Let `f(x)={{:(a,","x=(pi)/(2)),((sqrt(2x-pi))/(sqrt(9+sqrt(2x-pi))-b),","xgt(pi)/(2)):}`. If `f(x)` is continuous at `x=(pi)/(2)`, then the value of `(a^(2))/(5b)` is

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). The function is defined as follows:

\[ f(x) = \begin{cases} a & \text{if } x = \frac{\pi}{2} \\ \frac{\sqrt{2x - \pi}}{\sqrt{9 + \sqrt{2x - \pi}} - b} & \text{if } x > \frac{\pi}{2} \end{cases} \]

### Step 1: Set up the continuity condition
For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), we need:

\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = a \]

### Step 2: Calculate the limit
We need to evaluate the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi}}{\sqrt{9 + \sqrt{2x - \pi}} - b} \]

### Step 3: Substitute \( x = \frac{\pi}{2} \)
As \( x \) approaches \( \frac{\pi}{2} \), \( 2x - \pi \) approaches \( 0 \). Thus, we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi}}{\sqrt{9 + \sqrt{2x - \pi}} - b} \]

### Step 4: Rationalize the denominator
To simplify the limit, we will multiply the numerator and the denominator by the conjugate of the denominator:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi} \cdot (\sqrt{9 + \sqrt{2x - \pi}} + b)}{(\sqrt{9 + \sqrt{2x - \pi}} - b)(\sqrt{9 + \sqrt{2x - \pi}} + b)} \]

This simplifies to:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi} \cdot (\sqrt{9 + \sqrt{2x - \pi}} + b)}{9 + \sqrt{2x - \pi} - b^2} \]

### Step 5: Evaluate the limit
As \( x \to \frac{\pi}{2} \), \( \sqrt{2x - \pi} \to 0 \):

\[ = \frac{0 \cdot (\sqrt{9 + 0} + b)}{9 + 0 - b^2} = 0 \]

For the limit to equal \( a \), we need:

\[ a = 0 \]

### Step 6: Determine \( b \)
To ensure the limit exists, the denominator must not equal zero. Thus, we require:

\[ 9 - b^2 \neq 0 \implies b^2 \neq 9 \implies b \neq 3 \text{ and } b \neq -3 \]

### Step 7: Find values of \( a \) and \( b \)
We can choose \( b = 3 \) (or \( b = -3 \)) to simplify calculations. Let's take \( b = 3 \):

Substituting \( b = 3 \) into the limit:

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi} \cdot (\sqrt{9 + \sqrt{2x - \pi}} + 3)}{9 + \sqrt{2x - \pi} - 9} = \lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2x - \pi} \cdot (\sqrt{9 + \sqrt{2x - \pi}} + 3)}{\sqrt{2x - \pi}} \]

This cancels to:

\[ \sqrt{9 + 0} + 3 = 3 + 3 = 6 \]

Thus, \( a = 6 \).

### Step 8: Calculate \( \frac{a^2}{5b} \)
Now, we can find \( \frac{a^2}{5b} \):

\[ \frac{a^2}{5b} = \frac{6^2}{5 \cdot 3} = \frac{36}{15} = \frac{12}{5} = 2.4 \]

### Final Answer
Thus, the value of \( \frac{a^2}{5b} \) is:

\[ \boxed{2.4} \]