Methods Of Differentiation(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
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
Explanation:-
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
Explanation:-
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