What is the value of p for which the function f(x) = `p sin x +1/3 sin 3x` has an extreme at `x = pi/3` ?

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To find the value of \( p \) for which the function \( f(x) = p \sin x + \frac{1}{3} \sin 3x \) has an extreme at \( x = \frac{\pi}{3} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(p \sin x) + \frac{d}{dx}\left(\frac{1}{3} \sin 3x\right) \]

Using the derivative of \( \sin x \) which is \( \cos x \) and applying the chain rule for \( \sin 3x \):

\[ f'(x) = p \cos x + \frac{1}{3} \cdot 3 \cos 3x = p \cos x + \cos 3x \]

### Step 2: Set the derivative to zero at \( x = \frac{\pi}{3} \)
For the function to have an extreme at \( x = \frac{\pi}{3} \), we need to set the derivative equal to zero:

\[ f'\left(\frac{\pi}{3}\right) = 0 \]

Substituting \( x = \frac{\pi}{3} \):

\[ p \cos\left(\frac{\pi}{3}\right) + \cos\left(3 \cdot \frac{\pi}{3}\right) = 0 \]

### Step 3: Calculate the cosine values
Now, we calculate the cosine values:

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
\[ \cos\left(\pi\right) = -1 \]

Substituting these values into the equation:

\[ p \cdot \frac{1}{2} + (-1) = 0 \]

### Step 4: Solve for \( p \)
Now, we solve for \( p \):

\[ \frac{p}{2} - 1 = 0 \]

Adding 1 to both sides:

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

Multiplying both sides by 2:

\[ p = 2 \]

### Conclusion
The value of \( p \) for which the function \( f(x) \) has an extreme at \( x = \frac{\pi}{3} \) is:

\[ \boxed{2} \]