Let `f` be a differentiable function satisfying `int_(0)^(f(x))f^(-1)(t)dt-int_(0)^(x)(cost-f(t)dt=0` and `f((pi)/2)=2/(pi)`
The value of `int_(0)^(pi//2) f(x)dx` lies in the interval

Options

A. `(2/(pi),1)`
B. `(1,(pi)/2)`
C. `(3/2,(pi)/2)`
D. `(0,2/(pi))`

Answer

B

Explanation

`int_(0)^(f(x))f^(-1)(t)dt-int_(0)^(x)(cost-f(t)dt=0`
Differentiating w.r.t `x`
`f^(-1)(f(x)).f(x)-(cosx-f(x))=0`
`implies xf'(x)+f(x)=cosx`
`implies(xf(x))'=cosx`
`implies xf(x)=sinx+c`
Given `f((pi)/2)=2/(pi)`
`impliesf(x)=(sinx)/x`
`int_(0)^(pi//2) f(x)dx=int_(0)^(pi//2) (sinx)x dx`
Now `2/(pi)lt (sinx)/x lt 1`
`:. 1lt int_(0)^(pi//2) (sinx)/xdxlt (pi)/2`
`lim_(xto 0) (cosx)/(f(x))=lim_(xto0) x/(tan)=1`