1.

If f(x)=int_(0)^(1)(dt)/(1+|x-t|),x epsilonR the value of f'(1//2) is equal to

Answer»

`1//2`
`0`
`1`
`2`

Solution :`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):}`


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