test - 11 (code-a) (answers) all india aakash test series

Test - 11 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2018

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PHYSICS CHEMISTRY MATHEMATICS

31. (1)

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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.


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


5/8

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


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)


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


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


∫

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


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

⎡ ⎤⎢ ⎥⎣ ⎦


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)


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


∫

I =

20

0


∫

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


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


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

test - 11 (code-a) (answers) all india aakash test series

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