If `f'` is a differentiable function satisfying `f(x)=int_(0)^(x)sqrt(1-f^(2)(t))dt+1/2` then the value of `f(pi)` is equal to

Options

A. `-(sqrt(3))/2`
B. `-1/2`
C. `(sqrt(3))/2`
D. `1/2`

Answer

B

Explanation

`f(x)=int_(0)^(x)sqrt(1-f^(2)(t))dt+1/2`……………1
`implies f'(x)=sqrt(1-f^(2)(x))`
`implies int(f'(x))/(sqrt(1-f^(2)(x)))dx=int1dx`
`implies sin^(-1)(f(x))=x+c`
From 1 `f(0)=1/2`
`:.c=(pi)/6`
`implies f(x)=sin(x+(pi)/6)`
`impliesf(pi)=sin(-(pi)/6)=-1/2`