ct-1(jp) 26-05-2013

RESONANCE SOLJP*CT1260513 - 1

PAPER-1PART-I (Physics)

1. Two large vertical .................................

Sol. (C)

qE = mg

X = Ed

= mg

dq

X = 2�

19�

27�

101106.1

10106.1

X = 10�9

Volt

2. (A)Work done by external agent = q [Vfinal � Vinitial ]

Wext =

22 r

º135coskp�

r

º45coskpq

= 2r4

qp

2

1��

2

1 Wext = 2

0r4

qp2

3. (C)

4. (C)

2

2

1

1

rGm

rGm

= 43

; 21

1

r4

m

= 2

2

2

r4

m

m1 + m2 = m 2R4

m

= 2

1

1

r4

m

or =

1

1

rGm

R

Gm

= 35

Ans.

5. In an electric .................................

Sol. (B)

Rate of change of potential is increasing

v1 � v2/x1 < v2 � v3/x2

x1 > x2

6. The diagram .................................

Sol. (A)

HINTS & SOLUTIONS

DATE : 26-05-2013COURSE : VIJETA (JP)

CUMULATIVE TEST-1 (CT-1)

TARGET : JEE (MAIN+ADVANCED)-2014

(Easy) It can be seen from the diagram that only the sphere B

and sphere C repel. Hence they both must be of same type.

According to the fact that at least two spheres are positively

charged, therefore both spheres should be positively charged.

Since attraction occurs for two remaining pairs it can be

concluded that the sphere A is negatively charged.

7. (C)

m = uf

f

m1 = 1xf

f

m2 = 2xf

f

m1 = �m2

2f = (x1 + x2) f = 2

xx 21 .

8. An electric .................................

Sol. (A)

Work done on 2 C charge = rd.Eq

= q

3

1

Edr

[ r for (1, 1, 0) = 2 & r for (3,0, 0) = 3]

= 2 × area of E-r graph from r = 2 m to r = 3

= 2 ×

1(3 2)20

2

= 20 (3 2) J.

9. A thin prism .................................

Sol. (D)

a =

1

23

× A = 2A

W = A13/42/3

=

8A

water

air

=

14

10. A converging .................................

Sol. (A)

401

� 601

=

f1

....(1)

v1

� 601

=

f21

....(2)


v1

+ 601

= 21

.

601

401

v1

+ 601

= 21

. 6040

100

�v1

= 481

� 601

v = 240 cm

11. (A)The inner sphere is grounded, hence its potential is zero.The net charge on isolated outer sphere is zero. Let thecharge on inner sphere be q�.

Potential at centre of inner sphere is

= 0a'q

41

o

+ a4q2

41

0 = 0 q� =

4q2

= �2q

12. (B)

The electric field at B is = 20 x2

q.

41

towards left.

VC = VC � VA = dxx2

q4

12

a

a2 0

= a16q

o

13. The distance .................................

Sol. (C)

The image distance of A is vA = � xf

)xf( ( x < f)

The image distance of B is vB = x)xf2(]f)xf2[(

x x

A F B Cpole

Solving | vA | + | vB | = 4f we get x = 2f

required time t = u2f

ux

14. The magnitude .................................

Sol. (B)

At time t = 0, velocity of image of A and B are i�u and i�urespectively. Therefore magnitude of their relative velocity

is 2u.

15. The magnitude .................................

Ans. (C)

16. The square .................................

Sol. (D)

The direction of electric field is in x-y plane as shown in figure

E

60°

N

O y

x

The magnitude of electric field is

E = 2y

2x E E = 13 = 2V/m.

The direction of electric field is given by

= tan�1 x

y

E

E= tan�1

3

1 = 30°

Hence electric field is normal to square frame LMNO as shown

in figure.

electric flux = E

. A

= E A cos0 = 2 × 1 = 2 V/m

C is correct option of Q. 15.

Since flux is maximum at = 60° , rotation by 30° either way

would lead to decrease in flux.

D is correct option of Q.16.

Lines ON and NM are both normal to uniform electric field E

.

Hence work done by electric field as a point charge 1 C is

taken from O to M is zero.

17. (8)

� R

GMm �

R9GMm

+ 21

mv02 =

21

mv2 � 2R5

GMm

� R9

]10[GMm +

R5GMm2

= 21

mv2 � 21

mv02

R45

]1850[GMm =

21

mv2 � 21

mv02

� R4532GMm

+ 21

mv02 0

21

v02

R45Gm32

v0 R45

Gm64 n = 8

18. (6)

E =

22

2

1R

)2(k2

2

1Rk2

= 52

1.

Rk2

= R

k 10 = 9 × 109 × 3 × 10�6 = 18 × 103 = 1.8 × 104 .

Alternate Solution:

Enet = Rk

102

= 10

103109102

69

= 54 × 103 N/C.Ans. 6


19. (1)

1 2kq kqV

3R 3R

1 2k q q9000

3R

99 10 3 c 9000 3R

R 1m

20. A positive .................................

Sol. (4)

� q2q1

x

4 7

1 2q qx 4 4

= 01 2q q

x 7 7

= 0

1

2

qq

= x 4

4 1

2

qq =

x 77

x 44

= x 7

7

7x � 28 = 4x + 28

3x = 56

x = 563

1

2

qq

=

567

37

= 113

| q2 | = + 1211

c q1 = 1211

× 113

= 4 c

21. A particle is .................................

Ans. 9

22. The side of the .................................

Sol. (3)Only effective pairs will be body diagonals for this interaction

electrostatic potential energy is

= 3

kq4 2

.

23. A ray of light .................................

Sol. (3)Apply Snell's law on various surfaces one by one :

1 sin 90° = 1 sin r1 sin r1 = 2

1 r1 = 45°

1 cos r1 = 2 sin r2 sin r2 = 2

1

2 cos r2 = 3 sin r3 sin r3 = 3

rsin1 22

2

3 cos r3 = 1 = 3

122

sin2 r3 + cos2 r3 = 1

3

122

+ 31

= 1 22 = 3 2 = 3

PART-II (Chemistry)

24. The number of carbon ....................

Sol. (A)Diamond structure is like ZnS (Zinc blende).

Carbon forming CCP(FCC) and also occupying half of tetrahedral

voids.

Total no. of carbon atoms per unit cell

=

)TV()FC()Corners(

421

681

8 = 8

25. How many grams of concentrated ...................

Sol. (D)

2501000

63Mass

2

mass = 2

63gm

mass of acid × 2

6310070

mass of acid = 45 gm

26. A metal has a fcc lattice ..................

Sol. (B)

3AaN

ZMd

31023 )10404(1002.6

M472.2

7

3

104

)404(02.672.2M

= 26.99 = 27 gm mole�1

27. (A)

P × 1 =

60001

20001

× 0.082 × 300

P = 0.0164 atm.

28. Which of the following represents correctly ...................

Sol. (D)

Ideal solution Gmix

= � ve,

TSmix

= + ve, Hmix

= 0.


29. (B)

30. (A)

31. (B)

32. (B)Cyclic unsaturated structural isomers are :

33. (D)

Seniority of F.G is �COOR > � COX > �CN >

O||

R�C�

34. Vant Hoff factor ....................

Sol. (A)

i = 1 � 2

i = 1 � 23.0

= 0.85

35. For 1 M solution of HA .............................

Sol. (D)

HA H+ + A¯

i = 1 +

= (i � 1)

Now Ka =

1C 2

Ka = )1i(1

)1i( 2

Ka = i2)1i( 2

36. An aqueous solution of 0.1 ....................

Sol. (D)Tf = Kfm = 1.86 × 0.1 = 0.186 ; Freezing point

= 0 � Tf = � 0.186ºC.

37. When 250 mg of eugenol is ......................

Sol. (A)

0.62 = 100M10250 3

× 1000 × 39.7 × 1

M = 160 or 1.6 × 102 g/mol

38. (A)

39. (C)

40. 32 g of hydrated magnesium sulphate MgSO4.x H2O ...................

Sol. (7)molar mass of salt = 120 + 18x

mass of water present in the salt = g32x18120

x18

Molality of the solution =

x18120x3218

84

1000x18120

32

= x26112604000

Tf = 4.836 = 2 1.86 x26112604000

x = 6.9 7

41. The difference in height of the mercury column in two.................

Sol. (4)

2CaClM = 111 g

2CaCln = 111222

= 2 mole O2Hn = 18324

= 18 mole

Relative lowering in vapour pressure

R.L.V.P. = 0

S0

PPP

= 21

1

ninin

= 100

80100 =

182i2i

or 0.2 = 18i2i2

or 0.4i + 3.6 = 2i

i = 2.25.

So i = 2.25

For CaCl2 i = 1 + (n �1)

2.25 = 1 + (3 � 1)

= 225.1

= 0.625.

6.4 × 0.625 = 4.

42. 5

XN2 (gas) = 2520

= 54

PN2 =

54

50 = 40 atm

40 = 20108

K3

H

KH = 3108

2040

= 105 atm.

43. For the reaction : A + 2B C .........................

Sol. (4)A + 2B C

5 8

15

28

(B is L.R)

From mole�mole analysis

28

= 1

nC

44. 5

45. 4

46. 6

C�C�C�C�OCH3, , ,

, C�C�C�OC2H5,


PART-III (Mathematics)

47. If the roots of the equation..............

Sol. (D)Product of roots = 1

)cb(a)ba(c

= 1 b = )ca(

ac2

ab

= ca

c2

=

1ca

2

� (+) =

11

2

2= � 1 �

48. If x2 + ax + b is an integer...............

Sol. (A)Let f(x) = x2 + ax + b

Now it is integer for every integral value of x a, b I

Now roots of f(x) = 0 are

2a a 4b2

..........(1)

roots are rational a2 � 4b = k2where k I

Now k is even or odd according as a is even or odd

(1) roots must be integers.

49. The least value of |a|..............

Sol. (A)sin + cosec = � a

|sin + cosec |= |a|

|sin|1sin2

= |a|

|sin|1

+ |sin |= |a|

|a| 2

50. If p, q R and p2 + q2 � pq................

Sol. (C)p2 + q2 � pq � p � q + 1 0

2

)1q()1p()qp( 222

0

p = q = 1

Now =

1coscos

cos1cos

coscos1

= sin2 � cos2 + 2 cos cos cos � cos2

= 0

= 0

cos () = cos

or (cos cos � cos )2 = sin2 sin2

= (1 � cos2 ) (1 � cos2 )

51. The set of values of 'a' for ..............

Sol. (D)Let y = x2 � 4x + 5

y = (x � 2)2 + 1

sin�1 y, cos�1 y defined only at y = 1

at which x = 2

So given equation

4 + 2a + 2

= 0

8a

4

52. The domain of definition ..............

Sol. (A)for f(x) to be defined

025284 )1x(2)2x(

32

x

and 01xx

x2

52]221[2 24x2

and 0x

642 x2 3x

Hence domain is ,3

53. tan�1 1 + tan�1 .............

Sol. (C)tan�1 1 + (tan�1 2 + tan�1 3)

= tan�1 1 + + tan�1

6132

=

54. The period of sin [x]4

+............

Sol. (D)

period of sin [x]4

+ cos [x]2

+ cot 3

]x[ are 8, 4, 3 respec-

tively.

Hence L.C.M. of 8, 4, 3 i.e.,24 is required period.

55. The domain of the...........

Sol. (C)

)xx3x2(

)1x2(23

0 x (� , �1)

0,

21

1,

2

for log2 x, x > 0

for )x(logsin 21 , 0 log2 x 1

x [1, 2] ........(2)

So by (1) and (2) we get x [1, 2]


56. If the function f : [1, ) ...........

Sol. (A)y = f(x) = 2x(x � 1)

x(x � 1) = log2y

x2 � x � log2 y = 0

x = 21 1 4log y

2

x 1

x = 21 1 4log y

2

y = 21

2(1 1 4log x )

57. Minimum value of y = g(x)..........

Ans. (C)

58. The value of b2 .........

Ans. (B)

Sol. 57, 58( � ) = (( + ) � ( + ))

( + )2 � 4 = [( + )+ ( + )]2 � 4( + ) ( + )

(�b1)2 � 4c1 = (�b2)

2 � 4c2

D1 = D2 ..........(i)

minimum of y = f(x) is � 4

D1 = �

41

D1 = 1 D2 = 1

minimum of y = g(x) is � 4

D2 = �

41

minimum of y = g(x) occur at x = � 2

b2 =

27 b2 = �7

b22 � 4c2 = D2

49 � 4c2 = 1 4

48 = c2 c2 = 12

59. Which of the following .................

Ans. (D)

60. Which of the following.............

Ans. (D)

Sol. cos�1

2x

+ cos�1

3y

=

cos = 6xy

� 4x

12

9y

12

4x

12

9y

12

=

cos

6xy

4x

12

9y

12

=

2

cos6xy

1 � 4x2

� 9y2

+ 36

yx 22

= 36

yx 22

+ cos2 � 3cos xy

1 � cos2 = 4x2

+ 9y2

� 3cos xy

9x2 + 4y2 � 12 xycos = 36 sin2

N = 36

and (cos�1 x)2 � (sin�1 x)2 > 0

2

xsinxcos 11 > 0

x

2

1,1 [p, q)

p = � 1 , q = 2

1

59. N � 6 = 36 � 6 = 0

2

1,1

60. sec�1x is not defined at x = 0

61. If D is the set of .............

Sol. (B)

1 � 1�

x1

e > 0i.e. 1�

x1

e < 1

i.e.x1

� 1 < 0 i.e. x

1x > 0 i.e. x (� , 0) (1, )

62. f(x) is a/an..............

Sol. (A)

Here f(x) = 2

1�x1

x

e > 0 ;

f(x) is one-one

63. Sum of all elements...........

Ans. 5

Sol. For defining f(x)

(16 � x), (20 � 3x) N ; (2x � 1), (4x � 5) W

(16 � x) (2x � 1) ; (20 � 3x) (4x � 5)

Hence x = 2, 3

when x = 2, f(x) = 14C3 + 14C3 = 728

when x = 3, f(x) = 13C5 + 11C4 = 1617

Hence sum 728 + 1617 = 2345

64. If 1 lies between the roots.............

Ans. 3

Sol. 1 lies between the rootss

hence f (1) < 0

3 � 3 sin � 2 cos2 < 0

2 sin2 � 3 sin + 1 < 0

21

< sin < 1


2n + 6

< < 6

5+ 2n, n I

In [0, 2],

65

,6 �

2

= 1, 2

Required sum 1 + 2 = 3

65. If f(x) =

0x0

n,nex

21

eex

)1xxlog(x 2

I

.........

Ans.0

Sol. Let f(x) = )x(h

)x(g.)x(p

Function (M)where p(x) = x odd function

g(x) = log (x + 1x2 )

g(x) + g(�x) = log (x + 1x2 ) + log (�x + 1x2

)

= log (�x2 + x2 + 1)

= log (1)

= 0 g(x) is an odd function.

h(x) = 21

eex

=

ex

+ 21

h(x) + h(�x) =

ex

+ 21

+

ex

+ 21

=

ex

+

ex

+ 1

= �1 + 1 [x] + [�x] = �1 if x R � I

= 0

h(x) is an odd function

f(x) is an odd function f(1947) + f(� 1947) + f(0) = 0.

66. (7)

67. If sum of the series .............

Ans.0Sol. Tr = cot�1 (2r2)

S =

1r2

1�

1r

21�

r2

1tan)r2(cot

=

1r

1�

)1�r2)(1r2(1)1�r2(�)1r2(

tan

=

1r

1�1� )1�r2(tan�)1r2(tan

= (tan�13 � tan�11) + (tan�15 � tan�13) + (tan�17 � tan�15) +.... + tan�1

S = �4

+ 2

= 4

Given S = 4

= k k = 41

[k] = 0

68. Given equation |x2 � 5x + 4 + sinx|...............

Ans. 5Sol. |a + b| = |a| + |b|

ab 0 (x2 � 5x + 4) sinx 0 x [0, 1] [,

4] {2}

sum of all integral values is 0 + 1 + 4 = 5

69. If f(x) =

)�cos()�cos()�cos(

)xcos()xcos()xcos(

)xsin()xsin()xsin(

.............

Ans. 9

Sol. Using the idea of the differentiation of determinant, we get

f(x) =


)xcos()xcos()xcos(

)xcos()xcos()xcos(

+


)xsin(�)xsin(�)xsin(�

)xsin()xsin()xsin(

+

000

)xcos()xcos()xcos(

)xsin()xsin()xsin(

= 0 + 0 + 0

f(x) = 0 for all x

f(x) = a constant

But f(9) =

f(x) = for all x.

9

k 1

f(k)

= 9


PAPER-2PART-I (Physics)

1. (D)

2. A positively ....................................

Sol. (A)Electric field on surface of a uniformly charged sphere is given

by 0

30 3

R

R4

Q

Electric field at outside point is given by

E = 20

3

20 r3

R

r4

Q

|E| r =

0

0

3

r

�

20

0

30

2r3

3

2r

= 0

0

54

r17

leftt

3. (C)

Given 8 = 21

2

=

21 T2

T2

2

, TT1 = 24 hours for

earth. T2 = 12 hours (T2 being the time period ofsatellite, it will remain same as the distancefrom the centre of the earth remains constant).

T = 12

2

=

12 T2

T2

2

= 24 hours.

4. (B)

5. In the figure ....................................

Sol. (C)

F = 2

2

r

qk K =

41

C

2

2

2

Rmv

r

qk RC = 2

22

qk

rmv;

RC = 2

22

q

mrv4

6. If uniform electric ....................................

Sol. (B)A = (O, O), B = (xo, O)

B

A

nd.E = VA � VB = O � VB

Eo Xo = � VB

VB = � Eo xo

7. A triangular medium ....................................

Sol. (B)

Clearly, PM = 23

cm

37º > sin�1 )2/3(an1

0

53

>

2a3

n

1

0

3n0 + 2a9

> 5

2a9

> 1 a > 92

8. A uniform electric ....................................

Sol. (D)

j�Ei�EE yx

, V = �Ex x � Ey y

for A and B

16 � 4 = � Ex (� 2 � 2) � Ey (2 � 2)

Ex = 3 V/m

for B and C

12 � 16 = �Ex {2� (� 2)} � Ey (4 � 2)

Ey = �4 V/m.

m/V)j�4�i�3(E

9. A cubical region ....................................

Sol. (ACD)


Net flux through x = a2

& x = a

�2

will be equal

Net flux through y = a2

& y = a

�2

will be equal

y = a2

y = a

�2

innet

0 0 0

q �q 3q � q q

.

Net flux through Z = +a

�2

& x = +a

�2

will be equal.

Ans. A, C & D

10. (A,B,C,D)

11. Two identical ....................................

Sol. (BCD)

Potential r1 < r2

V+ = 1r

kq2

+ � + � x

r2

r1

y

V� =

2r

kq2

Vnet = V+ + V� = 2kq

21 r

1

r

1 0

Electric field :

+ � + � x

r2r1

E�E�

EEz

E = 21r

kq

E' = 22r

kq

E' < E, hence electric field for z > 0 is directed along +ve z-axis.

Similarly for z < 0, electric field is directed along �ve z-axis.

Hence electric field at origin is zero.

+ � + � x

z

E� E�

E E

12. (BCD)

13. A luminous point ....................................

Sol. (AC)

(A)

(C)

(D) Image is inverted It should be real

14. S1 : When a concave ....................................

Sol. (C)S1 : The focal length of a concave mirror depends only on its

radius of curvature.

S2 : f1

= (nrel � 1)

21 R1

�R1

nrel = gsurroundin

rel

n

n

nsurrounding fnrel

S3 : Since E = drdv�

: if E = 0

V = constant not necessarily equal to zero


15. Four particles are ....................................

Sol. (B)The potential of the centre is positive. From symmetry of the

problem B and D will move away on the line BD and A and C will

move away on the line AC.

Since masses are different, their speeds will be different.

16. (A)

17. (C)

18. In each figure ....................................

Ans. (A) � r ; (B) � p,q,r ; (C) � q ; (D) � s,t

19. In each situation ....................................

Ans. (A) � p,r, t ; (B) � p,q,r,s,t ; (C) � p,q,r,t ; (D) � p,s,tSol. The electric field due to one dipole at centre of other dipole

is parallel to that dipole. Hence torque on all given dipoles

is zero.

In case B and C the electric field at second dipole due to

first is along the second dipole. hence electrostatic potential

energy of second dipole is negative.

In case A and B x-axis is line of zero potential. In case B

and C electric field at origin is zero.

PART-II (Chemistry)

20. The olivine series of minerals consists of crystal in which Fe2+

and Mg2+ .........................

Sol. (C)According to given data :

O2� = 8 × 81

+ 6 × 21

= 4

Si4+ = 41

× octahedral void = 41

× 4 = 1

M++ = 41

× tetrahedral void = 41

× 8 = 2.

Forsterite = Mg2SiO4 Fayalite = Fe2SiO4

Let the forsterite is x% and fayalite is (100�x) % then

10034.4)x100(21.3x

= 3.88

x = 40.71% (Forsterite) and 59.29% (Fayalite).

21. What is the concentration of nitrate ions .........................

Sol. (C)

[NO3�] = V2

0V1.0 = 2

1.0 = 0.05 M

22. (C)For immiscible solution = PT = PA

0 + PB0

= 400 + 200 = 600

23. In the closest packing of atoms.........................

Sol. (B)

24. (C)�COOH is the prior F,G and there are four carboxylic acid

are possible.

25. (B)

26. (A)

27. (C)

28. For a binary ideal liquid solution ...............

Sol. (B)In (B) option component B is more volatile where as in (D) option

A is more volatile.

29. (D)

30. (A,C,D)

31. (C,D)Esters do not show chain and position isomerism.

32. (A,C)(a) and are structural

isomers.(b) These compounds are identicle(c) These are chain isomers(d) These are homologs.

33. (B)

34. S1 : The lowering of vapour pressure ...............

Sol. (D)S1 : Relative lowering = Xsolute

S2 : NaCl is electrolyte while sugar is not

S3 : Eqilibrium constant of an endothermic reaction decrease

with decrease in temperature.

S4 : PP

is a colligative property..

35. (B)

36. (B)

37. (A) 1 : 0.2 M KCl, 2 :0.2 ...............

Sol. (A) � r , s ; (B) � p ; (C) � r, s ; (D) � q, t

38. (A) - p ; (B) - p,q ; (C) - p,r ; (D) - s

PART-III (Mathematics)

39. The value of where...............

Sol. (D)

= sin�1

813

2

+ cos�1

23

+ sec�1 ( 2 )

= sin�1

22

13 + cos�1

23

+ sec�1 2


= 12

+ 6

+ 4

= 12

32 =

2

.

40. 2 cot�1 7 + cos�1 53

..............

Sol. (C)

2 cot�1 = 2 tan�1 71

= cos�1

49/1149/11

= cos�1 2524

Hence cos�1 2524

+ cos�1 53

= cos�1

54

.257

53

.2524

= cos�1 12544

= cot�1

11744

= cosec�1 117125

41. Let f : R R defined...........

Sol. (C) f ' (x) = 3x2 + 2x + 100 + 5 cos x

= 3x2 + 2x + 94 + (6 + 5 cos x) > 0

f is strictly increasing hence one one also

when x � , f(x) �

x , f(x) onto

42. The principle argument of z = x....................

Ans.(B)

43. The graph of |y| = | |x| � 1|....................

Sol. (C)

|x|

|x| � 1

||x| � 1|

|y| = ||x| � 1|

44. (B)

As |sin 4x| � |cos 4x| has period 4

But on taking x4cosx4sin as g(x)

we get

8xg =

x4

2cosx4

2sin

= x4sinx4cos = g(x)

period of g(x) is 8

. Now h(x) = cos (cos 6x)

then x6coscos6

xh

= cos (� cos 6x) = cos (cos 6x) Period is 6

Taking L. C. M of 6

, 8

we get 2

45. Period of f(x) = sin ...............

Sol. (D)

L.C.M.

3

2,

2

2

= For period L.C.M. is not possible

46. If A =

43

21.............

Sol. (D)

AB =

43

21

dc

ba =

d4b3c4a3

d2bc2a

BA =

dc

ba

43

21 =

d4c2d3c

b4a2b3a

if AB = BA, then a + 2c = a + 3b

2c = 3b

b 0

b + 2d = 2a + 4b

2a � 2d = � 3b

cb3d3a3

=

b23

b3

b29

= � 1


47. The value(s) of x...........

Sol. (A,B,C)

1 � 2

3)1x(log 2 =

21

3x5x

log 2/13

1 � 22

log3 |x + 1| = 22

log3

3x5x

log3

|1x|3

= log3

3x5x

|1x|3

= 3x5x

Case - I x + 1 > 0 x > � 1

3(x + 3) = (x + 1)(x + 5)

x2 + 3x � 4 = 0

x = � 4 or x = 1

x = � 4 rejected ( x > � 1)

x = 1

Case - II x + 1 < 0 x < � 1

3(x + 3) = � (x + 1)(x + 5)

x2 + 9x + 14 = 0

x = � 2 or x = �7

Set of value of x = {�7, �2, 1}

48. For the function f(x)...............

Sol. (A,C)

f(x) = ex (x2 + bx + c) > 0 iff D = b2 � 4c < 0

g(x) = ex (x2 + (b + 2)x + (b + c)) > 0 iff

D = b2 � 4c + 4 = D + 4 < 0

49. If A and B are two................

Sol. (B)(A + B)2 = (A + B). (A + B)

= A2 + AB + BA + B2 .........(1)

Since B = � A�1 BA

AB = A (�A�1 BA)

A(BC) = (AB) C

= �(AA�1) BA

AB = � BA

put in (1), we get

(A + B)2 = A2 � BA + BA + B2

(A + B)2 = A2 + B2

50. (B,C)

51. The maximum number...............

Sol. (B)The maximum number of distinct elements

= no. of positions in diagonal & above

= 50 + 49 + ...... + 1 = 1275

52. Consider the following statements....................

Sol. (D)

S1 : x2 + 3x + 3 = x2 + 3x + 43

49 =

2

23

x

+

43

least value of 3x3x2 is

23

3

sin�1 3x3x2

2

3x3xsin 21 = 1

n

3x3xsin 21 = 0

S2 : f(x) = tan�1

2�xtanxcot2

1 11

= tan�1

xcot1 1

Range of f(x) is

43

tan,41

tan 11

no integral value in the range

S3 : 2x

sin 1 < 2

xcos 1

i.e. 2x

sin 1 < cos�1

2x

which is true for �2 x < 2

integral value are

�2, �1, 0, 1

53. Consider the following statements.................

Sol. (B)S1 : f(x) = sin�1{x}

f(x + 1) = sin�1 {x + 1} = sin�1 {x} = f(x)

1 is a period and f(x) is periodic

S2 : Let h(x) = f(x) + |g(x)|

h(�x) = f(�x) + |g(�x)| = f(x) + |�g(x)| = f(x) +

|g(x)| = h(x)

h(x) is an even function

S3 : sin (sin�1x) = x and cos (cos�1x) = x for all x [�1, 1]

domains are equals

both are identical

54. Consider the following statements.............

Sol. (D)Obvious

55. (C)


56. Match the inequality in column-I with................

Ans. (A) - (r), (B) - s, (C) - q, (D) - pSol. (A) logsinx (log3 (log0.2 x)) < 0 = logsinx1

log3 (log0.2x) > 1 log0.2x > 3 = log0.2(0.2)3

0 < x < (0.2)3 0 < x < 125

1

(B) )1x(x)2x(sin)2xx)(3x2)(1e( 2x

0

)1x(x

)2/3x)(1e( x

0 x < � 1 or x

23

x (�, �1)

,

23

(C) |2 � | [x] � 1| | 2

||[x] � 1| � 2| 2 0 |[x] � 1| 4

� 3 [x] 5 x [�3, 6)

(D) |sin�1 (3x � 4x3)| 2

�2

sin�1 (3x � 4x3) 2

� 1 3x � 4x3 1 � 1 x 1

57. If f(x) = 2�x1�x2

.............

Ans. (A) - (r), (B) - (t), (C) - (q), (D) - (p)

Sol. (A) x 0 and g(x) 2 x 1

Domain is x (�, 0) (0, 1) (1, )

(B) x 2 and f(x) 0 x 1/2

Domain is x (�, 1/2) (1/2, 2) (2, )

(C) f(f(x)) = x and x 2

Range is (�, 2) (2, )

(D) g(g(x)) =1x1x2

and x 0

Range is (�, 1) (1, )

ct-1(jp) 26-05-2013

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