Let `f(x) ={(sin2x , if 0leqxleqpi/6),(ax+b , if pi/6ltxlt1):}` If `f(x)` and `f'(x)` are continuous then `a` & `b` are (A) `a=1,b=(1)/(sqrt( 2)) +(pi)/(6)` (B) `a=(1) /( sqrt(2)),b=(1)/(sqrt(2))` (C) `a=1,b=(sqrt(3))/(2)-(pi)/(6)` (D) None of these

Options

A. `a=1,b=(1)/(sqrt( 2)) +(pi)/(6)`
B. `a=(1) /( sqrt(2)),b=(1)/(sqrt(2))`
C. `a=1,b=(sqrt(3))/(2)-(pi)/(6)`
D. none of these

Answer

To solve the problem, we need to ensure that the function \( f(x) \) and its derivative \( f'(x) \) are continuous at the point \( x = \frac{\pi}{6} \).

### Step 1: Ensure Continuity of \( f(x) \)

The function \( f(x) \) is defined piecewise:
- For \( 0 \leq x \leq \frac{\pi}{6} \), \( f(x) = \sin(2x) \)
- For \( \frac{\pi}{6} < x < 1 \), \( f(x) = ax + b \)

To ensure continuity at \( x = \frac{\pi}{6} \), we need:
\[ \lim_{x \to \left(\frac{\pi}{6}\right)^-} f(x) = \lim_{x \to \left(\frac{\pi}{6}\right)^+} f(x) \]

Calculating the left-hand limit:
\[ \lim_{x \to \left(\frac{\pi}{6}\right)^-} f(x) = \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Calculating the right-hand limit:
\[ \lim_{x \to \left(\frac{\pi}{6}\right)^+} f(x) = a\left(\frac{\pi}{6}\right) + b \]

Setting these equal gives us our first equation:
\[ \frac{\sqrt{3}}{2} = a\left(\frac{\pi}{6}\right) + b \quad \text{(1)} \]

### Step 2: Ensure Continuity of \( f'(x) \)

Next, we need to ensure that the derivatives are also continuous at \( x = \frac{\pi}{6} \).

Calculating \( f'(x) \):
- For \( 0 \leq x < \frac{\pi}{6} \), \( f'(x) = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \)
- For \( \frac{\pi}{6} < x < 1 \), \( f'(x) = a \)

Calculating the left-hand derivative:
\[ \lim_{x \to \left(\frac{\pi}{6}\right)^-} f'(x) = 2\cos\left(2 \cdot \frac{\pi}{6}\right) = 2\cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \]

Setting this equal to the right-hand derivative gives us our second equation:
\[ 1 = a \quad \text{(2)} \]

### Step 3: Solve the Equations

From equation (2), we have:
\[ a = 1 \]

Substituting \( a = 1 \) into equation (1):
\[ \frac{\sqrt{3}}{2} = 1 \cdot \left(\frac{\pi}{6}\right) + b \]
\[ b = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \]

### Final Values

Thus, we have:
\[ a = 1, \quad b = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \]

### Conclusion

The correct answer is:
**(C) \( a = 1, b = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \)**