12th Grade > Mathematics
CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 30
| Page 3 of 3 pages
Answer: Option B. -> y2−x2y3−2xy−1
:
B
∴y=√x+√y+y
⇒(y2−x)=√2y
or(y2−x)2=2y
Differentiating both sides w.r.t. x, then
2(y2−x)(2ydydx−1)=2dydx
∴dydx=(y2−x)2y3−2xy−1
:
B
∴y=√x+√y+y
⇒(y2−x)=√2y
or(y2−x)2=2y
Differentiating both sides w.r.t. x, then
2(y2−x)(2ydydx−1)=2dydx
∴dydx=(y2−x)2y3−2xy−1
Answer: Option C. -> 2a2
:
C
Given,
√(x+y)+√(y−x)=a..........(i)
∴√x+y−√y−x=2xa.......(ii)
AddingEqs.(i)and(ii),then
2√x+y=a+2xa
Squaring,4x+4y=a2+4x2a2+4x
∴4+4dydx=0+8xa2+4
∴0+4d2ydx2=8a2
∴d2ydx2=2a2
:
C
Given,
√(x+y)+√(y−x)=a..........(i)
∴√x+y−√y−x=2xa.......(ii)
AddingEqs.(i)and(ii),then
2√x+y=a+2xa
Squaring,4x+4y=a2+4x2a2+4x
∴4+4dydx=0+8xa2+4
∴0+4d2ydx2=8a2
∴d2ydx2=2a2
Answer: Option B. -> 1
:
B
Letu=sin−1(2x1+x2)=2tan−1x
∴dudx=21+x2
andv=tan−1(2x1+x2)=2tan−1x
∴dvdx=21+x2
∴dudv=(dudx)(dvdx)=1
:
B
Letu=sin−1(2x1+x2)=2tan−1x
∴dudx=21+x2
andv=tan−1(2x1+x2)=2tan−1x
∴dvdx=21+x2
∴dudv=(dudx)(dvdx)=1
Answer: Option D. -> - 1
:
D
∵sin(x+y)=loge(x+y),
cos(x+y)(1+dydx)=1(x+y)(1+dydx)
⇒(1+dydx)(cos(x+y)−1x+y)=0
∵cos(x+y)−1(x+y)≠0
∴1+dydx=0
∴dydx=−1
:
D
∵sin(x+y)=loge(x+y),
cos(x+y)(1+dydx)=1(x+y)(1+dydx)
⇒(1+dydx)(cos(x+y)−1x+y)=0
∵cos(x+y)−1(x+y)≠0
∴1+dydx=0
∴dydx=−1
Answer: Option C. -> at x = y = 1, y’ = – 1
:
C
xcosy+ycosx=5
⇒ecosylogex+ecosxlogey=5
∴ecosylogex{cosyx−logex(siny)dydx}+ecosxlogey{cosxydydx−sinxlogey}=0
Putx=y=1,(cos1−0)+(cos1dydx−0)=0
∴dydx=−1
ory′=−1
:
C
xcosy+ycosx=5
⇒ecosylogex+ecosxlogey=5
∴ecosylogex{cosyx−logex(siny)dydx}+ecosxlogey{cosxydydx−sinxlogey}=0
Putx=y=1,(cos1−0)+(cos1dydx−0)=0
∴dydx=−1
ory′=−1
Answer: Option C. -> sin a
:
C
x=sinysin(a+y).....(i)
⇒dxdy=sin(a)sin2(a+y)
∴dydx=sin2(a+y)sin(a)=A1+x2−2xcosa
Putx=0,y=0,
thenA=sina
:
C
x=sinysin(a+y).....(i)
⇒dxdy=sin(a)sin2(a+y)
∴dydx=sin2(a+y)sin(a)=A1+x2−2xcosa
Putx=0,y=0,
thenA=sina
Answer: Option D. -> f(x,y)=x2y2
:
D
Putx3=sinθ,y3=sinϕ,
thencosθ+cosϕ=a(sinθ−sinϕ)
⇒2cos(θ+ϕ2)cos(θ−ϕ2) =2acos(θ+ϕ2)sin(θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2cot−1a
∴3x2√(1−x6)−3y2√(1−y6)dydx=0
⇒dydx=x2y2√(1−y61−x6)
∴f(x,y)=x2y2
:
D
Putx3=sinθ,y3=sinϕ,
thencosθ+cosϕ=a(sinθ−sinϕ)
⇒2cos(θ+ϕ2)cos(θ−ϕ2) =2acos(θ+ϕ2)sin(θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2cot−1a
∴3x2√(1−x6)−3y2√(1−y6)dydx=0
⇒dydx=x2y2√(1−y61−x6)
∴f(x,y)=x2y2
Answer: Option C. -> (−3ba3)cosec4θ cotθ
:
C
∵x=acosθ⇒dxdθ=−asinθandy=bsinθ⇒dydθ=bcosθ∴dydx=−bacotθ⇒d2ydx2=bacosec2θdθdx=−ba2cosec3θ∴d3ydx3=3ba2cosec2θ(−cosecθcotθ)dθdx=3ba2cosec3θcotθ(−1asinθ)=−3ba3cosec4θcotθ
:
C
∵x=acosθ⇒dxdθ=−asinθandy=bsinθ⇒dydθ=bcosθ∴dydx=−bacotθ⇒d2ydx2=bacosec2θdθdx=−ba2cosec3θ∴d3ydx3=3ba2cosec2θ(−cosecθcotθ)dθdx=3ba2cosec3θcotθ(−1asinθ)=−3ba3cosec4θcotθ
Answer: Option A. -> 0
:
A
f(x)=loge{u(x)v(x)}
=logeu(x)−logev(x)
∴f′(x)=u′(x)u(x)−v′(x)v(x)
f′(2)=u′(2)u(2)−v′(2)v(2)
=42−21
=2−2=0
:
A
f(x)=loge{u(x)v(x)}
=logeu(x)−logev(x)
∴f′(x)=u′(x)u(x)−v′(x)v(x)
f′(2)=u′(2)u(2)−v′(2)v(2)
=42−21
=2−2=0
Answer: Option D. -> 1
:
D
We have, y=cos−1(x−1−xx−1+x)⇒y=cos−1(1−x21+x2)
Now, Put x=tanθ
We get y=cos−1(1−tan2θ1+tan2θ)⇒y=cos−1(cos2θ)⇒y=2θ⇒y=2tan−1x∴dydx=21+x2∴dydx|x=−1=21+(−1)2=2
:
D
We have, y=cos−1(x−1−xx−1+x)⇒y=cos−1(1−x21+x2)
Now, Put x=tanθ
We get y=cos−1(1−tan2θ1+tan2θ)⇒y=cos−1(cos2θ)⇒y=2θ⇒y=2tan−1x∴dydx=21+x2∴dydx|x=−1=21+(−1)2=2
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