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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 15 - MCQExams.com

Ltxπ42cosxsinx(4xπ)2=?
  • 1162
  • 1322
  • 116
  • 18
limx01cos3xxsin2x=
  • 1/2
  • 3/2
  • 3/4
  • 1/4
The value of limx0((sinx)1/x+(1x)sinx) equals
  • 0
  • 1
  • -1
limx0ln(sin3x)ln(sinx) is equal to
  • 0
  • 1
  • 2
  • none of these
limx01e2tan(π4+x)1/x
  • 0
  • 1
  • -1
  • e
The value of θ,is
limoocos2{1cos2(1cos2.....(cos2{1cos2θ}))}sin(π(θ+42θ)
  • 24
  • 2
  • 1
  • 2
limxsinxsin(π3+x)sin(π3x)x=
  • 34
  • 14
  • 43
  • 0
If α,β, in(π2,0) such that (sinα+sinβ)+sinαsinβ=0 and (sinα+sinβ)sinαsinβ=1 and λ=limn1+(2sintheta)2n(2sinbeta)2n then
  • 2α+3β=5π6
  • λπ+α+β=5π6
  • αβ=π3
  • α+β=π3
ddx(sin5xsin5x)=
  • (sin4xsin5x)
  • 5(sin4xsin5x)
  • 5(sin4xsin6x)
  • 5(sin4xsin6x)
limx027x9x3x+121+cosx=
  • 0
  • 82(log3)2
  • 8(log3)2
  • 1
limh0sin(a+3h)3sin(a+2h)+3sin(a+h)sinah3 is equal to 
  • cosa
  • cosa
  • sina
  • sinacosa
Limx(sinx+1sinx)=
  • 2
  • -2
  • 0
  • 1
limx0{(sinxx)/x3)} equals: 
  • 1/3
  • 1/3
  • 1/6
  • 1/6
Solve 
Limx08x8(1cosx22cosx24+cosx22.cosx24)=
  • 116
  • 115
  • 132
  • 1
limxπ/2[xtanx(π2)secx] is equal to
  • 1
  • -1
  • 0
  • None of these
limx0sinxx=y
  • y>1
  • y<1
  • y1
  • y1
The value of Limitx0cos(sinx)cosxx4 is equal to 
  • 1/5
  • 1/6
  • 1/4
  • 1/2
limx08x8(1cosx22cosx24+cosx22.cosx24)=
  • 116
  • 115
  • 132
  • 1
limx0[100tanx.sinxx2] where [.] represents greatest integer function is 
  • 99
  • 100
  • 0
  • 98
The value of  limx122sin1xπ212x2  is equal to
  • 1
  • 0
  • 1
  • 2
The value of limx0sin3(x)ln(1+3x)(tan1x)2(e5(x)1)x is equal to
  • 15
  • 35
  • 25
  • 45
The value of limxπ2tan2(2sin2x+3sinx+4sin2x+6sinx+2) is equal to
  • 0
  • 111
  • 112
  • 18
If x sin (α+y)=sinyandy=m(x2+)nx+, then
  • m-n=1
  • m+n=1
  • m2+n2=1
  • m=n
limx(x2sin(1x)x1|x|)=
  • 0
  • 1
  • -1`
  • 2
Ltx1(1x)tanπx=
  • 1
  • 0
  • 1
  • 2
Let a(0,π2), then the value of
lima01a3a0n(1+tanatanx)dx is equal to 
  • 13
  • 12
  • 16
  • 1
Find:
limx01cos3xxsin2x=
  • 1/2
  • 3/2
  • 3/4
  • 1/4
Lim1cos2xxsin2x
  • 1/2
  • 3/2
  • 3/4
  • 1/4
The value of limx0sec5xsec3xsec3xsecx
  • -2
  • 1
  • 2
  • -1/2
y=sec(tan1x) then at x=1 value of dydx equals :
  • 12
  • 1
  • 2
  • 12
0:0:4


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