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1. Ans. (2)Required probability = 1 probability of nowicket keeper is selected.

1311

1511

C 331

C 35= - =

2. Ans. (2)General term (T

r+1) = 40C

r(2x1/2)r

for 'x' with integral power r I2

, wherer [0,40] r = 0, 2, 4.......40 21 terms.

3. Ans. (3)

( )x 2xx 0lim x e e .....

+

- -+

+ + x

x xx 0 x 0

xe xlim lim 1

1 e e 1+ +

-

-+ += = =- -

4. Ans. (1)24 24 24

2

r 1 r 1 r 1

S r(25 r) 25 r r= = =

= - = - 24.25 24.25.49

25. 26002 6

= - =

5. Ans. (2)

Put 2

x2 tan

2sin xx

1 tan2

=+

SOLUTION

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Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 7 2

PATTERN : JEE (Main)LEADER & ENTHUSIAST COURSESCORE-I

Date : 08 - 03 - 2014TARGET : JEE (Main) 2014

TEST # 03

2 2

22

x xsec dx sec dx

12 2I

x x 5 x 4 95tan 8tan 5 tan2 2 2 5 25

= =

+ + + +

1

x 4tan1 5 2 5. .2 tan c

35 35

-

+ = +

12 5 x 4tan tan c

3 3 2 3- = + +

2A

3= & 5B

3=

6. Ans. (3)2 2

2 2x sin x x 1 cos x.sec xdx sec xdxx 1 x 1

- - +=- -

2 1sec x dx tan x n | x 1 | cx 1

= + = + - + - l7. Ans. (3)

( )4 r 15 5

2

r 0 r 02 r 2

(x)dx (x)dx r 2-

= =- -= = -

= 4 + 1 + 0 + 1 + 4 + 9 = 198. Ans. (2)

( )x

0

g x (t)dt= g'(x) = (x) an odd function. g(x) is an even function.

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20A. 2 2 3 1 2 3 3 2 4 2 3 3 4 3 2 3 3 2 3 2Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40A. 4 1 3 1 4 2 2 1 3 1 1 4 2 4 1 2 4 1 2 2Q. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60A. 3 1 3 3 1 4 1 1 3 4 3 3 2 3 3 2 3 4 3 3Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80A. 1 3 1 1 2 2 3 1 3 2 1 4 2 4 2 1 2 3 4 2Q. 81 82 83 84 85 86 87 88 89 90A. 4 3 4 2 3 2 2 3 4 2

ANSWERKEY

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9. Ans. (4)

2 2 2 2 2 2n

n 1 n 2 n nlim ....

n 1 n 2 n n+ + + + + + + + +

n

2nr 1

r1

1 nlim

n r1

n

=

+ = +

=1

20

1 xdx

1 x++

1 1

2 20 0

dx 1 2xdx

1 x 2 1 x= ++ +

11 2

0

1 1tan x n 1 x n2

2 4 2- p = + + = +

l l

10. Ans. (2)

( )( )

2 2 2

22 2x 0 2

sin 6 tan x 6 tan x xlim . .

6 tan x x 2x x -( )( )

( )( )( )

22 2

2 2

2x x cos 2x x 1.

cos 2x x 1 n 1 cos 2x x 1

- - -- - + - -l

= 1.6.1.(2).1 = 1211. Ans. (3)

6

3

6

x ; x 1

(x) x ; 1 x 1

x ; x 1

< -= - >

Discontinuous at x =1 and non differentiableat x = 1

12. Ans. (3)

( )32x 3 27(x) 2 - +=

( ) ( )32 2x 3 27 2 '(x) 2 .3 x 3 .2x

- += -Q '(x) changes it's sign from negative topositive at x = 0

\ (x) will be minimum at x = 0 (0) = 113. Ans. (4)

Possible curve of (x) will be

a>0 a 118. Ans. (2)

If identical letters are symmetric about middleletter then middle letter should be 'A'.Now in first four letters there will be one 'A' &one 'M' and one 'A' & one 'M' will be in lastfour (symmetric to first four) this can be done in4P2.4! = 12.24 = 288 ways.

19. Ans. (3)

I =/ 4

2 20

tan xdx

1 tan x.sin x

p

+

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Path to success KOTA (RAJASTHAN)LEADER & ENTHUSIAST COURSE

/ 4

2 40

sin x cos xdx

cos x sin x

p= +Let sin2x = t sin2xdx = dt

1/ 2 1/ 2

22 20 0

1 dt 1 dtI

2 t t 1 2 1 3t

2 2

= =- + - +

( ) 1/ 21

0

12 t1 2tan3 3 6 3

-- p= =

20. Ans. (2)

For (x) to be continuous at x =2p-

2 2 2

- + p p p - = - = - sina cos2a = 2 .......(1)

and for continuous at x2p=

2 2 2

+- p p p = = cos2a + sina = 0 ........(2)by (1) & (2)

sina = 1 & cos2a = 15

, ,2 2 2p p 9pa =

21. Ans. (4)Statement-I : '(0) is not defined. Q x 0

22. Ans. (1)

Q

2 2 2a b ccosC

2ab+ -=

If 'C' is an obtuse angle, then cosC < 0

c2 > a2 + b2 = (a + b)2 2ab= 144 70 = 74.

23. Ans. (3)1.2.3 + 2.3.4 + .... + n(n + 1)(n + 2)

n(n 1)(n 2)(n 3)4

+ + +=and 1.2 + 2.3 + 3.4 + ...... + n(n + 1)

n(n 1)(n 2)3

+ +=

\n

1.2.3 2.3.4 .... n termslim

n(1.2 2.3 ..... n terms)+ ++ +

2n

3n(n 1)(n 2)(n 3) 3lim

4.n (n 1)(n 2) 4+ + += =

+ +

But 33n 5

nr 1 0

1 rlim (x)dx

n n

+

=

= .24. Ans. (1)

Q 1 + ar + a2r+.......+ a(n1)r0 ; r n

; In ; r n

l= l = l

\ ( )( )50 n 1 rr 2rr 0

1 54-

=+ a + a + a =

(Q equal to '9' for r = 0,9,18,27,36,45)25. Ans. (4)

g(x) = sinx(sin3x + sinx)\ g(x) = 2(x) sinx(sin3x + sinx) = sin2x sinx.sin3x = 0sinx = 0 x = 0, p, 2p .....(1)

2 4 5x 0, , , , , , 23 3 3 3p p p p= p p .....(2)

from (1) & (2), '7' solutions26. Ans. (2)

2 2sin cos (1 sin cos ) 1cosC

2sin cos 2a + a - + a a= = -a a

2C3p=

D = 1 2sin .cos .sin2 3

pa a 3 sin 28

= a

\ Dmaximum=3

827. Ans. (2)

Let (x) = ax3 + bx2 + cx 1 (Q (0) = 1)

Q (1) = 0 a + b + c = 1 .....(1)'(0) = 0 c = 0 .....(2)at x = 0 there is no extremum

\ ''(0) = 0 b = 0 .....(3)from (1), (2) & (3), a = 1

\ (x) = x3 1

\ 3(x)

dx x Cx 1

= +-

01CT313072KOTA / HS - 4

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28. Ans. (1)

A(t) = 2000 10t2(5 t / 8)e -

2 2t t5 5

8 8dA t10e 10t. e 0dt 4

- -= - + = , t = 2

Maximum of A(t) = max{A(0), A(2), A(10)}= A(0) = 2000 = M

Mn 10

199 = =

Area of polygon formed by nth root of unity

n 2sin

2 np =

= 5 5

5sin 36 58

- =

29. Ans. (3)

(x) =1 x xsin sin

3 3- =

& x (3, 3); x I

&x x

g(x) x (0,3); x I3 3

= = " /

\ required probability = 10 122 10

=

30. Ans. (1)

If P(x) is of degree 'n', then degree of 3 2

2

y d ydx

will be =3n n

2 2n 22 2

+ - = -

Now degree of Q(x) = 2n 3

Difference = 2n (2n 3) = 3

31. Ans. (1)

Sol.L

0BVdy = voltage induced

32. Ans. (4)Sol. Use the lens maker formula

-

-m

m=21m

L

R1

R11

f1

where L and m

represent the refractive indices of the liquid filledand the surrounding medium respectively. Note

that for a double concave lens

-

21 R1

R1 is

negative. Now, for the lens to act as a diverginglens, that is, for f to be negative, the first bracket

on the right hand side of the relation must bepositive. This requires L to be greater than m.

33. Ans. (2)

Sol. Sensitivity = KBNA

, by increasing B (stronger

magnet) and decreasing torsion constant, k(weaker springs)

34. Ans. (4)35. Ans. (1)Sol. DV

in = 50mV, DI

B = 50A, DI

C = 5mA

RL = 5kW, DV

0 = ?

0 L

i in

V RV R

D = bD, where b = C

B

II

DD & Rin =

in

B

VI

DD

\ 0V 5mA 5k 50A

10mV 50A 50mV

D W = DV

0 = 5V

36. Ans. (2)Sol. Dsinf = nl

Dsin45 = 150

(1)75

37. Ans. (4)Sol. In sample x no impurity level seen, so it is

undoped. In sample y impurity energy level liesbelow the conduction bond so it is doped withfifth group impurity.In sample z, impurity energy level lies abovethe valence band so it is doped with third groupimpurity.

38. Ans. (1)Sol. Before,

Voltage V2 =

10084V

100 200

+

V1 =

8084V

80 200

+39. Ans. (2)40. Ans. (2)41. Ans. (3)42. Ans. (1)43. Ans. (3)

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Sol. X = Y = 2f, ( )1 2

1 1 11

f R R

= m - -

44. Ans. (3)45. Ans. (1)46. Ans. (4)47. Ans. (1)

Sol. Imax

= IP + 0

I2

for q = 0

Imin = 0I2

for q = 90

Imin 4 = Imax (Given)

48. Ans. (1)49. Ans. (3)50. Ans. (4)51. Ans. (3)52. Ans. (3)53. Ans. (2)54. Ans. (3)Sol. (Q. No. 55 & 56)

Let the activity of the nuclide be A nuclei persecond. It means A b-particle are emitted persecond. If a spherical surface of radius r withcentre at position of nuclide be considered, thenduring elemental time interval dt a number (A.dt)of b-particles cross this surface. If the velocityof the b-particle be v, then (A . dt)b-particles are in space having the shape of aspherical shell of radius r and radial thickness(v . dt).Volume of this space = (4pr2)v dt.Concentration of b-particles at distance r fromthe nuclide is

h = 2A . dt

4 r (v dt)pActivity of the nuclide A = 4pr2vh.But A = lN

N = 24 r v T

0.6931p h

Kinetic energy of b-particle, E = 21 mv2

2Ev

m=

DP = mv 0 = mv

Force experienced by plateF = DP no. of particles striking/second,

= mv 20.6931 NS

4 r Tp

P = 2F 0.6931N

mvS 4 r T

=p

55. Ans. (3)Sol. P Bt =

r r

Bat center

= I

2Rm

56. Ans. (2)

Sol.

P

S1

( 1)tm - bl

O q

For maxima n

sindlq = q = sin1 (nl/d)

For 2 / 2-p q p 2d

n =l +1

For 2 / 2-p < q < p 2d

n =l 1

Total number of maxima = 4d l

57. Ans. (3)Sol. Let A be the work function of metal.

1

hcl = A + 2

mv21

2

hcl = A + 2

mv22

hc

l-l 2111

= ( )2221 vv2m - = [ ]2222 vv42m -=

2mv3 22

2mv22 =

l-l 2111

3hc

A = 2mvhc 22

2-l =

l-l-l 21211

3hchc

= 12 3

hc3

hc4l-l

=

l-l 2141

3hc

01CT313072KOTA / HS - 6

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Path to success KOTA (RAJASTHAN)TARGET : JEE (Main) 2014

=

-

--

-

99

834

103501

104504

31031062.6

\ A = 3.93 1019 J.58. Ans. (4)

Sol. 2x 2= q , 1y 2= q & q2 = mq1eye

\\\\\\\\\ \\\

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\

\\\\\

Dq '

q1

q 2

D'

x y

(with water)

eye\\\\\\

\\\\\\\ \\\ \\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\ \\\ \\\\\\\\\\\

\\\\\\

4m

D(without water)

x+y = D2

= ( )2 12 q + q = 222 qq + m

= 2m ( ) 21+ m q =

2m ( )1 2

Dq+ m = ( )D

2 1m

+ m

As 4D

Dq = Therefore

( )4

2D 4 2 8342 1 D 1 713

Dq m m= = = =Dq + m + m +

59. Ans. (3)

60. Ans. (3)

Sol. XL = X

C =

1 242 6

, R = 4W

I = L C

12X X R+ + 4

61. Ans. (1)

62. Ans. (3)

63. Ans. (1)

64. Ans. (1)

65. Ans. (2)

66. Ans. (2)

67. Ans. (3)

68. Ans. (1)

69. Ans. (3)

70. Ans. (2)

71. Ans. (1)

72. Ans. (4)

73. Ans. (2)

74. Ans. (4)

75. Ans. (2)

76. Ans. (1)

77. Ans. (2)

78. Ans. (3)

79. Ans. (4)

80. Ans. (2)

81. Ans. (4)

82. Ans. (3)

83 Ans. (4)

84. Ans. (2)

85. Ans. (3)

86. Ans. (2)

87. Ans. (2)

88. Ans. (3)

89. Ans. (4)

90 Ans. (2)