12th Grade > Mathematics
CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 30
| Page 2 of 3 pages
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 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 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 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 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 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 A. -> 0
:
A
dydx=(dxdy)−1⇒d2ydx2=−(dxdy)−2{ddx(dxdy)}⇒d2ydx2=−(dxdy)−2{ddy(dxdy)dydx}⇒d2ydx2=−(dydx)2{d2xdy2.dydx}⇒d2ydx2=−(dydx)3d2xdy2⇒d2ydx2+(dydx)3d2xdy2=0
:
A
dydx=(dxdy)−1⇒d2ydx2=−(dxdy)−2{ddx(dxdy)}⇒d2ydx2=−(dxdy)−2{ddy(dxdy)dydx}⇒d2ydx2=−(dydx)2{d2xdy2.dydx}⇒d2ydx2=−(dydx)3d2xdy2⇒d2ydx2+(dydx)3d2xdy2=0
Answer: Option C. -> 1
:
C
x2+y2=t−1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t2−2
Given that -
x4+y4=t2+1t2
∴x2y2=−1
⇒x2.2ydydx+y2.2x=0
⇒x3ydydx=−x2y2=1
:
C
x2+y2=t−1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t2−2
Given that -
x4+y4=t2+1t2
∴x2y2=−1
⇒x2.2ydydx+y2.2x=0
⇒x3ydydx=−x2y2=1
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. -> 1
:
C
Since,y=(1+x)(1+x2)(1+x4)⋯(1+x2n),(1−x)y=(1−x2)(1+x2)(1+x4)⋯(1+x2n)=(1−x4)(1+x4)⋯(1+x2n+1)∴y=1−x2n+1(1−x)∵dydx=(1−x)(−2n+1.x2n)−(1−x2n+1)(−1)(1−x)2∴dydx∣∣x=0=(1)(0)−(1)(−1)(1)2=1
:
C
Since,y=(1+x)(1+x2)(1+x4)⋯(1+x2n),(1−x)y=(1−x2)(1+x2)(1+x4)⋯(1+x2n)=(1−x4)(1+x4)⋯(1+x2n+1)∴y=1−x2n+1(1−x)∵dydx=(1−x)(−2n+1.x2n)−(1−x2n+1)(−1)(1−x)2∴dydx∣∣x=0=(1)(0)−(1)(−1)(1)2=1
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