# Free Objective Test 02 Practice Test - 11th and 12th

f(x)={4x3,x<1x2x1, then
limx1f(x)=

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

#### SOLUTION

Solution : A

limx1f(x)=limx1(4x3)=1
limx1+f(x)=limx1+x2=1
Since LHL = RHL
limx1f(x)=1

limn2.3n3.5n3.3n+4.5n=

A. 23
B. 34
C. 1
D. 0

#### SOLUTION

Solution : B

limn2.3n3.5n3.3n+4.5n
=limn5n(2(35)n3)5n(3(35)n+4)
As n,(35)n0
=34

limαβsin2αsin2βα2β2 is equal to

A. sin3β(4β)
B. sin2β(2β)
C. sin8β(7β)
D. sin6β(4β)

#### SOLUTION

Solution : B

limαβsin2αsin2βα2β2=limαβsin(αβ)sin(α+β)(αβ)(α+β)=sin2β(2β)

limx(2+x)40(4+x)5(2x)45

A. 1
B. 1
C. 16
D. 32

#### SOLUTION

Solution : A

limx(2+x)40(4+x)5(2x)45
=limxx45(2x+1)40(4x+1)5x45(2x1)45
=(1)40(1)5(1)45
=(1)45(1)45
=1

limxx+sinxxcosx=

A. 0
B. 1
C. 1
D. does not exist

#### SOLUTION

Solution : B

limxx+sinxxcosx
=limxx121+sinxxx121cosxx
We know, that for any value of x, sinx and cosx will be [-1,1]
So,  =limxSinxx=0
And  =limxCosxx=0
=x121+sinxx1+sinxx1+cosxx
=limx11
=1

limx2x2+x2x24 is equal to

A. 12
B. 1
C. 2
D. 0

#### SOLUTION

Solution : A

limx2x2+x2x24
=limx2(x2x+2x2+x2x24)
On rationalisation -
=limx2(1x+2+x2x24(x+2))
=limx21x+2+limx2x2x+2×1x+2
=12

The value of limx0xa[bx] is, ([.]G.I.F)

A. 0
B.
C. ba
D. does not exist

#### SOLUTION

Solution : C

limx0xa[bx]
=limx0xa(bx{bx})
Since {bx}ϵ[0,1)
=limx0xa.{bx}=0
limx0xa[bx]=limx0(xa)(bx)
=limx0.ba
=ba

If G(x) = - 25x2 then limx1G(x)G(1)x1 has the value

A. 124
B. 15
C. 24
D. 126

#### SOLUTION

Solution : D

limx125x2(24)x1
=limx12425x2x1×24+25x224+25x2
limx1x21(x1)[24+25x2]=2224=126

limx02sinxsin2xx3 is equal to

A. 1
B. 1
C. 0
D. does not exist

#### SOLUTION

Solution : A

limx02sinxsin2xx3
=limx02sinx(1cosx)(1+cosx)x3(1+cosx)
=limx02sin3xx3×11+cosx
=2×(1)3×11+1=1

If limx0((an)nxtanx)sinnxx2=0 where n is nonzero real number, then a is equal to

A. 0
B. n+1n
C. n
D. n+1n

#### SOLUTION

Solution : D

limx0((an)nxtanx)sinnxx2=0
limx0((an)n(tanxx)).sinnxnx.n=0
((an)n1).1.n=0
a=1n+n

The value of limx1(tanxπ4)tanπx2 is

A. e2
B. e1
C. e
D. 1

#### SOLUTION

Solution : B

limx1(tanxπ4)tanπx2(1from)
elimx1(tanxπ41)tanπx2
=elimx1(sinπx4cosπx4cosπx4)2sinπx4cosπx4cosπx2
=elimx12sin2πx42sinπx4cosπx4cosπx2
=elimx1(1cosπx2sinπx2cosπx2)
=elimx1(1sinπx2cosπx21)
=elimx1(cosπx21+sinπx21)
= e1

limx0(1+2x1+3x)1x2.e1x=

A. e52
B. e2
C. 2
D. 1

#### SOLUTION

Solution : A

limx0e1x2(log(1+2x)log(1+3x))+1xelimx0(log(1+2x)log(1+3x))+xx2
Using the expansion we get -
=e52

limnnr=1cot1(r2+34) is

A.

0

B.

tan11

C.

tan12

D.

tan112

#### SOLUTION

Solution : C

limnnr=1tan1(44r2+3)

limnnr=1tan1(1r214+1)

limnnr=1tan1((r+12)(r12)1+(r+12)(r12))

limnnr=1tan1{tan1(r+12)tan1(r12)}

limn{tan1(n+12)tan1(12)}

=  π2tan112

=  tan12

The value of limxπ2[sin1sinx],[x] is the greatest integer function of x, is

A.

1

B.

π2

C.

0

D.

12

#### SOLUTION

Solution : A

limxπ2[sin1sinx]
= limxπ2[x]
= 1

If f(x)={x23,2<x<32x+5,3<x<4, the equation whose roots are limx3f(x) and limx3+f(x) is

A. x212x+36=0
B. x226x+66=0
C. x217x+66=0
D. x222x+121=0

#### SOLUTION

Solution : C

f(x)={x23,2<x<32x+5,3<x<4
limx3f(x)=limx3(x23)=6
and limx3+f(x)=limx3+(2x+5)=11
Hence, the required equation will be
x2 - (sum of roots)x + (Products of roots) = 0