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f is a real valued function from R to R ...

f is a real valued function from R to R such that `f(x)+f(-x)=2`, then `int_(1-x)^(1+X)f^(-1)(t)dt=`

A

`-1`

B

0

C

1

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Given `f(x)+f(-x)=2`
Let f(x) = t
`therefore" "f(-x)=2-t`
`therefore" "f^(-1)(t)+f^(-1)(2-t)=0`
`I=int_(1-x)^(1+x)f^(-1)(t)dt=int_(1-x)^(1+x)f^(-1)(2-t)dt`
`therefore" "2I=int_(1-x)^(1+x)(f^(-1)(t)+f^(-)(2-t))dt`
I = 0
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