Given that for each `a epsilon(0,1),lim(hto 0^(+)) int_(h)^(1-h)t^(-a)(1-t)^(a-1)dt` exists. Let this limit be `g(a)`. In addition it is given the function `g(a)` is differentiable on`(0,1)`.
The value of `g(1/2)` is a. `(pi)/2` b. `pi` c.`-(pi)/2` d. `0`

Options

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

Explanation

To solve the problem, we need to evaluate the limit:

\[ g\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \int_h^{1-h} t^{-\frac{1}{2}} (1-t)^{\frac{1}{2}-1} dt \]

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ g\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \int_h^{1-h} \frac{1}{\sqrt{t}} \cdot \frac{1}{\sqrt{1-t}} dt \]

### Step 2: Change of Variables
To simplify the evaluation, we can use a substitution. Let \( t = \sin^2(x) \), then \( dt = 2\sin(x)\cos(x)dx = \sin(2x)dx \). The limits change as follows:
- When \( t = h \), \( x = \arcsin(\sqrt{h}) \)
- When \( t = 1-h \), \( x = \frac{\pi}{2} - \arcsin(\sqrt{h}) \)

Thus, the integral becomes:

\[ g\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \int_{\arcsin(\sqrt{h})}^{\frac{\pi}{2} - \arcsin(\sqrt{h})} \frac{1}{\sqrt{\sin^2(x)}} \cdot \frac{1}{\sqrt{\cos^2(x)}} \cdot \sin(2x) dx \]

### Step 3: Simplify the Integral
The integral simplifies to:

\[ g\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \int_{\arcsin(\sqrt{h})}^{\frac{\pi}{2} - \arcsin(\sqrt{h})} 2 \, dx \]

### Step 4: Evaluate the Limit
As \( h \to 0^+ \), \( \arcsin(\sqrt{h}) \to 0 \). Thus, the limits of integration approach:

\[ g\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \int_0^{\frac{\pi}{2}} 2 \, dx = 2 \cdot \frac{\pi}{2} = \pi \]

### Final Answer
Thus, the value of \( g\left(\frac{1}{2}\right) \) is:

\[ \boxed{\pi} \]