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

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

is continuous for \( x \in [0, \pi] \), we need to ensure continuity at the points where the function changes its definition, specifically at \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{2} \).

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

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

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting the two limits equal for continuity:

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

Rearranging gives us our first equation:

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

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

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

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting the two limits equal for continuity:

\[ b = -a - b \]

Rearranging gives us our second equation:

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

### Step 3: Solving the equations

Now we have the system of equations:

1. \( a - b = \frac{\pi}{4} \)
2. \( a + 2b = 0 \)

From Equation 2, we can express \( a \) in terms of \( b \):

\[ a = -2b \]

Substituting this into Equation 1:

\[ -2b - b = \frac{\pi}{4} \]

This simplifies to:

\[ -3b = \frac{\pi}{4} \]

Thus,

\[ b = -\frac{\pi}{12} \]

Now substituting \( b \) back into the expression for \( a \):

\[ a = -2\left(-\frac{\pi}{12}\right) = \frac{\pi}{6} \]

### Final Values

The values of \( a \) and \( b \) that ensure the function is continuous are:

\[ a = \frac{\pi}{6}, \quad b = -\frac{\pi}{12} \]