4)
18. (4)
19. (2)
20. (1)
21. (4)
22. (3)
23. (4)
24. (2)
25. (2)
26. (2)
27. (4)
28. (2)
29. (3)
30. (2)
PHYSICS CHEMISTRY MATHEMATICS
31. (1)
32. (3)
33. (1)
34. (4)
35. (2)
36. (4)
37. (4)
38. (4)
39. (2)
40. (3)
41. (2)
42. (2)
43. (2)
44. (1)
45. (3)
46. (2)
47. (3)
48. (1)
49. (2)
50. (1)
51. (3)
52. (1)
53. (1)
54. (3)
55. (1)
56. (3)
57. (3)
58. (2)
59. (2)
60. (1)
61. (4)
62. (4)
63. (1)
64. (4)
65. (3)
66. (1)
67. (4)
68. (2)
69. (2)
70. (3)
71. (4)
72. (2)
73. (1)
74. (2)
75. (3)
76. (1)
77. (3)
78. (2)
79. (2)
80. (2)
81. (3)
82. (4)
83. (3)
84. (1)
85. (2)
86. (3)
87. (4)
88. (2)
89. (2)
90. (1)
Test Date : 29/03/2018
ANSWERS
TEST - 11 (Code-B)
All India Aakash Test Series for JEE (Main)-2018
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-B) (Hints)
2/8
1.
2r
F Ic
8
22 250.7 0.18 N
7 3 10
2.1
1
4TP
r , 2
2
4TP
r
r1 < r
2
P1 > P
2
air flows from end (1) to end (2) and volume of
bubble at end-1 will decrease.
3. Decrease in surface energy = Heat required in
vaporization
T ds = L dm
T × 2 × 4r dr = L × 4r2 dr ×
2T
rL
4. For an ideal gas PV = nRT
PV = nRT
V nR nR V
nRTT P T
V
1V
V T T
or = 1
T
5.
3
4
5
6. = D
d
d is halved and D is doubled
fringe width will become four times.
7.1 1 1
v u f
2 2
–1 1– 0dv du
v u
2
2
vdv du
u
22
2–
v f
u fu
⎛ ⎞ ⎜ ⎟⎝ ⎠
Image size =
2
–
fl
u f
⎛ ⎞⎜ ⎟⎝ ⎠
8. KE = –U = Ui – U
f = 0 – U
f
9. Activity of S1 =
1
2 activity of S
2
1N
1 =
1
2(
2N
2) or
1 2
2 12
N
N
PART - A (PHYSICS)
1 1
2 2
2T N
T N
1/2
ln2T⎛ ⎞⎜ ⎟⎝ ⎠
Given N1 = 2N
2,
1
2
T
T = 4
11. VA – 5 + 15 + 5 = V
B
VA – V
B = –15 V
VB – V
A = 15 V
12. For forward bias; p-side must be a higher potential
than n-side.
13. E0 = B
0c = 20 × 10–9 × 3 × 108 = 6 V/m
15. r2 = 90° – 60° = 30°
30° = A
A
B C
i2
60°
i1 = 90°
r2
2
2
sin
sin
i
r ,
2sin
2sin30
i
sini2 =
12
2 i
2 = 45°
= i2 – r
2 = 45° – 30° = 15°
16. K = eV = 21
2mv
v = 2eV
m
So, v|| =
2coseV
m
T = 2mBe
Pitch P = V||T =
2 2cos
m eV
Be m
= 2
22 cos
mV
eB
17. C = 0A
d
C0
= 0
1 2
1 2
1 2
– ( )
A
t td t t
k k
⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0
–3 3 3(2) 3(3)
A
d d d dd
⎛ ⎞ ⎜ ⎟⎝ ⎠
Test - 11 (Code-B) (Hints) All India Aakash Test Series for JEE (Main)-2018
3/8
18. For R1
Frequency received by R1 from wall
1 0–
v
v u
⎛ ⎞ ⎜ ⎟⎝ ⎠
and frequency received by R1, when source moves
away from R1
2 0
v
v u
⎛ ⎞ ⎜ ⎟⎝ ⎠
SR1
R2
19. XL = L = 110
XC =
1
C = 145
R = 20
Z = 2 2
LX R = 50 5 145
1.350 5
CX
Z
21. PC – P
D = ax ...(i)
PA – P
C= –gh = –103 × 10 × 6
= –60 × 103 N/m2
PA – P
D= (P
A – P
C) + (P
C – P
D)
= –(60 × 103) + 4 × 103 = –5.6 × 104 N/m2
22. l
x
x0
mg – 2T0cos = 0
mg – 2T0cos = ma
mg – 2T0
0
2
x x
l
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
= ma
mg – 2T0
022 xx
l l
⎛ ⎞⎜ ⎟⎝ ⎠
= ma
0
0 0
22– 2 – 2
xxmg T T
l l = ma
Acceleration = 0
4–
Tx
ml
= 0
4 2T
ml T
204T
ml
l = 0
2
4
T
m
23.
g
v0
v
0,v v gt
�� �
2
0
1
2
� ��
r v t gt
L r p �
� �
= 2
0 0
1
2m v t gt v gt
⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
� �� �
2
0cos ˆ(– )2
mv g tL k
�
2
0cos ( )
2
� mv g tL =
2
0 0
2
cos
2
mv g v
g
=
3
0cos
2
mv
g
26.21
2mv = as
2 mv2 = 2as
2
2 22
n
mv asF
R R
2
2 2asv
m
2a
v s
m
2dv a
vdt m
= at
F = 2 2
t nF F =
22
2 2(2 )
asas
R
⎛ ⎞ ⎜ ⎟⎝ ⎠
27. For minimum separation, v1 = v
2
A B
a1 = 2 m/s
2
u1 = 4 m/s
a2 = 4 m/s
2
u2 = 2 m/s
10 mu
1 + a
1t = u
2 + a
2t
4 + 2t = 2 + 4t
2 = 2t t = 1 s
S1 =
2
1 1
1
2u t a t =
14 1 2 1
2 = 5
S2 =
2
2 2
1
2u t a t =
12 1 4 1
2 = 4 m
S1 – S
2 = (5 – 4) = 1m
Minimum separation = (10 –1) = 9 m
28. We know that q = CV = (20 × 240) mc = 4.8 mc
Now 100C
C
= 1
10020
= 5%
100V
V
= 6
100240
= 2.5%
100 100 100 q C V
q C V = 5 + 2.5 = 7.5%
The error in q is 7.5
4.8100
q⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0.36 mC
Thus q = (4.8 ± 0.36) mC
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-B) (Hints)
4/8
PART - B (CHEMISTRY)
31. g eq. of KMnO4 = g eq. of Fe2+
NV
1000 =
W
E
M n V W
1000 E
0.02 5 12 W
1000 56
W = 0.067 g
∵ Here n-factor of KMnO4 is 5 and n-factor of Fe2+ is 1.
Now, according to the stoichiometry of reaction
4 × 56 g of Fe2+ = 2 × 33 g of NH2OH
0.067 g Fe2+ = 2 33 0.067
4 56
= 0.0198 g NH
2OH
Now, 0.0198 g NH2OH is present in 50 mL solution
1000 mL of solution will contain NH2OH
= 0.0198 1000
50
= 0.396 g of NH
2OH
Now strength = Weight of solute
1000Volume of solution
= 0.396
100010
= 39.6 gL–1
33.
N|H
+ Br—BrBr
2
N|H
H
Br+
N|H
Br
Attack at 3 position is favoured for indole.
34. CH3
H
H H
One hydrogen atomis antiperiplanar
(I)
I
CH3
H
HCH
3
No hydrogen is antiperiplanar
(II)
I
CH3
H
H
Two hydrogen atomsare antiperiplanar
(III)
I
H
Hence order of ease of E2 will be III > I > II.
35. Mg(s)
+ Cl2(g)
MgCl2(s)
; H1 = –642
Mg(s)
+ 1
2Cl
2(g) MgCl
(s); H
2 = –125
2MgCl disproportionation MgCl
2 + Mg; H = ?
H = H1 – 2H
2 = (–642) – (2 × –125) = –392 kJ mol–1
36. Anomers have different configuration on C-1 carbon
while epimer have different configuration on any other
carbon atom besides C-1.
38.
CH OH2
H CO3
CH OH2
>
CH OH2
I
>
CH OH2
O N2
>
39. SN1 and S
N2 both substitutions are not possible on
bridgehead carbon.
43. 2 2 3 2 2 2(X)
6Cl 2Ba(OH) Ba(ClO ) 5BaCl 6H O
3 2 2 4 3 4(Y)
Ba(ClO ) H SO 2HClO BaSO
3 2 2 2
12HClO 2ClO H O O
2
45. t 0
4
9
t for [CoCl
4]2– =
418000
9 cm–1 = 8000 cm–1
29. P = 1 mm, N = 50
LC = P
N =
1mm
50 = 0.02 mm
If 44th division on the circular scale conicides with
the reference line, then the zero of the scale should
lie above the reference line, thus the instrument has
a negative zero error.
e = –(50 – 44) × 0.02 = –0.12 mm
Main scale reading is 3 mm
Circular scale reading is 26 × 0.02 = 0.52 mm
Observed reading R0
= 3 mm + 0.52 mm
= 3.52 mm
True reading Rt
= R0 – e
= 3.52 – (–0.12) = 3.64 mm
30. l1M = l
2M
1 M
1 =
1
2
lM
l
l1M
2 = l
2M M
2 =
2
1
lM
l
Now with this balance, the salesman gives (M1 + M
2)
instead of M + M = 2M.
Loss of the salesman = M1 + M
2 – 2M
or Loss = 1 2
2 1
– 2
Ml MlM
l l
=
2 2
1 2 1 2
1 2
– 2l l l lM
l l
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
=
2
1 2
1 2
( – )M l l
l l
⎡ ⎤⎢ ⎥⎣ ⎦
Test - 11 (Code-B) (Hints) All India Aakash Test Series for JEE (Main)-2018
5/8
PART - C (MATHEMATICS)
61. Let, –1
–3cos
5
⎛ ⎞ ⎜ ⎟⎝ ⎠
–3
cos5
; also 2
2sin 1– cos =
91–
25 =
4
5
sin(2) = 2sin.cos = 4 (–3)
25 5
= –24
25
62. Being coplanar :
1 1
1 1 1 0
1 –1
m
m
m
1[m + (m + 1)] – 1[m – (m + 1)] + m[–1 – 1] = 0
2m + 1 + 1 – 2m = 0 2 = 0 (Not possible)
63. Plane through the given line can be taken as
(2x – y + 3z + 1) + (x + y + z + 3) = 0
(2 + )x + ( – 1)y + ( + 3)z + (3 + 1) = 0
46. – 2–2I2 –2 4
(Red ppt) (Colourless)
Hg 2I HgI Hg I
2 2 –4
From cobalt thiocyonate (Blue ppt)
Hg Co 4SCN Hg Co(SCN)
50. Conductivity of solution of Co2[Fe(CN)
6]
k = 2.06 × 10–6 – 4.1 × 10–7
= 1.65 × 10–6 ohm–1 cm–1
Now, 2 4–2 6 6
m Co [Fe(CN) ] m (Co ) m [Fe(CN) ]2
= 6165 cm2 mol–1
Now, m
= k 1000
M
616 =
–6
1.65 10 1000
S
S = 2.7 × 10–6 (solubility)
Ksp
= 4s3 = 4 (2.7 × 10–6)3 = 7.87 × 10–17 M3
53. 3 3 2
0100 0
2080 20
CH COOH NaOH CH COONa H O
a
[salt]pH pK log
[acid]
20pH 4.7447 log
80 = 4.14
54.
2 – 2sp 2
2 2–sp 3 3
K (SrF ) [Sr ] [F ]
K (SrCO ) [Sr ] [CO ]
–10 – 2
–10 2–
3
7.9 10 [F ]
7.0 10 [CO ]
[F–] = 3.674 × 10–2 3.7 × 10–2 M
55. HC CH—CH —C—OH2
C OO C H2 5
O
1234
4-ethoxy carbonyl but 3-ene 1-oic acid.
56. From PV = nRT or PV = W
RTM
2 2N O
W W
57. (1) Cr = [Ar] 4s13d
5
(2) Wavelength of -rays is of the order of 10–11 m
(3) In hydrogen, the energy depends only on the
principal quantum number.
59.
O
+ Ph MgBrH O
3
+
CH3 H
OH Ph
CH3 H
Ph OH
+
(Geometrical isomer)
60. N2 + 3H
2 2NH
3
1 mole nitrogen is combining with 6 atoms of hydrogen.
2
2
xy
6
2 moles of NH3 are formed by 6 moles of atoms of
hydrogen
1
1
2xy
6
Thus 1 2 1 2
1 2
2x x 2x – xy – y –
6 6 6
⎛ ⎞ ⎜ ⎟⎝ ⎠
∵ It is parallel to 1 2 3
x y z
1.(2 + ) + 2( – 1) + 3( + 3) = 0
–3
2
Plane will be x – 5y + 3z = 7
64.3( 2 )
dyx y y
dx
x dy + 2y3dy = y dx
2y dy = 2
–y dx xdy xd
yy
⎛ ⎞ ⎜ ⎟⎝ ⎠
d(y2) =
xd
y
⎛ ⎞⎜ ⎟⎝ ⎠
An integrating,
C + y2 = x
y x = y (C + y2)
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-B) (Hints)
6/8
65.Y
C(4, 1)
B
(1, 2)A(0, 1)
O(0, 0)
X
2y
x
1y x
4
xy
Required region’s area = Area of region OABCO
= 1 4
0 1
21 – . – .
4 4
x xx dx dx
x ∫ ∫
=
1 42 2
0
1
2 2.– –
13 8 8
2
x x x x xx
⎛ ⎞⎜ ⎟⎝ ⎠
= 11
3 sq. unit
66. I = 20
–20
sin sinx x dx
∫ =
20
–20
sin(– ) sin(– )x x dx
∫
2I = 20
–20
sin sin – sinx x x dx
∫
I =
20
0
sin sin – sinx x x dx
∫
I =
20
0
– sinx dx
∫ = –40
67. Clearly g(x) is inverse of f(x).
y = f(x) =
–
–
2
x xe e
2–1
2
x
x
ey
e
e2x – 2y e
x – 1 = 0
2
2 4 4
2
xy y
e
= 2
1y y
g(x) = 2log 1x x
Now, 22
1 2( ) 1
2 11
xg x
xx x
⎡ ⎤ ⎢ ⎥⎣ ⎦
= 2
10,
1
x Rx
Also, g(x) is always real.
Tangent can’t be horizontal or vertical and since
g(x) > 0 therefore, it has no extremum.
68. 2 2 2( –3 4 – 4) ( – 5)x x x can be interpreted
as distance of (5, –4) from (x, y); where
2–3 4 –y x x
y2 = –3 + 4x – x2 x2 + y2 – 4x + 3 = 0
A
O
(2, 0)B
C
(5, –4)
Required value = (CA)2 = (CO + OA)2
= 22 2(5 – 2) 4 1 = (5 + 1)2
= 36
69. Let L =
11 { }{ }
0
(1 { })lim
x
x
x
x
e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
logL =
1
{ }
0
1 (1 { })lim log
{ }
x
x
x
x e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
LHL = –
1
{ }
0
1 (1 { })lim log
{ }
x
x
x
x e
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 0
1 log(1 (1 ))lim – log
(1 ) (1 )x
xe
x x
⎡ ⎤⎢ ⎥ ⎣ ⎦
= log2 – loge = 2
loge
⎛ ⎞⎜ ⎟⎝ ⎠
For L; LHL = 2
e
RHL =
1
0
1lim log(1 ) – logx
x
x e
x
⎡ ⎤⎢ ⎥⎣ ⎦
= 0
1 log(1 )lim – 1x
x
x x
⎡ ⎤⎢ ⎥⎣ ⎦
=
2 3
0
– ... –1 2 3
limx
x xx x
x x
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
= –1
2
For L; RHL = e–1/2 = 1
e
LHL RHL Limit does not exist.
70. f(x) is continuous when 5x = x2 + 6
i.e., x2 – 5x + 6 = 0 x = 2, 3
71. Let equation of circle be
(x – r)2 + (y – 4)2 = r2
and solving with
(y – 4)2 = 4(x + 2)
we get 2 2 2r 72. Required circle is passing through intersection of
point circle (x – 1)2 + (y – 2)2 = 0 and line
4x + 3y – 10 = 0.
It will be,
(x – 1)2 + (y – 2)2 + (4x + 3y – 10) = 0
Test - 11 (Code-B) (Hints) All India Aakash Test Series for JEE (Main)-2018
7/8
x2 + y2 – 2x – 4y + 1 + 4 + (4x + 3y – 10) = 0
x2 + y2 + 2(2 – 1)x + (3 – 4)y + (5 – 10) = 0
Its radius = 5
2
2 3 – 4(2 – 1) – (5 – 10 ) 25
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
4(42 – 4 + 1) + 92 + 16 – 24 – 20 + 40 = 100
252 = 100 = ±2
Circles are : x2 + y
2 + 6x + 2y – 15 = 0 and
x2 + y2 – 10x – 10y + 25 = 0
73. (0, 0)
( , )h k ( , )a bline l
xcos + ysin = p, acos + bsin = p
– 0 –cos
– 0 sin
k
h
⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = –1 k cos = h sin ...(i)
Also, h cos + k sin = p = a cos + b sin (h – a) cos + (k – b) sin = 0 ...(ii)
Combining (i), (ii) to eliminate cos, sinWe get, h(h – a) + k (k – b) = 0
h2 + k2 – ah – bk = 0
74. ~ (~ ) ( ) ( (~ )
(Tautology)
p q q p q p q p q p q
T T F T F T
T F T F T T
F T F T T T
F F T T T T
75.
22
2–
10 10
i i
x
x x⎛ ⎞ ⎜ ⎟
⎝ ⎠
∑ ∑
2
24 – 5
10
ix
∑ 2
290i
x ∑22
2–
20 20
i iy
y y⎛ ⎞ ⎜ ⎟
⎝ ⎠
∑ ∑
2680
iy ∑
2 2
2( )
– (5)30
i ix y
∑ =
970 25–
30 1 =
22
3
76. In ABC, tan60° = 150
y y =
150m
3
60° 45°B
CD
A
150 m
y(m) x(m)
In ABD, tan 45° = 150
x y x + y = 150
x = 150
150 –3
m
Speed =
1150 1–
3
2
⎛ ⎞⎜ ⎟⎝ ⎠
m/min
= 150 ( 3 –1) 60
2 13
m/hr. = 4500( 3 –1)
3
= 4500 ( 3 –1) ( 3 1)
3
3 3 ( 3 1)
= 4500 2 3
3 ( 3 1)
= 3000 3
( 3 1) m/hr.
77. Probability of getting ‘‘15’’ = 1
20
Probability of getting token with value less than or
equal to 15 =
15
1
20
1
3
4
C
C
Required probability =
1 3
4
1
1 3 27
20 4 320C
⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
78. 1400 = 7 × 200, 1000 = 5 × 200
Common divisor of both will be all possible divisors
of 200.
200 = 23 × 52
Number of such divisors = (3 + 1) × (2 + 1) = 12
79.1
( ) ; 11–
f x xx
1 1–( ( ))
1 (1– ) – 11–
1–
xf f x
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 1– –1
; 0–
x xx
x x
1( ( ( )))
– 1 – ( – 1)1–
xf f f x x
x x x
x
⎛ ⎞⎜ ⎟⎝ ⎠
For f(f(f(x))) to be real; f(f(x)), f(x) must be real.
Therefore, x 0, 1.
80. ∵ AB = BA
1 1 1 1
0 1 0 1
a b a b
c d c d
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a a b a c b d
c c d c d
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
a = a + c
a + b = b + d c = 0; a = d
c = c
c + d = d
81. |x| + y + sin(z) = 3, |x| + 3y + 2sin(z) = 6,
|x| + y + 3sin(z) = 1
All India Aakash Test Series for JEE (Main)-2018 Test - 11 (Code-B) (Hints)
8/8
� � �
1 1 1
1 3 2 4
1 1 3
D , 1
3 1 1
6 3 2 8
1 1 3
D
2
1 3 1
1 6 2 8
1 1 3
D , 3
1 1 3
1 3 6 – 4
1 1 1
D
|x| = 1 8
24
D
D x = 2, –2
y = 8
4 = 2 and sin(z) =
–4
4
= –1
∵ x, y, z [–, 2]
x = 2, –2; y = 2; z = – 3
,2 2
82. A =
1 3
2 –2
⎡ ⎤⎢ ⎥⎣ ⎦
; adj(A) = –2 –2 –2 –3
–3 1 –2 1
T
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
(adj(A))2 =
–2 –3 –2 –3 10 3
–2 1 –2 1 2 7
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦83. Coefficient of x10 in (1 – x2)10 = 10
C5(1)5 (–1)5 = –10
C5
Term independent of x in
102
–x
x
⎛ ⎞⎜ ⎟⎝ ⎠
=
5
10 5
5
–2
C xx
⎛ ⎞⎜ ⎟⎝ ⎠
= –10C
5 25
Required ratio =
10
5
10 5
5
– 1
32– .2
C
C
84. For real and distinct roots; D > 0
D = b2 – 4a > 0 b2 > 4a
For a = 1, b2 > 4 b = 3, 4
For a = 2, b2 > 8 b = 3, 4
For a = 3, b2 > 12 b = 4
For a = 4, b2 > 16 b is nothing
5 equations can be possible.
85. z3 = 8i
1
32
28 ; 0, 1, 2i k
z e K
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
3 62. ; 0, 1, 2
ki
z e K
⎛ ⎞⎜ ⎟⎝ ⎠
35
26 62. ; 2 ; 2.
i ii
z e e e
⎛ ⎞⎜ ⎟⎝ ⎠
(0, 2)
– 3, 1 3, 1
3 ; – 3 ; – 2z i i i
Required area = 1
2 3 32 = 3 3 sq. units.
86. Number of elements till end of 20 rows
= 1 + 2 + 3 + ... + 20 = 20 21
2
= 210
Sum of elements of 21st row
= 211 + 212 + ... 21 terms
= 21
2 [2 × 211 + 20 × 1] = 4641
87. Number of permutation in which no 2 ‘‘C’’ are together
=
L L O E T
2 can be arrangedC =
6
2
5!
2!C = 900
Number of permutation in which 2 ‘‘L’’ are together
but no 2 ‘‘C’’ are together
=
LL O E T
2 can be arrangedC
= 4! × 5C
2 = 240
Required number of permutations = 900 – 240 = 660
88. At any x, distance (vertical) between curves is
difference of values of y.
2 + sinx – cos2x = 2 + sinx – (1 – 2sin2x)
= 2sin2x + sinx + 1 =
2 12 sin sin 1
2x x
⎡ ⎤ ⎢ ⎥⎣ ⎦
=
21 1
2 sin 1–4 8
x⎛ ⎞ ⎜ ⎟⎝ ⎠
=
21 7
2 sin4 8
x⎛ ⎞ ⎜ ⎟⎝ ⎠
Its minimum value = 7
8
89.
2
2 –[(–6) – 4 5 2]
5 – 6 2
4 5
x x
5x2 – 6x + 2
1
5
– < log1/5
(5x2
– 6x + 2) 1
–1 2
1/5– tan log (5 – 6 2)2 4
x x
90. [sinx] 0 0 sinx 1
Clearly sinx = 0, 1 does not satisfy the equation
when 0 < sinx < 1, then [sinx] = 0
LHS = 1
For solution, RHS = [2 cosx] = 1
1 2 cosx < 2 1
2 cosx < 1
0, 2 , 23 3
x ⎛ ⎤ ⎛ ⎤ ⎜ ⎜⎥ ⎥
⎝ ⎦ ⎝ ⎦(∵ 0 < sinx < 1)
Integral x = 1, 7