Question

If the function \( f(x) = \mu \sin x + \frac{1}{3} \sin 3x \) has its derivative equal to zero at \( x = \frac{\pi}{3} \), then the value of \( \mu \) is __________.

• A. \( 0 \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

While differentiating \( \sin(ax) \), remember derivative becomes \( a\cos(ax) \)This cancels coefficients nicely in such problems.

Answer 1

The Correct Option is D

Step 1: Differentiate the function.
\[ f(x) = \mu \sin x + \frac{1}{3}\sin 3x \]
\[ f'(x) = \mu \cos x + \frac{1}{3} \cdot 3\cos 3x \]
\[ f'(x) = \mu \cos x + \cos 3x \]

Step 2: Use the condition \( f'(x)=0 \).
\[ \mu \cos x + \cos 3x = 0 \]

Step 3: Substitute \( x = \frac{\pi}{3} \).
\[ \mu \cos \frac{\pi}{3} + \cos \pi = 0 \]

Step 4: Use standard values.
\[ \cos \frac{\pi}{3} = \frac{1}{2}, \quad \cos \pi = -1 \]

Step 5: Substitute values.
\[ \mu \cdot \frac{1}{2} - 1 = 0 \]

Step 6: Solve for \( \mu \).
\[ \frac{\mu}{2} = 1 \]
\[ \mu = 2 \]

Step 7: Final conclusion.
\[ \boxed{2} \]