If `f(x)=int_(0)^(1)(dt)/(1+|x-t|),x in R`. The value of `f'(1//2)` is equal to

Options

A. `f(x)` is decreasing for `xgt1`
B. `f(x)` is increasing for `xlt1`
C. `f(1)=log_(e)2`
D. `f(1//2)=log_(e)(3//2)`

Answer

D

Explanation

`f(x)=int_(0)^(1)(dt)/(1+|x-t|)`
`={(int_(0)^(1)(dt)/(1+|x-t|),xlt0),(int_(0)^(1)(dt)/(1+|x-t+),0lexle1),(int_(0)^(1)(dt)/(1+|x-t|),xgt1):}`
`={(int_(0)^(1)(dt)/(1+t-x),xlt0),(int_(0)^(x)(dt)/(1+x-t)+int_(x)^(1)(dt)/(1+t-x),0lexle1),(int_(0)^(1)(dt)/(1+x-t),xgt1):}`
`={([log_(e)(1+t-x)]_(0)^(1),xlt0),([-log_(e)(1+x-t)]_(0)^(1)+[log_(e)(1+t-x)]_(x)^(1),0lexle1),([-log_(e)(1+x-t)]_(0)^(1),xgt1):}`
`={("log"_(e)(2-x)/(1-x),xlt0),("log"_(e)(1+x)(2-x),0lexle1),("log"_(e)(1+x)/x,xgt1):}`