Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f(x), the x-axis, and the ordinates `x=(pi)/(4) and x=betagt(pi)/(4)" is "beta sin beta +(pi)/(4)cos beta +sqrt(2)beta.` Then `f'((pi)/(2))` is

Options

A. `((pi)/(2)-sqrt(2)-1)`
B. 1-π/4
+sqrt(2)
C. `-(pi)/(2)`
D. `(1-(pi)/(2)-sqrt(2))`

Answer

C

Explanation

`int_(pi//4)^(beta)f(x)dx=beta sin beta +(pi)/(4)cos beta +sqrt(2)beta`
Differentiating both sides w.r.t`beta,` we get
`therefore" "f(beta)=beta cos beta + sin beta -angle(pi)/(4)sin beta +sqrt(2)`
`rArr" "f'(beta)=-beta sin beta + cos beta+ cos beta-(pi)/(4)cos beta`
`rArr" "f'((pi)/(2))=-(pi)/(2)`