Consider the function `f(theta)=int_(0)^(1)(|sqrt(1-x^(2))-sintheta|)/(sqrt(1-x^(2)))dx`, where `0le theta le (pi)/2`, then

Options

A. `f_("min")=sqrt(2)-1`
B. `f_("min")=sqrt(2)+1`
C. `f_("max")=1`
D. `f_("max")=(pi)/2-1`

Answer

A, B, C, D

Explanation

`f(theta)=int_(0)^(1)(|sqrt(1-x^(2))-sin theta|)/(sqrt(1-x^(2)))dx`
Put `x=cos phi`
`:.f(theta)=int_(0)^((pi)/2)|sinphi-sin theta|d phi`
`=int_(0)^(theta)(sin theta-sin phi)d phi +int_(theta)^((pi)/2)(sin phi-sin theta)d phi`
`=[phi sin theta+cos phi]_(0)^(theta)+[-cos phi-phi sin theta]_(theta)^(pi//2)`
`=[theta sin theta+cos theta-0-1]-0-(pi)/2 sin theta +cos theta + theta sin theta`
`=2(theta sin theta +cos theta)-(pi)/2 sin theta-1`
`f'(theta)=2(theta cos theta+sin theta -sin theta)-(pi)/2 cos theta`
`=2 (theta- (pi)/4) . cos theta`
`f'(theta)=0`
`:. theta =(pi)/4`, which is point of minima.
`:. f((pi)/4)=2((pi)/4 1/(sqrt(2))+1/(sqrt(2)))-(pi)/2 1/(2sqrt(2))-1=sqrt(2)-1`
`:. f(x)=1` and `f((pi)/2)=(pi)/2-1`
So `f_("min")=sqrt(2)-1` and `f_("max")=1`