11th And 12th > Mathematics
METHODS OF DIFFERENTIATION MCQs
Methods Of Differentiation
Total Questions : 55
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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 A. -> −12
:
A
y=tan−1
⎷(1−cos(π2+x)1+cos(π2+x))=tan−1∣∣tan(π4+x2)∣∣⋯)(i)
Now,π2<x<π
∴π4<x2<π2
orπ2<π4+x2<3π4
∴∣∣tan(π4+x2)∣∣=−tan(π4+x2)(∴insecondquadrant)
=tan{π−(π4+x2)}
FromEq.(i),
y=tan−1tan{π−(π4+x2)}
=π−(π4+x2)
=3π4−x2
(∵principalvalueoftan−1xin−π2toπ2)
∴dydx=−12
:
A
y=tan−1
⎷(1−cos(π2+x)1+cos(π2+x))=tan−1∣∣tan(π4+x2)∣∣⋯)(i)
Now,π2<x<π
∴π4<x2<π2
orπ2<π4+x2<3π4
∴∣∣tan(π4+x2)∣∣=−tan(π4+x2)(∴insecondquadrant)
=tan{π−(π4+x2)}
FromEq.(i),
y=tan−1tan{π−(π4+x2)}
=π−(π4+x2)
=3π4−x2
(∵principalvalueoftan−1xin−π2toπ2)
∴dydx=−12
Answer: Option C. -> −2ae−π/2
:
C
√(x2+y2)=a.etan−1(y/x)
⇒12√x2+y2(2x+2yy′)=a.etan−1(y/x)×11+y2x2×xy′−yx2.....(i)
⇒x+yy′√x2+y2=√(x2+y2)×xy′−yx2+y2
[fromEq.(i)]
∴x+yy′=xy′−y⇒y′=x+yx−y
∴y"=2(xy′−y)(x−y)2
y"(0)=2(0−y(0)){0−y(0)}2=−2aeπ/2=−2ae−π/2
:
C
√(x2+y2)=a.etan−1(y/x)
⇒12√x2+y2(2x+2yy′)=a.etan−1(y/x)×11+y2x2×xy′−yx2.....(i)
⇒x+yy′√x2+y2=√(x2+y2)×xy′−yx2+y2
[fromEq.(i)]
∴x+yy′=xy′−y⇒y′=x+yx−y
∴y"=2(xy′−y)(x−y)2
y"(0)=2(0−y(0)){0−y(0)}2=−2aeπ/2=−2ae−π/2
Answer: Option C. -> ln x(1+ln x)−2
:
C
Since,xy=ex−y⇒ylnx=x−y∴y=x1+lnx∴dydx=lnx(1+lnx)2
:
C
Since,xy=ex−y⇒ylnx=x−y∴y=x1+lnx∴dydx=lnx(1+lnx)2
Answer: Option B. -> 2
:
B
y=√(1+cos2θ1−cos2θ)
=|cotθ|=−cotθ(θ=3π4)
∴dydθ=cosec2θ
dydθ|θ=3π/4=(√2)2=2
:
B
y=√(1+cos2θ1−cos2θ)
=|cotθ|=−cotθ(θ=3π4)
∴dydθ=cosec2θ
dydθ|θ=3π/4=(√2)2=2
Answer: Option D. -> 100
:
D
∵x2+4=(secθ−cosθ)2+4=(secθ+cosθ)2⋯(i)Similarly,y2+4=(sec10θ+cos10θ)2⋯(ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)anddydθ=10sec9θsecθtanθ−10cos9θ(−sinθ)=10tanθ(sec10θ−cos10θ)⇒dydx=(dydθ)(dxdθ)=10tanθ(sec10θ+cos10θ)tanθ(secθ+cosθ)∴(dydx)2=100(sec10θ+cos10θ)(secθ+cosθ)2=100(y2+4)(x2+4)or(x2+4)(dydx)2=100(y2+4)
On comparing with the expression given we get k = 100
:
D
∵x2+4=(secθ−cosθ)2+4=(secθ+cosθ)2⋯(i)Similarly,y2+4=(sec10θ+cos10θ)2⋯(ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)anddydθ=10sec9θsecθtanθ−10cos9θ(−sinθ)=10tanθ(sec10θ−cos10θ)⇒dydx=(dydθ)(dxdθ)=10tanθ(sec10θ+cos10θ)tanθ(secθ+cosθ)∴(dydx)2=100(sec10θ+cos10θ)(secθ+cosθ)2=100(y2+4)(x2+4)or(x2+4)(dydx)2=100(y2+4)
On comparing with the expression given we get k = 100
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. -> 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
Answer: Option D. -> |x|x
:
D
∵f(x)={x,x≥0−x,x<0∵f′(x)=⎧⎨⎩1,x>0,i.e,|x|x,x>0−1,x>0,i.e,|x|x,x>0=|x|x,x≠0
:
D
∵f(x)={x,x≥0−x,x<0∵f′(x)=⎧⎨⎩1,x>0,i.e,|x|x,x>0−1,x>0,i.e,|x|x,x>0=|x|x,x≠0
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
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