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KCET-2020 (Code-C3)

MATHEMATICS

Subject No. Of

Questions Max. Marks

Maximum Time For Answering

Mathematics 60 60 70 Minutes

01. If ( ) nP n : 2 n!

1. Then the smallest positive integer for which

P(n) is true if

(A) 4 (B) 5

(C) 2 (D) 3

Sol. Answer (A)

n2 n!

By hit & trial

n 4= & 5 are to be true

So we require small n 4 =

2. If z x iy= + , then the equation z 1 z 1+ = −

represents

(A) x-axis (B) y-axis

(C) a circle (D) a parabola

Sol. Answer (B)

z 1 z 1+ = −

Locus of z is perpendicular bisector of line

segment joining -1 and 1

3. The value of 16 16 16 16

9 10 6 7C C C C+ − − is

(A) 17

10C (B) 17

3C

(C) 0 (D) 1

Sol. Answer (C)

+ − − =16 16 16 16

9 10 6 7C C C C 0

−

=n n

r n rC C

4. The number of terms in the expansion of

( )10

x y z+ + is

(A) 11 (B) 110

(C) 66 (D) 142

Sol. Answer (C)

( )10

x y z+ +

no. of terms in the expansion (x1+x2+....xr)n is

n r 1

r 1C+ −

−

10 3 1

3 1C+ −

−=

12

2

12 11C 66

2

= = =

5. If the sum of n terms of an A.P. is given by

2

nS n n,= + then the common difference of

the A.P. is

(A) 2 (B) 6

(C) 4 (D) 1

Sol. Answer (A)

2

nS n n= +

= + =1 1S 1 1 T

2

2 1 2S T T 2 2 6= + = + =

1T 2 a= =

( )+ = + + =1 2T T a a d 6

2a d 6 + = 4 d 6 d 2 + = =

Choose the correct answer:

30/07/2020

KCET-2020 Mathematics (Code-C3)


6. The two lines lx my n+ = and l x m y n + =

are perpendicular if

(A) lm l m 0 + = (B) lm ml 0 + =

(C) ll mm 0 + = (D) lm ml =

Sol. Answer (C)

lx my x+ =

l x m y n + =

Product of their slopes is -1

− −

= −

l l1

m m

ll mm ll mm 0 = − + =

7. If the parabola 2x 4ay= passes through the

point (2, 1), then the length of the latusrectum

is

(A) 2 (B) 8

(C) 1 (D) 4

Sol. Answer (D)

2x 4ay=

It passes through (2, 1)

22 4.a.1 =

a 1 =

So parabola 2x 4y=

length of latusrectum is 4

8. x 0

tanxlim

2x 4 2→

+ − is equal to

(A) 4 (B) 6

(C) 2 (D) 3

Sol. Answer: (C)

x 0

tanxlim

2x 4 2→

+ −

( )

( )( )x 0

2x 4 2tanxlim

2x 4 2 2x 4 2→

+ +

+ − + +

( )

→

+ +=

+ −x 0

tanx 2x 4 2lim

2x 4 4

( )x 0

tanxlim 2x 4 2

2x→= + +

0

tanlim 1→

=

( )1 14 2 4 2

2 2= + = =

9. The negation of the statement “For all real

numbers x and y, x y y x+ = + is

(A) for some real numbers x and y,

+ +x y y x

(B) for some real numbers x and y,

x y y x− = −

(C) for all real numbers x and y, x y y x+ +

(D) for some real numbers x and y,

x y y x+ = +

Sol. Answer (A)

For some real numbers x and y, x y y x+ +

10. The standard deviation of the data 6,7,8,9,10

is

(A) 2 (B) 10

(C) 2 (D) 10

Sol. Answer (C)

6,7,8,9,10

6 7 8 9 10

X 85

+ + + += =

( )

2

ix x

N

− =

( ) ( )

2 2 2 22 1 0 1 2

5

− + − + + +=

10

5=

2=

11. If A, B, C are three mutually exclusive and

exhaustive events of an experiment such that

P(A)=2 P(B)=3 P(C), then P(B) is equal to

(A) 3

11 (B)

4

11

(C) 1

11 (D)

2

11

Mathematics (Code-C3) KCET-2020


Sol. Answer (A)

Let ( ) ( ) ( )P A 2.P B 3.P C x= = =

Events are mutually exclusive and exhaustive

( ) ( ) ( )P A P B P C 1+ + =

x x

x 12 3

+ + =

6x 3x 2x

16

+ + =

11x 6

1 x6 11

= =

( )x 3

P B2 11

= =

12. If a relation R on the set 1,2,3 be defined

by ( ) R 1,1= , then R is

(A) Symmetric and transitive

(B) Only symmetric

(C) Reflexive and symmetric

(D) Reflexive and transitive

Sol. Answer (A)

( ) R 1,1=

It is only symmetric & transitive

13. Let ) →f : 2, R be the function defined by

( ) 2f x x 4x 5,= − + then the range of f is

(A) ( )1, (B) )5,

(C) ( ),− (D) )1,

Sol. Answer (D)

( ) 2f x x 4x 5= − +

( ) ( )2

f x x 2 1= − +

( ) )f x 1,

14. If A a,b,c= , then the number of binary

operations on A is

(A) 33 (B)

93

(C) 3 (D) 63

Sol. Answer (B)

A a,b,c=

no. of binary operation 2nn of set is having in

elements = 93

15. The domain of the function defined by

( ) 1f x cos x 1−= − is

(A) 1,1− (B) 0,1

(C) 1,2 (D) 0,2

Sol. Answer (C)

( ) 1f x cos x 1−= −

( )x 1 0 x 1 1− →

1 x 1 1− −

0 x 1 1 −

x 1 1−

x 1 1−

( )x 2 2 →

From (1) & (2)

x 1,2

16. The value of 1 1cos sin cos3 3

− − +

is

(A) -1 (B) Does not exist

(C) 0 (D) 1

Sol. Answer (B)

For arc sin x, x should lie between -1 and 1

17. If

0 0 1

A 0 1 0

1 0 0

=

, then 4A is equal to

(A) I (B) 4A

(C) A (D) 2A

Sol. Answer (A)

0 0 1

A 0 1 0

1 0 0

=

2

0 0 1 0 0 1

A 0 1 0 0 1 0

1 0 0 1 0 0

=

2

1 0 0

A 0 1 0 I

0 0 1

= =

=2A I, 4 2 2A A .A I.I I= = =



18. If A and B are square matrices of same order

and B is a skew symmetric matrix, then A BA

is

(A) Diagonal matrix

(B) Skew symmetric matrix

(C) Symmetric matrix

(D) Null matrix

Sol. Answer (B)

Let TP A BA=

( )T

T TP A BA=

( )T

T T TA B A=

( )T T T TPQR R Q P =

( )TA B A= −

T TP A B A= = −

TP P= −

Skew symmetric

19. If 2 1 1 0

A ,3 2 0 1

=

then the matrix A is

(A) 2 1

3 2

−

− (B)

2 1

3 2

−

(C) 2 1

3 2

(D) 2 1

3 2

− −

Sol. Answer (D)

2 1 1 0

A3 2 0 1

=

1

1 2 1A

3 2

−

− =

2 11

3 24 3

− =

−−

1 2 1A

3 2

−−

= −

20. If ( )

− + +

= − − +

− −

3

2

x x a x b x

f x x a x x c x ,

x b x c 0

then

(A) ( )f 0 0= (B) ( )f 1 0− =

(C) ( )f 1 0= (D) ( )f 2 0=

Sol. Answer (A)

( )

− + +

= − − +

− −

3

2

x x a x b x

f x x a x x c x

x b x c 0

( ) = − =

− −

0 a b

f 0 a 0 c 0

b c 0

Determinant of skew symmetric matrix of odd

order = 0

21. If 1 2 3 9a ,a ,a ,......,a are in A.P, then the value

of

1 2 3

4 5 6

7 8 9

a a a

a a a

a a a

is

(A) ( )e elog log e (B) 1

(C) ( )1 9

9a a

2+ (D) 1 9a a+

Sol. Answer (A)

Apply 1 1 3 2C C C 2C→ + −

22. If A is a square matrix of order 3 and |A|=5,

then A adjA is

(A) 25 (B) 625

(C) 5 (D) 125

Sol. Answer (D)

= nA.adjA A .I

3A.adjA A .I=

3 3A 5= =

23. If ( )

1 cosKx, if x 0

xsinxf x

1, if x 0

2

−

= =

is continuous at x 0,= then the value of K is

(A) 2 (B) 1

(C) 1

2 (D) 0

Sol. Answer (B)

( )

−

= =

1 coskx; x 0

xsinxf x

1; x 0

2



x 0

1 coskxlim

x.sinx→

−

→

2

x 0

kx2sin

2limx.sin x

→

2

x 02

kxsin

kx 12lim2 . .kx 2 sinx

x2 x

2

2 2

2x 0

kxsin

2.k x 12lim . .kx sinx4.x

2 x

→

=

2k

2=

Since ( )f x is continuous at x=0

So ( )f 0 LHL RHL= =

2

21 kk 1 k 1

2 2= = =

24. The right hand and left hand limit of the

function ( )

1/ x

1/ x

e 1, if x 0

f x e 1

0 , if x 0

−

= + =

are

respectively

(A) -1 and -1 (B) -1 and 1

(C) 1 and 1 (D) 1 and -1

Sol. Answer (D)

RHL

Put x 0 u= +

1

u

1u 0u

e 1lim

e 1→

−

+

Take 1

ue common

1

u

1u 0u

1 elim 1

1 e

−

→ −

−=

+

LHL, Put x 0 u= −

1

u

1u 0 1u

e 1lim 1

e

−

→ − +

−= −

25. If x y x y2 2 2 ,++ = then dy

dx is

(A) x y2 − (B) y

x

2 1

2 1

−

−

(C) y x2 − (D) y x2 −−

Sol. Answer (D)

x y x y2 2 2 ++ =

x y x y2 2 2 .2+ =

x y2 2 1− − + =

Differentiating with respect to ‘x’

( ) ( )x y dy2 .log2. 1 2 .log2. 1 0

dx

− −− + − =

y

x

dy 2

dx 2

−=

26. If ( ) 1

2

2xf x sin

1 x

− =

+ , then ( )f 3 is

(A) 1

3 (B)

1

3−

(C) 1

2− (D)

1

2

Sol. Answer (C)

1

1 1

2

1

2tan x ; x 12x

sin 2tan x ; 1 x 11 x

2tan x; x 1

−

− −

−

−

= − +

− − −

( )

− +

= − +

− −

+

2

2

2

2;x 1

1 x

2f x ; 1 x 1

1 x

2;x 1

1 x

( ) − = = −

+

2 1f 3

1 3 2

27. If ( )y xxe e ,= then

dy

dx is

(A) ( )

logx

1 logx+ (B)

( )

xe

x y 1−

(C) ( )

2

logx

1 logx+ (D)

( )2

1

1 logx+

Sol. Answer (C)



( )y xxe e=

Take log on both sides

( )e ey.ln xe x.ln e=

( )ey. ln x lne x+ =

x

yln x 1

=+


( )

( ) ( )2 2

11. lnx 1 .x

lnxxylnx 1 lnx 1

+ − = =

+ +

28. If n 1

n

3y 2x ,

x

+= + then 2

2

2

d yx

dx is

(A) dy

x ydx

+ (B) y

(C) ( )6n n 1 y+ (D) ( )n n 1 y+

Sol. Answer (D)

+= +n 1

n

3y 2.x

x

( ) − − = + −n n 1y 2. n 1 .x 3nx

( ) ( )− − − = + + +n 1 n 2y 2.n n 1 .x 3n n 1 .x

Now according to question, multiply 2x on

both sides

( ) ( )− − − = + + +2 2 n 1 2 n 2x .y 2.n n 1 .x .x 3n n 1 .x .x

( ) ( )n 1

n

12.n n 1 .x 3n n 1 .

x

+= + + +

( )( ) = +2y .x n n 1 y

29. If the curves 22x y= and 2xy K= intersect

perpendicularly, then the value of 2K is

(A) 2 (B) 8

(C) 4 (D) 2 2

Sol. Answer (B)

Curves are ⊥ , so product of the slopes = -1

2

1

12x y 2 2y.y y m

y = = = =

k

2xy k y2x

= =

−

= = 22

ky m

2x

1 2m m 1= −

2

1 k. 1

y 2x

− = −

2k 2x y =

Put 2xy k= in 2k 2x y=

22xy 2x y=

( )xy x 1 0 − =

x 0,y 0= = or x 1=

Put x 1= in equation (1), to get y

22x y= = = 2y 2 y 2

Now =2xy k

(Squaring on both sides)

2 2 2k 4x y 4.1.2 8= = =

30. The maximum value of elog x,

x if x 0 is

(A) 1

e (B)

1

e−

(C) e (D) 1

Sol. Answer (A)

elog xy

x=


e

2

1x. log x

xyx

− =

e

2

1 log xy

x

− =

Maxima occurs at x = e as dy

dx in the NBD

changes from +ve to –ve

( ) elog e 1f e

e e= =



31. If the side of a cube is increased by 5%, then

the surface area of a cube is increased by

(A) 6% (B) 20%

(C) 10% (D) 60%

Sol. Answer (C)

dx

100 5x =

ds

100 ?s =

2s 6x=

logs log6 2.logx= +

1 dx

.ds 2s x

=

ds dx

2.s x

=

ds dx

100 2. 100 2 5s x = = =10%

32. The value of 4

6

1 xdx

1 x

+

+ is

(A) 1 1 31tan x tan x C

3

− −− +

(B) 1 1 21tan x tan x C

3

− −+ +

(C) 1 1 3tan x tan x C− −+ +

(D) 1 1 31tan x tan x C

3

− −+ +

Sol. Answer (D)

( )

4

6

1 xdx

1 x

+

+

( )( )( )( )

4 2

6 2

1 x 1 xdx

1 x 1 x

+ +

+ +

( ) ( )( )( )

6 2 2

6 2

1 x x 1 xdx

1 x 1 x

+ + +

+ +

2

2 6

dx x dx

1 x 1 x+

+ +

( )

2

2 23

dx 1 3xdx

31 x 1 x+

+ +

1 1 31tan x .tan x C

3

− −= + +

33. If ( )( )( )

3x 1dx

x 1 x 2 x 3

+

− − −

= − + − + − +Alog x 1 Blog x 2 Clog x 3 Constant ,

then the values of A, B and C are respectively

(A) 5,-7,5 (B) 2,-7,5

(C) 5,-7,-5 (D) 2,-7,-5

Sol. Answer (B)

( )( )( )

3x 1dx

x 1 x 2 x 3

+

− − −

( )( )( )

3x 1 A B C

x 1 x 2 x 3 x 1 x 2 x 3

+= + +

− − − − − −

Take LCM

( )( ) ( )( )+ = − − + − − +3x 1 A x 2 x 3 B x 1 x 3

( )( )C x 1 x 2− −

Put x 1=

( )( )4 A 1 2 A 2 = − − =

Put x=2

( )( )7 B 1 1 B 7= − = −

Put x=3

9 1 C.2.1 C 5+ = =

( )( )( )

3x 1 dx dx dxdx 2 7 5

x 1 x 2 x 3 x 1 x 2 x 3

+= − +

− − − − − −

2ln x 1 7.ln x 2 5.ln x 3 C= − − − + − +

A=2, B=-7, C=5

34. The value of sinxe sin2xdx is

(A) ( )sinx2e cosx 1 C+ +

(B) ( )sinx2e cosx 1 C− +

(C) ( )sinx2e sinx 1 C− +

(D) ( )sinx2e sinx 1 C+ +

Sol. Answer (C)

sinxe .2.sinx.cosx.dx

Put sinx = t



=cos x dx dt

t2 e .t dt

−

t tdt

2 t e dt e dtdt

= − + t t2 e .t e C

( )t2e t 1 C= − +

Replace ‘t’ by sinx

( )sinx2e sinx 1 C= − +

35. The value of −

−

1

21

1

2

cos xdx is

(A) 1 (B) 2

2

(C) (D) 2

Sol. Answer (D)

( )1

121

2

I cos xdx 1−

−= →

( ) ( )b b

a af x dx f a b x dx= + −

( )1

121

2

I cos x dx−

−= −

( ) ( )1

121

2

I cos x dx 2−

−= − →

Add both (1) and (2)

( )− −

−= + −

11 12

1

2

2I cos x cos x dx

1

21

2

2I dx−

=

1 1

2I .2 2

= +

I2

=

36. The value of

−

+2

x

2

cosxdx

1 eis

(A) 1 (B) -2

(C) 2 (D) 0

Sol. Answer (A)

( )/ 2

x/ 2

cosxI dx 1

1 e

−= →

+

Use ( ) ( )b b

a af x dx f a b x dx= + −

x

cosxI dx

1 e−=

+

( )x

x

e .cos xI dx 2

1 e= →

+

Add (1) and (2)

2

2

2I cosxdx

−

=

( )/ 2

/ 22I sin x

−=

2I 1 1 I 1= + =

37. The value of ( )+

+1

2

0

log 1 xdx

1 x is

(A) 1

2 (B) log2

8

(C) log22

(D) log2

4

Sol. Answer (B)

( )1

20

log 1 xdx

1 x

+

+

Put x tan=

( )

+ =/ 4

0log 1 tan d log2

8

38. The area of the region bounded by the curve

2y 8x= and the line y 2x= is

(A) 3

4sq. units (B)

8

3 sq. units

(C) 16

3 sq. units (D)

4sq.units

3

Sol. Answer (D)



2y 8x=

y=2x

substitute ( )2

2x 8x =

24x 8x=

x 0,2=

( )−2

08x 2x dx

1

2 22

0 02 2 x dx 2 xdx−

−

23

22

0

2 2.x 2.x

3 2

2

3

224 2

.2 23

= −

24.2 4

43 3

= − =

39. The area of the region bounded by the line

y 2x 1,= + x-axis and the ordinates x 1= −

and x 1= is

(A) 5

2 (B) 5

(C) 9

4 (D) 2

Sol. Answer (A)

y=2x+1

( )1,0−

1,0

2

−

( )1,0

= ( ) ( )

−

−−

− + + +

1

12

11

2

2x 1 dx 2x 1 dx

=1

12 22

11

2

x x x x−

− − − + + +

( ) ( )1 1 1 1

1 1 1 14 2 4 2

= − − − − + + − −

1 1 5

24 4 2

= + + =

40. The order of the differential equation obtained

by eliminating arbitrary constants in the family

of curves ( ) 4x c

1 2 3c y c c e += + is

(A) 3 (B) 4

(C) 1 (D) 2

Sol. Answer (C)

( ) 4x C

1 2 3C y C C .e += +

( ) 4C x

1 2 3C y C C .e .e= +

( )

42 3 C x

1

C Cy .e .e

C

+=

xy C.e=

Only 1 arbitrary constant

Order = 1

41. The general solution of the differential

equation − =2 4x dy 2xydx x cosxdx is

(A) 2y sinx cx= +

(B) 2y cosx cx= +

(C) 2 2y x sinx cx= +

(D) 2y x sinx c= +

Sol. Answer (C)

2 4x dy 2xy.dx x .cosxdx− =

2dy 2.y x .cosx

dx x− =

I.F −

2dx

xe

2.lnx

2

1e

x

−= =

Now solution

2

2 2

1 1y. x .cos x. dx

x x=



= 2

ycosx dx

x

2

ysinx C

x= +

42. The curve passing through the point (1, 2)

given that the slope of the tangent at any

point ( )x,y is 2x

y represents

(A) Ellipse (B) Hyperbola

(C) Circle

(D) Parabola

Sol. Answer (B)

dy 2x

dx y=

ydy 2xdx=

Variable separable

2

2yx C

2= +

Passes through (1, 2)

4

1 C C 12= + =

2

2yx 1

2 = +

2

2 yx 1

2 − = − , Which is hyperbola

43.. The two vectors i j k+ + and i 3j 5k+ +

represent the two sides AB and AC

respectively of a ABC. The length of the

median through A is

(A) 7

(B) 14

(C) 14

2 (D) 14

Sol. Answer (B)

Let A be origin

D is midpoint of B & C

( ) ( )+ + + + +

=i j k i 3j 5k

AD2

= + +AD i 2j 3k

= + + =AD 1 4 9 14

44. If a and b are unit vectors and is the

angle between a and b , then sin2

is

(A) a b

2

− (B) a b−

(C) a b+

(D) a b

2

+

Sol. Answer (A)

− = + − 2 2

a b a b 2 a b .cos

1 1 2cos= + −

( )2 1 cos= −

a b 2.sin2

− =

45. If the vectors − + + −2i 3j 4k, 2i j k and

i j 2k − + are coplanar, then the value of

is

(A) -6 (B) 5

(C) 6 (D) -5

Sol. Answer (C)

−

− =

−

2 3 4

2 1 1 0

1 2

=6

46. If 2 2

a b a b 144 + = and =a 6 , then b

is equal to

(A) 2

(B) 4

(C) 6

(D) 3

Sol. Answer (A)

+ =2 2 2 2

a b a b a b , we get =b 2



47. The point (1, -3, 4) lies in the octant

(A) Fourth (B) Eighth

(C) Second (D) Third

Sol. Answer (A)

Conceptual

48. If a line makes an angle of 3

with each of x

and y-axis then the acute angle made by z-

axis is

(A) 3

(B)

2

(C) 4

(D)

6

Sol. Answer (C)

, , be angle with axes

2 2 2cos cos cos 1+ + =

2 2

2cos cos cos 13 3

+ + =

21 1cos 1

4 4+ + =

2 1cos 1

2 = −

1

cos2

=

=4

49. The distance of the point (1, 2, -4) from the

line x 3 y 3 z 5

2 3 6

− − += = is

(A) 293

49 (B)

293

49

(C) 293

7 (D)

293

7

Sol. Answer (D)

( )P 1,2, 4−

Q

− − +

= = = x 3 y 3 z 5

2 3 6

General point on line is ( )2 3,3 3,6 5 + + −

Direction ratio of PQ ( ) + + −2 2,3 1,6 1

Since line & PQ are perpendicular

So, product of their direction ratios will be 0.

So ( ) ( ) ( )2 2 .2 3 1 3 6 1 .6 0 + + + + − =

4 4 9 3 36 6 0 + + + + − =

1

49 1 049

+ = = −

Direction ratio PQ ( ) + + −2 2,3 1,6 1

− − −

+ + −

1 1 12 2, 3 1, 6 1

49 49 49

− =

96 46 55, ,

49 49 49

Distance

2 2 296 46 55 14537

49 49 49 492

− = + + =

293 293

49 7=

50. The sine of the angle between the straight

line − − −

= =−

x 2 3 y z 4

3 4 5 and the plane

2x 2y z 5− + = is

(A) 4

5 2 (B)

2

10

(C) 3

50 (D)

3

50

Sol. Answer (B)

x 2 y 3 z 4

3 4 5

− − −= = & 2x 2y z 5− + =

Angle between line & plane

( ) ( )

( )22 2 2 2 2

3.2 4 2 5 1sin

3 4 5 2 2 1

+ − + =

+ + + − +

6 8 5

sin5 2.3

− + =

3 1 2

103.5 2 5 2= = =



51. The feasible region of an LPP is shown in the

figure. If Z = 11x + 7y, then the maximum

value of Z occurs at

Y

X

x y 5+ =

x 3y 9+ =

0

(A) (5, 0) (B) (3, 2)

(C) (0, 5) (D) (3, 3)

Sol. Answer (B)

Solve

+ =

+ =

− = −

x y 5

x 3y 9

2y 4

y 2 x 3= =

x 3y 9+ = meets y axis at (0, 3)

x y 5+ = meets at (0, 5)

Now check z 11x 7y= +

At (0,3) z 0 7.3 21 = + =

At (0,5) z 0 7.5 35 = + =

At (3,2) z 11.3 7.2 47 = + =

So, Z is maximum at (3,2)

52. Corner points of the feasible region

determined by the system of linear

constraints are (0, 3), (1, 1) and (3, 0). Let z =

px + qy, where p, q > 0. Condition on p and q

so that the minimum of z occurs at (3, 0) and

(1, 1) is

(A) p 3q= (B) p q=

(C) p 2q= (D) q

p2

=

Sol. Answer (D)

= +z px qy

Minimum occurs at (3,0)

z 3p=

Also at z p q = +

So 3p= p + q 2p q =

53. If A and B are two events such that

( ) ( )1 1

P A ,P B3 2

= = and ( )1

P A B ,6

= then

( )P A /B is

(A) 1

2 (B)

1

12

(C) 2

3 (D)

1

3

Sol. Answer (C)

( )1

P A3

=

( )1

P B2

=

( )1

P A B6

=

Since ( ) ( ) ( )P A B P A .P B =

So events are independent

Now ( )A

P P AB

=

1 2

13 3

= − =

54. A die is thrown 10 times, the probability that

an odd number will come up atleast one time

is

(A) 11

1024 (B)

1013

1024

(C) 1

1024 (D)

1023

1024

Sol. Answer (D)

1 – P(no odd number comes) 10

10 10

3 1 10231 1

10246 2= − − =

55. The probability of solving a problem by three

persons A, B and C independently is 1 1

,2 4

and 1

3 respectively. Then the probability of

the problem is solved by any two of them is

(A) 1

24 (B)

1

8

(C) 1

12 (D)

1

4

Sol. Answer (D)



Solved by two implies one person should fail

( ) ( ) ( )P A B C P A B C P A B C= + +

Events are independent

1 1 2 1 3 1 1 1 1

2 4 3 2 4 3 2 4 3= + +

6 1

24 4= =

56. Events 1E and 2E form a partition of the

sample space S. A is any event such that

( ) ( )1 2

1P E P E ,

2= = ( )2

1P E / A

2= and

( )2

2P A /E

3= , then ( )1P E / A is

(A) 1 (B) 1

4

(C) 1

2 (D)

2

3

Sol. Answer (C)

Using Baye’s theorem

( )

( ) ( )

( ) ( ) ( ) ( )=

+

2 2

2

1 1 2 2

P E .P A /EP E / A

P E P A /E P E .P A /E

`

1 2.

1 2 31 1 22

.x .2 2 3

=

+

1 2/3 2 4

x22 3 3

x3

= + =

+

( )= = 1

2x P A /E

3

Now use again Baye’s theorem

( )( ) ( )

( ) ( ) ( ) ( )=

+

2 2

2

1 1 2 2

P E .P A /EP E / A

P E P A /E P E .P A /E

1 2.

2 31 2 1 2

. .2 3 2 3

=

+

( ) =1

1P E / A

2

57. If A 1,2,3,4,5,6= , then the number of

subsets of A which contain atleast two

elements is

(A) 57 (B) 58

(C) 64 (D) 63

Sol. Answer (A)

6 6 6 6 6

2 3 4 5 6C C C C C + + + +

6 6 6

0 12 C C= − −

64 1 6 57= − − =

58. If n(A)=2 and total number of possible

relations from set A to set B is 1024, then

n(B) is

(A) 10 (B) 5

(C) 512 (D) 20

Sol. Answer (B)

No.of relations (maximum) = mn2

mn 102 1024 2= =

( ) 2.m 10n A 2 2 2 m 5= = =

59. The value of 2 0 2 0sin 51 sin 39+ is

(A) 0sin12 (B)

0cos12

(C) 1 (D) 0

Sol. Answer (C)

2 2sin cos 1+ =

60. If tan A + cot A = 2, then the value of

4 4tan A cot A+ =

(A) 4 (B) 5

(C) 2 (D) 1

Sol. Answer (C)

tan A cot A 2+ =

1

tanA 2tanA

+ =

2tan A 2tanA 1 0 + + =

( )2

tanA 1 0 tanA 1− = =

cot A 1=

So, 4 4tan A cot A 1 1 2+ = + =

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