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Q.

 Let f:0,π2[0,1] be a differentiable function such that f(0)=0,fπ2=1 then 

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a

f'(α)=1-[f(a)]2 for atleast one a0,π2

b

f'(α)=2π for atleast one α0,π2

c

f(α)f'(α)=1π for atleast one α0,π2

d

f'(α)=8απ2 for atleast one α0,π2

answer is A, B, C, D.

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Detailed Solution

 Consider g(x)=sin-1f(x)-x

g(0)=0,gπ2=0

 A) By Rolle's theorem there is atleast one value of α0,π2 such that 

g'(α)=0f'(α)1-f(α)2-1=0f'(α)=1-(f(α))2

 It is not true for all values of a

 B) Consider g(x)=f(x)-2xπ

g(0)=0,gπ2=0

g'(x)=0 at least for one value of α0,π2

f'(α)=2π for atleast one value of α0,π2

 C) Consider g(x)=[f(x)]2-4x2π2

g(0)=0,gπ2=0

g'(x)=0 for atleast one value of α0,π2

f(α)f'(α)=1π( true )

 D) Consider g(x)=f(x)-4x2π2,g(0)=0 and gπ2=0

f'(α)=8απ2 atleast for one value of α0,π2

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