Given that for each `a in (0,1)lim_(x to 0) int_(h)^(-h) t^(-a)(1-t)^(a-1)dt` exits and is equal to g(a).If g(a) is differentiable in (0,1), then the value of g'((1)/(2))`, is

Options

A. `pi//2`
B. `pi`
C. `-pi//2`
D. 0

Answer

D

Explanation

We have,
`g(a)=underset(h to 0)lim underset(h)overset(1-h)intt^(a-1)dt" "`......(1)
`rArr g(1-a)=underset(h to 0)limoverset(1-h)underset(h)int t^(a-1)(1-t)^(-a)dt`
`rArr g(1-a)=underset(h to 0)lim underset(h)overset(1-h)int (1-t)^(a-1){1-(1-t)}^(-a)dt["Using"underset(a)overset(b)intf(x)=underset(a)overset(b)int dx=underset(a)overset(b)int f(a+b-x)dx]`
`g(1-a)=underset(h to 0)lim underset(h)overset(1-h)int t^(-a)(1-t)^(a-1)dt" "`......(ii)
From (i)and (ii), we get
`g(a)=g'(1-a)` for all `a in (0,1)`
Differentiating both sides with respect to a, we get
`g'(a)=-g'(1-a)`for all `a in (0,1)`
Putting `a=(1)/(2)`,we get
`g'((1)/(2))=-g''((1)/(2))=-g'((1)/(2))rArr g'((1)/(2))=0`