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A function is defined by f(x) = int0^pi ...

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

A

`(pi)/4cosx`

B

`(pi)/2cosx`

C

`(-pi)/2cosx`

D

`(-pi)/4cosx`

Text Solution

Verified by Experts

The correct Answer is:
C
`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`.
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