11th And 12th > Mathematics
METHODS OF DIFFERENTIATION MCQs
Methods Of Differentiation
Total Questions : 55
| Page 2 of 6 pages
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. -> (−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. -> xn−1yn−1
:
A
Putxn=sinθandyn=sinϕthen,(cosθ+cosϕ)=a(sinθ−sinϕ)⇒2cos(θ+ϕ2)cos(θ−ϕ2)=2acos(θ+ϕ2)sin(θ−ϕ2)⇒cot(θ−ϕ2)=a⇒(θ−ϕ2)=cot−1a⇒θ−ϕ=2cot−1aorsin−1xn−sin−1yn=2cot−1aDifferentiatingbothsides,wehavenxn−1√(1−x2n)−nyn−1√(1−y2n)dydx=0∴√(1−x2n1−y2n)dydx=xn−1yn−1
:
A
Putxn=sinθandyn=sinϕthen,(cosθ+cosϕ)=a(sinθ−sinϕ)⇒2cos(θ+ϕ2)cos(θ−ϕ2)=2acos(θ+ϕ2)sin(θ−ϕ2)⇒cot(θ−ϕ2)=a⇒(θ−ϕ2)=cot−1a⇒θ−ϕ=2cot−1aorsin−1xn−sin−1yn=2cot−1aDifferentiatingbothsides,wehavenxn−1√(1−x2n)−nyn−1√(1−y2n)dydx=0∴√(1−x2n1−y2n)dydx=xn−1yn−1
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. -> 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 B. -> 78
:
B
5f(x)+3f(1x)=x+2........(i)
Replacingxby1x
∴5f(1x)+3f(x)=1x+2...(ii)
FromEq.(i).
25f(x)+15f(1x)=5x+10....(iii)
andfromEq.(ii)
9f(x)+15f(1x)=3x+6.....(iv)
SubtractingEq.(iv)from(iii),weget
∴16f(x)=5x−3x+4
∴xf(x)=5x2−3+4x16=y
∴dydx=10x+416
∴dydx|x=1=10+416
=78
:
B
5f(x)+3f(1x)=x+2........(i)
Replacingxby1x
∴5f(1x)+3f(x)=1x+2...(ii)
FromEq.(i).
25f(x)+15f(1x)=5x+10....(iii)
andfromEq.(ii)
9f(x)+15f(1x)=3x+6.....(iv)
SubtractingEq.(iv)from(iii),weget
∴16f(x)=5x−3x+4
∴xf(x)=5x2−3+4x16=y
∴dydx=10x+416
∴dydx|x=1=10+416
=78
Answer: Option B. -> cos x2y−1
:
B
y=√sinx+y⇒y2−y=sinx
∴(2y−1)dydx=cosx
:
B
y=√sinx+y⇒y2−y=sinx
∴(2y−1)dydx=cosx
Answer: Option C. -> 3(d2ydx2)2
:
C
∵y=(ax+bcx+d)orcxy+dy=ax+b
Differentiating both sides w.r.t.x, then
c{xdydx+y.1}+ddydx=aorxdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t.x, then
orxd2ydx2+dydx+dydx+(dc)d2ydx2=0orx+2dydx(d2ydx2)+dc=0
Again differentiating both sides w.r.t.x, then
1+2(d2ydx2.d2ydx2−dydx.d3ydx3)(d2ydx2)2+0=0∴2dydx.d3ydx3=3(d2ydx2)2
:
C
∵y=(ax+bcx+d)orcxy+dy=ax+b
Differentiating both sides w.r.t.x, then
c{xdydx+y.1}+ddydx=aorxdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t.x, then
orxd2ydx2+dydx+dydx+(dc)d2ydx2=0orx+2dydx(d2ydx2)+dc=0
Again differentiating both sides w.r.t.x, then
1+2(d2ydx2.d2ydx2−dydx.d3ydx3)(d2ydx2)2+0=0∴2dydx.d3ydx3=3(d2ydx2)2
Answer: Option A. -> - 1
:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
Answer: Option C. -> 12
:
C
f(x)=(logcotxtanx)(logtanxcotx)+tan−1(x√(4−x2))
f(x)=1+tan−1(x√(4−x2))
Now Put x=2sinθ, we get
f(x)=1+tan−1(2sinθ√4−4sin2θ)⇒f(x)=1+tan−1(2sinθ2cosθ)⇒f(x)=1+tan−1(tanθ)⇒f(x)=1+θ⇒f(x)=1+sin−1(x2)⇒f′(x)=0+1√1−(x2)2⋅12⇒f′(0)=12
:
C
f(x)=(logcotxtanx)(logtanxcotx)+tan−1(x√(4−x2))
f(x)=1+tan−1(x√(4−x2))
Now Put x=2sinθ, we get
f(x)=1+tan−1(2sinθ√4−4sin2θ)⇒f(x)=1+tan−1(2sinθ2cosθ)⇒f(x)=1+tan−1(tanθ)⇒f(x)=1+θ⇒f(x)=1+sin−1(x2)⇒f′(x)=0+1√1−(x2)2⋅12⇒f′(0)=12
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