Let f(x) be defined as
`f(x) = {{:(sin 2x, 0 lt x lt pi/6),(px + q, pi/6 lt x lt 1):}`, If f and f' are continuous then [p,q) is equal to

Options

A. `(1/sqrt(2), 1/sqrt(2))`
B. `(1,1/sqrt(2) + pi/6)`
C. `(1, 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 \( f(x) \) at \( x = \frac{\pi}{6} \)

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

To ensure continuity at \( x = \frac{\pi}{6} \), we set the left-hand limit equal to the right-hand limit:

\[ \lim_{x \to \left(\frac{\pi}{6}\right)^-} f(x) = \lim_{x \to \left(\frac{\pi}{6}\right)^+} f(x) \]

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:

\[ \lim_{x \to \left(\frac{\pi}{6}\right)^+} f(x) = p \cdot \frac{\pi}{6} + q \]

Setting these equal gives us the first equation:

\[ \frac{\sqrt{3}}{2} = p \cdot \frac{\pi}{6} + q \tag{1} \]

### Step 2: Ensure continuity of \( f'(x) \) at \( x = \frac{\pi}{6} \)

Next, we find the derivatives of both pieces of the function:

- For \( 0 < x < \frac{\pi}{6} \), \( f'(x) = 2 \cos(2x) \)
- For \( \frac{\pi}{6} < x < 1 \), \( f'(x) = p \)

To ensure continuity of \( f'(x) \) at \( x = \frac{\pi}{6} \), we set the left-hand derivative equal to the right-hand derivative:

\[ \lim_{x \to \left(\frac{\pi}{6}\right)^-} f'(x) = \lim_{x \to \left(\frac{\pi}{6}\right)^+} f'(x) \]

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 the second equation:

\[ 1 = p \tag{2} \]

### Step 3: Solve the equations

From equation (2), we have:

\[ p = 1 \]

Substituting \( p = 1 \) into equation (1):

\[ \frac{\sqrt{3}}{2} = 1 \cdot \frac{\pi}{6} + q \]

Rearranging gives:

\[ q = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \]

### Final Result

Thus, the values of \( p \) and \( q \) are:

\[ (p, q) = \left(1, \frac{\sqrt{3}}{2} - \frac{\pi}{6}\right) \]