If ` f(x)={(x+a sqrt(2) sinx"," ,0 lt x lt (pi)/(4)),(2x cotx+b",",(pi)/(4) le x le (pi)/(2)),(a cos 2x-b sinx",", (pi)/(2) lt x le pi):}` is continuous at `x=(pi)/(4)`, then a - b is equal to

Options

A. `(pi)/(2)`
B. 0
C. `(1)/(4)`
D. `(pi)/(4)`

Explanation

To determine the value of \( a - b \) for the function \( f(x) \) to be continuous at \( x = \frac{\pi}{4} \), we need to ensure that the left-hand limit (LHL) and the right-hand limit (RHL) at this point are equal.

### Step-by-Step Solution:

1. **Define the function segments**:
\[ f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{for } 0 < x < \frac{\pi}{4} \\ 2x \cot x + b & \text{for } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

2. **Calculate the left-hand limit as \( x \) approaches \( \frac{\pi}{4} \)**:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{4}^-} f(x) = \frac{\pi}{4} + a \sqrt{2} \sin\left(\frac{\pi}{4}\right) \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \):
\[ \text{LHL} = \frac{\pi}{4} + a \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{\pi}{4} + a \]

3. **Calculate the right-hand limit as \( x \) approaches \( \frac{\pi}{4} \)**:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{4}^+} f(x) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b \]
Since \( \cot\left(\frac{\pi}{4}\right) = 1 \):
\[ \text{RHL} = 2 \cdot \frac{\pi}{4} \cdot 1 + b = \frac{\pi}{2} + b \]

4. **Set LHL equal to RHL for continuity**:
\[ \frac{\pi}{4} + a = \frac{\pi}{2} + b \]

5. **Rearranging the equation**:
\[ a - b = \frac{\pi}{2} - \frac{\pi}{4} \]
\[ a - b = \frac{\pi}{4} \]

6. **Final answer**:
\[ a - b = \frac{\pi}{4} \]