# Free Objective Test 01 Practice Test - 11th and 12th

If limx0(1+asinx)cosecx=3, then a is

A. In 2
B. In 3
C. In 4
D. e3

#### SOLUTION

Solution : B

3=limx0(1+asinx)cosecx
[1form]limx0ecosecx.asinx=ea
ea=3a=loge3=ln3.

limx0sinxx+x36x5 is equal to

A. 1120
B. 1160
C. 12
D. 0

#### SOLUTION

Solution : A

limx0sinxx+x36x5 ....... (1)

We know expansion of Sinx.
sin x=xx33!+x55!+ ........ (2)

Substituting (2) in (1) we get

limx0 x55!.x5

=15!

=1120

limnan+bnanbn, where a>b>1, is equal to

A. 1
B. 1
C. 0
D. ab

#### SOLUTION

Solution : B

limnan+bnanbn

a>b>1

=nn1+(ba)n1(ba)n

=1

limxπ2cotxcosx(π2x)3 is equal to

A. 1
B. π2
C. 116
D. 0

#### SOLUTION

Solution : C

limxπ2cotxcosx(π2x)3

Let x=π2+t

If x π2,t0

limt0sinttant8t3

=limt0(tt33!+t55!+)(t+t33+2t515+)8t3

=116

We can put  x=π2t and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is =116

limx{x4sin(1x)+x21+|x|3} is equal to

A. 2
B. 1
C. 1
D. does not exist

#### SOLUTION

Solution : C

limx[x4sin(1x)+x21+|x|3]

limxx31+|x|3[xsin(1x)+1x]

As x

x31+|x|3=1

x sin(1x)+1x=1

limxx31+|x|3(x sin1x+1x)=1

If f(9) = 9, f'(9) = 4, then limx9f(x)3x3 equals

A. 2
B. 4
C. 6
D. 0

#### SOLUTION

Solution : B

Given that f(9) = 9, f’(9) = 4

limx9f(x)3x3

On rationalisation we get -

=limx9(f(x)3x3)(x+3x+3)(f(x)+3f(x)+3)

=limx9 f(x)f(9)x966

=limx9 f(x)f(9)x9

=f(9)

=4

If f(x) = {sinxxnπ,n=0,±1,±2...2,otherwise and g(x) = x2+1,x0,24,x=05,x=2, then limx0g{f(x)} is

A. 2
B. 0
C. 3
D. 1

#### SOLUTION

Solution : D

Given that,
f(x) = {sinxxnπ,n=0,±1,±2...2,otherwise
and
g(x) = x2+1,x0,24,x=05,x=2
Then limx0g[f(x)]=limx9g(sinx)
=limx0(sin2x+1)=1

limnn.cos(π4n).sin(π4n) is equal to

A. π4
B. π6
C. π9
D. π3

#### SOLUTION

Solution : A

limxn.cos(π4n)sin(π4n)

=limnn2 sinπ2n

=π4

limx01cos 3xx(3x1)=

A. 92
B. 9(2log3)
C. 9log32
D. 1

#### SOLUTION

Solution : B

limx01cos3xx(3x1)

cos 3x=19x22!+

1cos 3x=9x22!+  ......... (1)

limx03x1x=log 3 .......... (2)

limx01cos 3xx2(3x1x)

From (1) and (2) we get

=92 log 3

Which of the following limit is not in the indeterminant form ?

A.

limxax3a3xa

B.

limxasin x cosec x

C.

limx0xx

D.

limx00x

#### SOLUTION

Solution : D

limxax3a3xa=(00 inderminant form)

limxasin x cosec x=(0× inderminant form)

limxaxx=(00 inderminant form)

limxa0x=0 (not inderminant form)

Let f(x)={x2k(x24)2xwhen x is an int eger otherwise then limx2f(x)

A.

exists only when k=1

B.

exists for every real k

C.

exists for every real k = 1

D.

does not exist

#### SOLUTION

Solution : B

limx2+f(x)=limx2+k(x24)2x      =limx2+k(2x)=4klimx2+f(x)=limx2+k(x24)2x=limx2k(2x)=4k

Hence limits exists for every real value of k.

limx2x2+x2x24is equal to

A.

1/2

B.

1

C.

2

D.

-1

#### SOLUTION

Solution : A

=limx2{1x+2+x2x24}Limit=12+limx2x2x+21(x+2)(x2)=12+limx21x+2x2x+2=12

The value of limx1[sin sin1x] is

A.

1

B.

does not exist

C.

π2

D.

0

#### SOLUTION

Solution : A

Since x1 means x1-, sin-1 x is not defined for x>1
limx1[sin sin1x]=limx1[sin sin1x]=1

The value oflimx01cos3xxsin xcosx

A.

2/5

B.

3/5

C.

3/2

D.

3/4

#### SOLUTION

Solution : C

limx01cos3xxsin xcosx=limx0(1cos x)(1+cos x+cos2x)xsin xcos x=limx02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cos x=limx0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cos x=12×3=32

If f(x)={sinx,xnπ,nϵI2,otherwise and g(x)=x2+1,x0,24,x=05,x=2, then limx0g{f(x)}

A. 5
B. 6
C. 7
D. 1

#### SOLUTION

Solution : D

limx0g{f(x)}=limx0(f(x))2+1,f(x)0,24,f(x)=05,f(x)=2=limx0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪sin2x+1,sinx0,2and xnπ4,sinx=0andxnπ5,sinx=2 and xnπ5,20,2and xnπ4,2=0and x=nπ5,2=2and x=nπ=limx0{1+sin2x,xnπ5,x=nπ=1+sin20=1+0=1