If the function `f(x) = a sin x + 1/3 sin 3x ` is maximum at `x= pi/3 ` a=?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( a \) such that the function \( f(x) = a \sin x + \frac{1}{3} \sin 3x \) has a maximum at \( x = \frac{\pi}{3} \).

### Step 1: Differentiate the function
First, we need to find the derivative of the function \( f(x) \).

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

Using the derivative rules, we have:

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

This simplifies to:

\[ f'(x) = a \cos x + \cos 3x \]

### Step 2: Set the derivative to zero at \( x = \frac{\pi}{3} \)
Since the function is maximum at \( x = \frac{\pi}{3} \), we set the derivative equal to zero at this point:

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

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

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

### Step 3: Calculate the cosine values
We know:

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

Substituting these values into the equation gives:

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

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

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

Adding 1 to both sides:

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

Multiplying both sides by 2:

\[ a = 2 \]

### Conclusion
Thus, the value of \( a \) is \( 2 \).

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