Determine the constants 'a' and 'b' so that the function 'f' defined below is continuous everywhere :
`f(x)={{:(5a", "xle0),(asinx+cosx", "0ltxltpi/2),(b-pi/2", "xgepi/2):}`.

Answer

To determine the constants 'a' and 'b' so that the function \( f \) is continuous everywhere, we need to ensure that the function is continuous at the points where the definition of the function changes, specifically at \( x = 0 \) and \( x = \frac{\pi}{2} \).

The function is defined as follows:
\[ f(x) = \begin{cases} 5a & \text{if } x \leq 0 \\ a \sin x + \cos x & \text{if } 0 < x < \frac{\pi}{2} \\ b - \frac{\pi}{2} & \text{if } x \geq \frac{\pi}{2} \end{cases} \]

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

For \( f \) to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Calculating these limits:
- \( f(0) = 5a \)
- \( \lim_{x \to 0^+} f(x) = a \sin(0) + \cos(0) = 0 + 1 = 1 \)

Setting these equal for continuity:
\[ 5a = 1 \]
Solving for \( a \):
\[ a = \frac{1}{5} \]

### Step 2: Check continuity at \( x = \frac{\pi}{2} \)

For \( f \) to be continuous at \( x = \frac{\pi}{2} \), we need:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = f\left(\frac{\pi}{2}\right) \]

Calculating these limits:
- \( f\left(\frac{\pi}{2}\right) = b - \frac{\pi}{2} \)
- \( \lim_{x \to \frac{\pi}{2}^-} f(x) = a \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = a \cdot 1 + 0 = a \)

Setting these equal for continuity:
\[ a = b - \frac{\pi}{2} \]

Substituting the value of \( a \):
\[ \frac{1}{5} = b - \frac{\pi}{2} \]

Solving for \( b \):
\[ b = \frac{1}{5} + \frac{\pi}{2} \]

### Final Values
Thus, the constants are:
\[ a = \frac{1}{5}, \quad b = \frac{1}{5} + \frac{\pi}{2} \]