A function f is defined by `f(x) = int_0^pi cos t cos(x-t)dt,0 <= x <= 2 pi` then which of the following.hold(s) good?

Options

A. `(pi)/4`
B. `(pi)/2`
C. `(-pi)/2`
D. `(-pi)/4`

Answer

C

Explanation

`f(x)=int_(0)^(pi)cost cos(x-t)dt`………..1
`=int_(0)^(pi)cos(pi-t)cos(x-(pi-t))dt`
`:.f(x)=int_(0)^(pi)cot.cos(x+t)dt` ………………2
From 1+2 gives
`2f(x)=int_(0)^(pi) cost(2cosx.cost)dt`
`:. f(x)=cosx int_(0)^(pi)cos^(2)tdt=2cos int_(0)^(pi//2) cos^(2)t dt`
`:.f(x)=(picosx)/2`
Hence minimum value of `f(x)` is `(-pi)/2`.