If the function
`f(x)={(x+a^(2)sqrt(2)sinx",", 0 le x lt (pi)/(4)),(x cot x+b",",(pi)/(4) le x lt (pi)/(2)),(b sin 2x-a cos 2x",", (pi)/(2) le x le pi):}`
is continuous in the interval `[0,pi]` then the values of (a, b) are

Options

A. `(0,0)`
B. `(0,(1)/(2))`
C. `(0,1)`
D. `(-1,1)`

Explanation

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

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

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

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

For continuity at \( x = \frac{\pi}{4} \), we need:

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

\[ \lim_{x \to \frac{\pi}{4}^+} f(x) = \frac{\pi}{4} \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{4} \cdot 1 + b = \frac{\pi}{4} + b \]

Setting the two limits equal gives:

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

Thus, we have:

\[ a^2 = b \quad \text{(1)} \]

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

For continuity at \( x = \frac{\pi}{2} \), we need:

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting the two limits equal gives:

\[ b = a \quad \text{(2)} \]

### Step 3: Solving the equations

From equation (1) \( a^2 = b \) and equation (2) \( b = a \), we can substitute \( b \) from equation (2) into equation (1):

\[ a^2 = a \]

Rearranging gives:

\[ a^2 - a = 0 \]

Factoring out \( a \):

\[ a(a - 1) = 0 \]

Thus, \( a = 0 \) or \( a = 1 \).

### Step 4: Finding corresponding \( b \)

1. If \( a = 0 \):
\[ b = 0 \]
So one solution is \( (a, b) = (0, 0) \).

2. If \( a = 1 \):
\[ b = 1 \]
So another solution is \( (a, b) = (1, 1) \).

### Final Answer

The values of \( (a, b) \) are \( (0, 0) \) and \( (1, 1) \).

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