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Off(x)=int0^x("cos"(sint)+"cos"(cost)dt ...

`Off(x)=int_0^x("cos"(sint)+"cos"(cost)dt ,t h e nf(x+pi)i s` `f(x)+f(pi)` (b) `f(x)+2(pi)` `f(x)+f(pi/2)` (d) `f(x)+2f(pi/2)`

A

`f(x)+f(pi)`

B

`f(x)+2f(pi)`

C

`f(x)+f((pi)/2)`

D

`f(x)+2f((pi)/2)`

Text Solution

Verified by Experts

The correct Answer is:
A, D

`f(x+pi)=int_(0)^(x+pi)(cos(sint)+cos(cost))dt`
`=int_(0)^(pi)(cos(sint)+cos(cost))dt`
`+int_(pi)^(x+pi)(cos(sint)+cos(cost)dt`
`=f(pi)+int_(0)^(x)(cos(sint)+cos(cost))dt`
`=f(pi)+int_(0)^(1)(cos(sint)+cos(cost))dt`
`[ :' "for" g(x)=cos(sinx)+cos(cosx),f(x+pi)=f(x)]`
`=f(pi)+f(x)`
`=f(pi)+2f((pi)/2) [ :' g(x)` has period `pi//2`]
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