The function `f(x) = {{:(x + asqrt(2) sin x, 0 le x lt pi//4),(2x cotx +6, pi//4 le x le pi//2),(a cos 2x-b sin x, pi//2 lt x le pi):}` is continuous for `0 le x le pi` then a, b are

Options

A. `pi/6, pi/12`
B. `pi/3, pi/6`
C. `pi/6, -pi/12`
D. none of these

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous over the interval \( [0, \pi] \). The function is defined piecewise, and we need to check the continuity at the points where the definition changes, specifically at \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{2} \).

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

We need to ensure that:
\[ \lim_{x \to \frac{\pi}{4}^-} f(x) = \lim_{x \to \frac{\pi}{4}^+} f(x) \]

**Left-hand limit:**
For \( 0 \leq x < \frac{\pi}{4} \):
\[ f(x) = x + a\sqrt{2} \sin x \]
Thus,
\[ \lim_{x \to \frac{\pi}{4}^-} f(x) = \frac{\pi}{4} + a\sqrt{2} \sin\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + a\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} + a \]

**Right-hand limit:**
For \( \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \):
\[ f(x) = 2x \cot x + 6 \]
Thus,
\[ \lim_{x \to \frac{\pi}{4}^+} f(x) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + 6 = \frac{\pi}{2} + 6 \]

Setting the two limits equal for continuity:
\[ \frac{\pi}{4} + a = \frac{\pi}{2} + 6 \]
Rearranging gives:
\[ a = \frac{\pi}{2} + 6 - \frac{\pi}{4} = \frac{\pi}{4} + 6 \]

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

We need to ensure that:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) \]

**Left-hand limit:**
For \( \frac{\pi}{4} < x < \frac{\pi}{2} \):
\[ f(x) = 2x \cot x + 6 \]
Thus,
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = 2\left(\frac{\pi}{2}\right) \cot\left(\frac{\pi}{2}\right) + 6 = 0 + 6 = 6 \]

**Right-hand limit:**
For \( \frac{\pi}{2} < x \leq \pi \):
\[ f(x) = a \cos(2x) - b \sin x \]
Thus,
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = a \cos(\pi) - b \sin\left(\frac{\pi}{2}\right) = -a - b \]

Setting the two limits equal for continuity:
\[ 6 = -a - b \]

### Step 3: Solve the equations

Now we have two equations:
1. \( a = \frac{\pi}{4} + 6 \)
2. \( 6 = -a - b \)

Substituting the first equation into the second:
\[ 6 = -\left(\frac{\pi}{4} + 6\right) - b \]
\[ 6 = -\frac{\pi}{4} - 6 - b \]
\[ b = -\frac{\pi}{4} - 12 \]

### Final Values
Now we can summarize our results:
- \( a = \frac{\pi}{4} + 6 \)
- \( b = -\frac{\pi}{4} - 12 \)