Question

Given that for each $a\in (0,1),$
$\underset{h\to 0^{+}}{lim}\underset{h}{\overset{1-h}{\int }}{t}^{-a}(1-t{)}^{a-1}dt$
exists. Let this limit be $g\left(a\right)$. In addition, it is given that the function $g\left(a\right)$ is differentiable on $(0,1)$.

The value of $g^{\prime }\left(\frac{1}{2}\right)$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $-\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is D $0$

$g\left(a\right)=\underset{h\to 0^{+}}{lim}\underset{h}{\overset{1-h}{\int }}{t}^{-a}(1-t{)}^{a-1}dt$
Substituting $t\to 1-t$
$\Rightarrow g\left(a\right)=\underset{h\to 0^{+}}{lim}\underset{h}{\overset{1-h}{\int }}(1-t{)}^{-a}(t{)}^{a-1}dt$
Substituting $a\to 1-a$
$\Rightarrow g(1-a)=\underset{h\to 0^{+}}{lim}\underset{h}{\overset{1-h}{\int }}(1-t{)}^{a-1}(t{)}^{-a}dt\Rightarrow g\left(a\right)=g(1-a)\Rightarrow g^{\prime }\left(a\right)=-g^{\prime }(1-a)$
Putting $a=\frac{1}{2}$
$\Rightarrow g^{\prime }\left(\frac{1}{2}\right)=-g^{\prime }\left(\frac{1}{2}\right)\Rightarrow g^{\prime }\left(\frac{1}{2}\right)=0$