Explanation

(c)
Given, $f(x) = a\sin x + \frac{1}{3}\sin{3x}$ has minimum value at $x = \frac{\pi}{3}$, then $f^{'}(x) = 0$ at $x = \frac{\pi}{3}$
$\Rightarrow \ \ f^{'}(x) = a\cos x + \cos{3x}$
$\Rightarrow f^{'}\left( \frac{\pi}{3} \right) = a\cos\frac{\pi}{3} + \cos\left( 3 \times \frac{\pi}{3} \right) = 0$
$\Rightarrow \ \ \ a \times \frac{1}{2} + \cos{(\pi)} = 0$
$\Rightarrow \ \frac{a}{2} - 1 = 0\ \Rightarrow \ a = 2$