The values of a and b so that the function `f(x) = {{:(x + a sqrt(2) sin x",",0 le x lt pi//4),(2 x cot x + b",",pi//4 le x le pi//2),(a cos 2 x - b sin x",",pi//2 lt x le pi):}` is continuous for `x in [0, pi]`, are

Options

A. `a = (pi)/(6),b=-(pi)/(6)`
B. `a = -(pi)/(6), b = (pi)/(12)`
C. `a = (pi)/(6), b = -(pi)/(12)`
D. None of these

Answer

To find the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{for } 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b & \text{for } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

is continuous on the interval \([0, \pi]\), we need to check the continuity at the points \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{2} \).

### Step 1: Continuity at \( x = \frac{\pi}{4} \)

1. **Calculate \( f\left(\frac{\pi}{4}\right) \)**:
\[ f\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\pi}{4} \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b \]

2. **Calculate \( \lim_{x \to \frac{\pi}{4}^-} f(x) \)**:
\[ \lim_{x \to \frac{\pi}{4}^-} f(x) = \lim_{h \to 0} \left( \frac{\pi}{4} + a \sqrt{2} \sin\left(\frac{\pi}{4} - h\right) \right) \]
Since \( \sin\left(\frac{\pi}{4} - h\right) \approx \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) as \( h \to 0 \):
\[ = \frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} + a \]

3. **Calculate \( \lim_{x \to \frac{\pi}{4}^+} f(x) \)**:
\[ \lim_{x \to \frac{\pi}{4}^+} f(x) = \lim_{h \to 0} \left( 2\left(\frac{\pi}{4} + h\right) \cot\left(\frac{\pi}{4} + h\right) + b \right) \]
As \( h \to 0 \), \( \cot\left(\frac{\pi}{4} + h\right) \approx 1 \):
\[ = \frac{\pi}{2} + b \]

4. **Set the limits equal for continuity**:
\[ \frac{\pi}{2} + b = \frac{\pi}{4} + a \]
Rearranging gives us Equation (1):
\[ a - b = \frac{\pi}{4} \quad \text{(1)} \]

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

1. **Calculate \( f\left(\frac{\pi}{2}\right) \)**:
\[ f\left(\frac{\pi}{2}\right) = 2 \cdot \frac{\pi}{2} \cot\left(\frac{\pi}{2}\right) + b = b \quad \text{(since } \cot\left(\frac{\pi}{2}\right) = 0\text{)} \]

2. **Calculate \( \lim_{x \to \frac{\pi}{2}^-} f(x) \)**:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = b \]

3. **Calculate \( \lim_{x \to \frac{\pi}{2}^+} f(x) \)**:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{h \to 0} \left( a \cos(2(\frac{\pi}{2} + h)) - b \sin(\frac{\pi}{2} + h) \right) \]
Since \( \cos(\pi + 2h) \approx -1 \) and \( \sin(\frac{\pi}{2} + h) \approx 1 \):
\[ = -a - b \]

4. **Set the limits equal for continuity**:
\[ b = -a - b \]
Rearranging gives us Equation (2):
\[ a + 2b = 0 \quad \text{(2)} \]

### Step 3: Solve the system of equations

From Equation (1):
\[ a - b = \frac{\pi}{4} \]

From Equation (2):
\[ a + 2b = 0 \implies a = -2b \]

Substituting \( a = -2b \) into Equation (1):
\[ -2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12} \]

Now substituting back to find \( a \):
\[ a = -2\left(-\frac{\pi}{12}\right) = \frac{\pi}{6} \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{\pi}{6}, \quad b = -\frac{\pi}{12} \]