Find the values of a and b so that the function
`f(x)={{:(x+asqrt2 sin x", "0 le x le pi//4),(2x cot x + b", "pi//4 le x le pi//2),(a cot 2x - b sin x", "pi//2 lt x le pi):}`
is continuous for ` 0 le x le pi`.

Explanation

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

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

is continuous for \( 0 \leq x \leq \pi \), we need to ensure that the function is continuous at the points where the definition changes, which are \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{2} \).

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

To ensure continuity at \( x = \frac{\pi}{4} \), we equate the left-hand limit and the right-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{4}\right)^-} f(x) = \lim_{x \to \left(\frac{\pi}{4}\right)^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{4}\right)^-} f(x) = \frac{\pi}{4} + a\sqrt{2} \sin\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + a\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} + a \]

Calculating the right-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{4}\right)^+} f(x) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b \]

Setting the two limits equal gives us the first equation:

\[ \frac{\pi}{4} + a = \frac{\pi}{2} + b \]

Rearranging this, we have:

\[ a - b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \quad \text{(Equation 1)} \]

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

Next, we ensure continuity at \( x = \frac{\pi}{2} \):

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

Calculating the left-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{2}\right)^-} f(x) = 2\left(\frac{\pi}{2}\right) \cot\left(\frac{\pi}{2}\right) + b = 0 + b = b \]

Calculating the right-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{2}\right)^+} f(x) = a \cot(2 \cdot \frac{\pi}{2}) - b \sin\left(\frac{\pi}{2}\right) = a \cdot 0 - b \cdot 1 = -b \]

Setting the two limits equal gives us the second equation:

\[ b = -b \]

This implies:

\[ 2b = 0 \quad \Rightarrow \quad b = 0 \quad \text{(Equation 2)} \]

### Step 3: Substitute \( b \) into Equation 1

Substituting \( b = 0 \) into Equation 1:

\[ a - 0 = \frac{\pi}{4} \quad \Rightarrow \quad a = \frac{\pi}{4} \]

### Final Values

Thus, the values of \( a \) and \( b \) that make the function continuous are:

\[ \boxed{a = \frac{\pi}{4}, \quad b = 0} \]