Given that for each $a \in(0,1)$, $\lim\limits_{h \rightarrow 0} \int\limits_h^{1-h} t^{-a}(1-t)^{a-1} d t$ exits and is equal to $g(a)$. If $g(a)$ is differentiable in $(0,1)$, then the value of $g'\left(\frac{1}{2}\right)$, is

0

We have,

$g(a)=\lim\limits_{h \rightarrow 0} \int\limits_h^{1-h} t^{-a}(1-t)^{a-1} d t$            ....(i)

$\Rightarrow g(1-a)=\lim\limits_{h \rightarrow 0} \int\limits_h^{1-h} t^{a-1}(1-t)^{-a} d t$

$\Rightarrow g(1-a)=\lim\limits_{h \rightarrow 0} \int\limits_h^{1-h}(1-t)^{a-1}\{1-(1-t)\}^{-a} d t$    $\left[\right.$ Using $\left.\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x\right]$

$\Rightarrow g(1-a)=\lim\limits_{h \rightarrow 0} \int\limits_h^{1-h} t^{-a}(1-t)^{a-1} d t$        .....(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 $a=\frac{1}{2}$, we get

$g'\left(\frac{1}{2}\right)=-g'\left(\frac{1}{2}\right) \Rightarrow g'\left(\frac{1}{2}\right)=0$