`f(x)={bsin^-1((x+c)/2),-1/2 lt xlt0 and 1/2 , x=0 and e^(((ax)/2)-1)/x , o lt x lt 1/2` If `f(x)` is differentiable at `x=0 and |c|<1/2`, then find thevalue of a and prove that `64b^2=(4-c^2)`.

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at \( x = 0 \). The function is defined piecewise, and we will analyze the limits from both sides of \( x = 0 \).

### Step 1: Check Continuity at \( x = 0 \)

The function is given as:
\[ f(x) = \begin{cases} b \sin^{-1}\left(\frac{x+c}{2}\right) & \text{for } -\frac{1}{2} < x < 0 \\ \frac{e^{\frac{ax}{2}} - 1}{x} & \text{for } 0 < x < \frac{1}{2} \\ \end{cases} \]
We need to check that:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

### Step 2: Calculate \( f(0) \)

From the definition, we can assume:
\[ f(0) = \frac{1}{2} \]

### Step 3: Calculate the Left-Hand Limit

For \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} b \sin^{-1}\left(\frac{x+c}{2}\right) \]
As \( x \to 0 \):
\[ \sin^{-1}\left(\frac{0+c}{2}\right) = \sin^{-1}\left(\frac{c}{2}\right) \]
Thus,
\[ \lim_{x \to 0^-} f(x) = b \sin^{-1}\left(\frac{c}{2}\right) \]

### Step 4: Calculate the Right-Hand Limit

For \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{\frac{ax}{2}} - 1}{x} \]
Using the limit property:
\[ \lim_{x \to 0} \frac{e^{\frac{ax}{2}} - 1}{x} = \frac{a}{2} \]
Thus,
\[ \lim_{x \to 0^+} f(x) = \frac{a}{2} \]

### Step 5: Set the Limits Equal for Continuity

Setting the left-hand limit equal to the right-hand limit:
\[ b \sin^{-1}\left(\frac{c}{2}\right) = \frac{a}{2} = \frac{1}{2} \]
From this, we can deduce:
\[ b \sin^{-1}\left(\frac{c}{2}\right) = \frac{1}{2} \]
This gives:
\[ b = \frac{1}{2 \sin^{-1}\left(\frac{c}{2}\right)} \]

### Step 6: Differentiate and Set Derivatives Equal

To ensure differentiability, we need to find the derivatives from both sides.

**Right-Hand Derivative:**
\[ f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = \lim_{x \to 0^+} \frac{\frac{e^{\frac{ax}{2}} - 1}{x} - \frac{1}{2}}{x} \]
Using L'Hôpital's Rule:
\[ f'(0^+) = \frac{a}{4} \]

**Left-Hand Derivative:**
\[ f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x} = \lim_{x \to 0^-} \frac{b \sin^{-1}\left(\frac{x+c}{2}\right) - \frac{1}{2}}{x} \]
Using L'Hôpital's Rule:
\[ f'(0^-) = b \cdot \frac{1}{\sqrt{1 - \left(\frac{c}{2}\right)^2}} \cdot \frac{1}{2} \]

Setting the derivatives equal:
\[ \frac{a}{4} = b \cdot \frac{1}{\sqrt{1 - \left(\frac{c}{2}\right)^2}} \cdot \frac{1}{2} \]

### Step 7: Substitute \( a = 1 \)

From earlier, we found \( a = 1 \):
\[ \frac{1}{4} = b \cdot \frac{1}{\sqrt{1 - \left(\frac{c}{2}\right)^2}} \cdot \frac{1}{2} \]
This leads to:
\[ b = \frac{1}{2} \sqrt{1 - \left(\frac{c}{2}\right)^2} \]

### Step 8: Prove \( 64b^2 = 4 - c^2 \)

Substituting \( b \):
\[ 64b^2 = 64 \left(\frac{1}{2} \sqrt{1 - \left(\frac{c}{2}\right)^2}\right)^2 = 16(1 - \frac{c^2}{4}) = 16 - 4c^2 \]
Thus:
\[ 64b^2 = 4 - c^2 \]

### Final Result
We have found \( a = 1 \) and proved that \( 64b^2 = 4 - c^2 \).