answerkey(1) mts 2

8/12/2019 AnswerKey(1) MTS 2

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PAPER CODE 0TM

Path to success KOTA (RAJASTHAN )1 C T 1 1 3 0 5 2

ANSWER KEY

01CT113052

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A. 3 1 3 3 1 2 3 1 4 3 4 4 2 3 4 1 3 4 2 4

Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

A. 2 1 3 2 4 4 3 3 2 1 2 3 3 3 3 2 2 2 3 1

Q. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

A. 2 2 4 2 2 2 4 2 1 3 2 2 2 2 3 1 4 4 4 1

Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

A. 2 3 1 3 3 4 2 3 1 4 2 1 2 4 3 3 4 2 4 1

Q. 81 82 83 84 85 86 87 88 89 90

A. 3 3 4 3 2 3 3 2 4 3

KOTA

NURTURE COURSE : TARGET - JEE (Main + Advanced) 2015MAJOR TEST # 02 Date : 13 - 02 - 2014 Pattern : JEE (Main)


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Corporate Office : ALLEN CAREER INSTITUTE, SANKALP, CP-6, INDRA VIHAR, KOTA-324005PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac. in KOTA / HS - 1

SOLUTION

PAPER CODE 0TM

Path to success KOTA (RAJASTHAN )1 C T 1 1 3 0 5 2

1. Ans. (3)Both roots are common

\ b ab a

-+

=10 15 55

+ = --

2. Ans. (1)

Number of into function4C1 3

5 4C2.25 + 4C1.1

5 = 784.3. Ans. (3)

(x) = ex(|x| + x2)

(x) = ex

(|x| + x2

)(x) = ex(|x| + x2)\ (x) (x) and (x) (x)

4. Ans. (3)

1 1tan tan- - a b - a+ b b + a

Q 0a >b &

0b - a >b + a

& 1

a 3\ sec1x 3\ x ( , sec3] [1, )

11. Ans. (4)No of ways

52 2 2

5! 5!C 3! 15003!(1!) .2! (2!) .1! 2!

= + = 12. Ans. (4)

[x2 (x + 2)]5Then relevent terms are= 5C3(x

2)2((x + 2))3 + 5C4(x2) ((x + 2))4

+ 5C5((x + 2))5

5C3x4(x + 2)3 + 5C4x2(x + 2)4 5C5(x + 2)5 5C3.3C2.22 + 5C4.4C1 2 5C5 1 10 3 4 + 5 4 2 1= 120 + 40 1 = 80 1 = 81

NURTURE COURSE : TARGET - JEE (Main + Advanced) 2015MAJOR TEST # 02 Date : 13 - 02 - 2014 Pattern : JEE (Main)


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01CT113052KOTA / HS - 2

13-02-2014TARGET : JEE (Main + Advanced) 2015TM

Path to success KOTA (RAJASTHAN )

13. Ans. (2)

'(x) = 3x3 + 2x + 100 + 4cosx = 3x2 + 2x + 94 + (6 + 4cosx) > 0

& (x) includes odd degree polynomial.\ Range = R\ 11 and onto

14. Ans. (3)

2xlog 0

[x] x 1

[x]

x [x] 0

[x]-

{x} 0[x]

[[x] 0, x [0, 1)]

[x] > 1

\ x [1, ) I {0}Q {x} = 0 for x = I

|x| [1, )15. Ans. (4)

Q A B/Three possibilities.

A B A BA

B

only (4) option satisfies all the three possibilities.16. Ans. (1)

Let the circle bex2 + y2 + 2gx + 2y = 0\ g + 2 = 1

4g + 6 = 3

g = 6 and = 7

2\ Circle is x2 + y2 12x + 7y = 017. Ans. (3)

Tr =15Cr1(x

4)15r+1r 1

31x

- - = 15Cr1(1)

r1x677r

67 7r = 17 7r = 84r = 12

18. Ans. (4)

O

305 a

2

a = (2cot30) 2 = 4 319. Ans. (2)

1sin x , 02

- p - Let sin1x = a x = sina

2

xtan

1 xa =

-

1

2

xtan

1 x

- a = - 20. Ans. (4)

Inverse function of any bijective function, is alsoa bijective function.

21. Ans. (2)

1 1 1 11 11 7 11tan tan tan cos3 4 11 170

- - - -+ = =

22. Ans. (1)

A BDtan2 AD

= = rational number

\ sinA =2

A2tan2A1 tan2

+ & cosA =

2

2

A1 tan2A1 tan2

-

+

are also irrational.23. Ans. (3)

O I(4,6) (2,2)M

M is (3, 2)

Slope of OI = 8 42

= - .

\ Equation of mirror is 1y 2 (x 3)4+ = -4y x + 11 = 0x 4y 11 = 0

24. Ans. (2)

11 2

1 cos(sin x)- =

- 1 1cos(sin x)

2- =

1sin x3

- p= 3x2

= .


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KOTA / HS - 301CT113052

NURTURE COURSETM

Path to success KOTA (RAJASTHAN )13-02-2014

25. Ans. (4)

Reflexive : (p, q) R(p, q) pq qp = 0 (True)Symmetric : (r, s) R(p, q) rq sp = 0 (True)Transitive : (p, q) R (r,s) ps = qr ....(i)

(r, s) R (t, u) ru = st ....(ii)(i) (ii)pu = qt (p,q) R (t, u) (True)

26. Ans. (4)

1 sin1 1tancos1

- - 1 1 sin1tan

cos1- - = -

1

11 tan 12tan 1 2 41 tan2

-

- p= - = - +

27. Ans. (3)

z = x + y x2 + xy y2

2z = 2x2 + 2y2 2xy 2x 2y= (x2 + y2 2xy) + (x2 2x) + (y2 2x)

2 2z = (x y)2 + (x 1)2 + (4 1)2

Least value of RHS 0 at x = y = 1

2 2z 0z 1

28. Ans. (3)

2 sinx 1 0 1 (x) 14

(x) is bounded & statement-2 is false forexample tanx.

29. Ans. (2)

O

p 2p

sin (sinx) 1

30. Ans. (1)2cot1x 3 0

cot1x > 32

3x cot2

- <

31. Ans. (2)

Sol.54l

= 10

A' = 2 sin k (l /8) = 232. Ans. (3)

Sol. mv2 dvds

= P

2

1

v s2

v 0

v dv P ds= 33. Ans. (3)

Sol. Kinetic energy =( )212 2

2 4 J

34. Ans. (3)

Sol. By using reduced mass concept Here relu 0= ,

relv ?= ,1 2

1 2

m mm m

m =+ =

16 216 2

+ =

169 kg

12 kx

2 = 2rel1 v2

m vrel =2kx

m

= ( )2100 1 / 4

16/9 = 3016 ms

1 = 1.875 ms1

= 1.88 ms1

35. Ans. (3)Sol. Impulse = change in momentum; moment of

impulse about C.M. = initial angular momentum;the ball moves like a projectile.

36. Ans. (2)Sol. Rate of change of angular momentum

= external torque

dLdt = mg R

37. Ans. (2)Sol. for net acceleration to be at 45 with radius vector

at = an and an =( )22 ta tv

r r=


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KOTA / HS - 501CT113052

NURTURE COURSETM

Path to success KOTA (RAJASTHAN )13-02-2014

\ relative acceleration of boy and B is 23 m / s

10relative velocity after 5 s is 1.5 m/s

47. Ans. (4)Sol. F1cos 30 + F2 cos 60 = mg

F1sin 30 = F2 sin 60

on solving F2 =mg2

3060

F 1 F2

Mg

\ a =g2 = 5 m/s.

2

1

x 10 3x 200 / 30 2

= =

48. Ans. (2)

Sol. Process is isochoric and CV for monoatomic is3 R2 . For piston to lift up P in = 2Patm

49. Ans. (1)Sol. We have taken x = 0 as node so that

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

x=0

N

l /2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

N

l /2 l /2 l /2

x= l

Amplitude = (Amax sin kx)2

K p=

l where

maxmax

32 4sin sin 32 3 3 2

2

A A A A

p p l = = = = =

l ll

50. Ans. (3)Sol. Potential energy depends on location and

orientation in the space.51. Ans. (2)Sol. Since motion is non-uniform thus tangential is

not zero52. Ans. (2)Sol. Stress is maximum at upper end and decreases

downwards.53. Ans. (2)Sol. Acceleration decreases (magnitude) through out

the motion.

54. Ans. (2)Sol. I = DP = 2mv55. Ans. (3)Sol. Angular impulse = change in angular

momentum56. Ans. (1)

Sol. The allowed standing-wave frequencies are f m= m(v/2L), so the mode number of a standingwave of frequency f is m = 2L0f/v. QuadruplingT, increases the wave speed v by a factor of 2.The initial mode number was 2, so the new modenumber is 1.

57. Ans. (4)

58. Ans. (4)

Sol. 17 32 12 4.2 = m 2.3 103 + (3 x) 4.2 (100 32)

59. Ans. (4)

Sol. Closed and represents antinode of pressure60. Ans. (1)

Sol. DU = WW is positiveSo DU = negative

61. Ans.(2)2Al + 2NaOH + H2O 2NaAlO2 + 3H2

0.27 gm32 0.01 moles

0.01 moles

VH2 =2Hn R T 3 0.01 0.0821 300

P 2 1 =

VH2 = 0.3694l = 369.4 ml

62. Ans.(3)

Uavg. =5 3

38 10 10 8 10

2 32 10

-

-p

p = 103 m/sec.

63. Ans.(1)

64. Ans.(3)PCl5 PCl3 + Cl2

t = 0 1 molet=eq (1a) a a

d =Mass

ol =1 MnRT

P

d =M P PM 1

(1 ) RT dRT a = -

+ a


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01CT113052KOTA / HS - 6

13-02-2014TARGET : JEE (Main + Advanced) 2015TM

Path to success KOTA (RAJASTHAN )

65. Ans.(3)

2A + B C + Dt = 0 4 2 6 2t=eq 2y y 8 4

KC =12

28 4 10

(2y) y

=

y = 2 104

2y = 4 104

66. Ans. (4)

T =

p r V

T T n3 /Z2

= =

67. Ans.(2)

68. Ans.(3)

69. Ans.(1)

AlCl3 + 3AgNO3 3AgCl + Al(NO3)345 0.2 milli moles

1

3

45 0.2 milli mol

13 0.45 0.2 = 0.2 V

V = 15 ml

70. Ans. (4)71. Ans. (2)

72. Ans. (1)

73. Ans. (2)

74. Ans. (4)

75. Ans. (3)

76. Ans. (3)

77. Ans. (4)

78. Ans. (2)

79. Ans. (4)

80. Ans. (1)

81. Ans. (3)

82. Ans. (3)

83. Ans. (4)

84. Ans. (3)

85. Ans. (2)

86. Ans. (3)

87. Ans. (3)

88. Ans. (2)

89. Ans. (4)

90. Ans. (3)

answerkey(1) mts 2

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