If the function f(x) = a sin(x) + frac{1}{3} | vedclass

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(C) Given function: $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$.First,find the derivative $f'(x)$:$f'(x) = a \cos(x) + \frac{1}{3} \cos(3x) \cdot 3 = a

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$2$

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(C) Given function: $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$.First,find the derivative $f'(x)$:$f'(x) = a \cos(x) + \frac{1}{3} \cos(3x) \cdot 3 = a \cos(x) + \cos(3x)$.Since the function attains its maximum at $x = \frac{\pi}{3}$,the first derivative must be zero at this point:$f'\left(\frac{\pi}{3}\right) = 0$.Substituting $x = \frac{\pi}{3}$ into the derivative:$a \cos\left(\frac{\pi}{3}\right) + \cos\left(3 \cdot \frac{\pi}{3}\right) = 0$.$a \cdot \frac{1}{2} + \cos(\pi) = 0$.Since $\cos(\pi) = -1$,we have:$\frac{a}{2} - 1 = 0$.$\frac{a}{2} = 1$.$a = 2$.

If the function $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$ attains its maximum value at $x = \frac{\pi}{3}$,then $a$ equals:

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