11th And 12th > Mathematics
METHODS OF DIFFERENTIATION MCQs
Methods Of Differentiation
Total Questions : 55
| Page 4 of 6 pages
Answer: Option B. ->
1
:
B
Let u=sin−1(2x1+x2)=2 tan−1x
∴dudx=21+x2
and v=tan−1(2x1+x2)=2 tan−1x
∴dvdx=21+x2
∴dudv=(dudx)(dvdx)=1
:
B
Let u=sin−1(2x1+x2)=2 tan−1x
∴dudx=21+x2
and v=tan−1(2x1+x2)=2 tan−1x
∴dvdx=21+x2
∴dudv=(dudx)(dvdx)=1
Answer: Option A. ->
0
:
A
f(x)=loge{u(x)v(x)}
=loge u(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)}
=loge u(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 C. ->
sin a
:
C
x=sin ysin(a+y).....(i)
⇒dxdy=sin(a)sin2(a+y)
∴dydx=sin2(a+y)sin(a)=A1+x2−2x cos a
Put x=0,y=0,
then A=sin a
:
C
x=sin ysin(a+y).....(i)
⇒dxdy=sin(a)sin2(a+y)
∴dydx=sin2(a+y)sin(a)=A1+x2−2x cos a
Put x=0,y=0,
then A=sin a
Answer: Option C. ->
2a2
:
C
Given,
√(x+y)+√(y−x)=a..........(i)
∴√x+y−√y−x=2xa.......(ii)
Adding Eqs.(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)
Adding Eqs.(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 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
xcos y+ycos x=5
⇒ecos y logex+ecos x logey=5
∴ecos y loge x{cosyx−loge x(sin y)dydx}+ecosx logey{cos xydydx−sin x logey}=0
Putx=y=1, (cos 1−0)+(cos 1dydx−0)=0
∴dydx=−1
or y′=−1
:
C
xcos y+ycos x=5
⇒ecos y logex+ecos x logey=5
∴ecos y loge x{cosyx−loge x(sin y)dydx}+ecosx logey{cos xydydx−sin x logey}=0
Putx=y=1, (cos 1−0)+(cos 1dydx−0)=0
∴dydx=−1
or y′=−1
Answer: Option D. ->
f(x,y)=x2y2
:
D
Put x3=sin θ,y3=sin ϕ,
then cosθ+cosϕ=a(sin θ−sinϕ)
⇒2 cos(θ+ϕ2) cos(θ−ϕ2) =2a cos (θ+ϕ2) sin (θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2 cot−1 a
∴3x2√(1−x6)−3y2√(1−y6)dydx=0
⇒dydx=x2y2√(1−y61−x6)
∴f(x,y)=x2y2
:
D
Put x3=sin θ,y3=sin ϕ,
then cosθ+cosϕ=a(sin θ−sinϕ)
⇒2 cos(θ+ϕ2) cos(θ−ϕ2) =2a cos (θ+ϕ2) sin (θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2 cot−1 a
∴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=−a sin θ and y=b sin θ ⇒dydθ=b cos θ∴dydx=−ba cot θ⇒d2ydx2=ba cosec2θ dθdx=−ba2cosec3θ∴d3ydx3=3ba2 cosec2θ(−cosec θ cot θ)dθdx=3ba2 cosec3θ cot θ (−1a sin θ)=−3ba3 cosec4θ cot θ
:
C
∵x=acosθ⇒dxdθ=−a sin θ and y=b sin θ ⇒dydθ=b cos θ∴dydx=−ba cot θ⇒d2ydx2=ba cosec2θ dθdx=−ba2cosec3θ∴d3ydx3=3ba2 cosec2θ(−cosec θ cot θ)dθdx=3ba2 cosec3θ cot θ (−1a sin θ)=−3ba3 cosec4θ cot θ
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 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