Discussion Forum

1)

Given that for each   $a \epsilon (0,1), \lim_{h \rightarrow 0}$ $\int_{h}^{1-h} t^{-a}(1-t)^{a-1}dt$  exists. Let this limit be g(a)  . In addition , it is given that the function g(a) differentiable (0,1)

 The value  of   $g(\frac{1}{2})$   is 

Answer:

Option A

Explanation:

Plan   $\int_{}^{} \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\frac{x}{a}\right)+c$

 As g(a)  is defined in the question, first use the numerical value of 'a' given in the question and then proved.

 Given, g(a) $=\lim_{h \rightarrow 0}\int_{h}^{1-h}t^{-a} (1-t)^{a-1}dt$

  $\therefore$   g(1/2)

     $=\lim_{h \rightarrow 0+}\int_{h}^{1-h} t^{-1/2}(1-t)^{-1/2}dt$

  $= \int_{0}^{1} \frac{dt}{\sqrt{t-t^{2}}}=\int_{0}^{1} \frac{dt}{\sqrt{\frac{1}{4}-\left(t-\frac{1}{2}\right)^{2}}}$

   = $\sin^{-1}\left[\left(\frac{t-1/2}{1/2}\right)\right]^{1}_{0}$

       = $\sin^{-1}1-\sin^{-1}(-1)=\pi$