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

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Aicharya · Accepted Answer

$f(x)=a \sin x+\frac{1}{3} \sin 3 x$ $ \begin{aligned} & 	herefore f^{\prime}(x)=a \cos x+\frac{1}{3} \cos 3 x \cdot(3)\left\{\because \frac{d}{d x} \sin x=\cos xight\} \ & f^{\prime}(x)=a \cos x+\cos 3 x \ & \end{aligned} $ $\because$ At $x=\frac{\pi}{3}, f(x)$ is maximum. $ f^{\prime}(x)=0 	ext {, at } x=\frac{\pi}{3} $ Hence, $a \cos x+\cos 3 x=0$ $ \begin{gathered} a \cos \frac{\pi}{3}+\cos \left(3 \cdot \frac{\pi}{3}ight)=0 \ a \cdot \frac{1}{2}+(-1)=0 \ \frac{a}{2}=1 \Rightarrow a=2 \end{gathered} $

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

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