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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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Verified Answer
The correct answer is:
2
$f(x)=a \sin x+\frac{1}{3} \sin 3 x$
$$
\begin{aligned}
& \therefore f^{\prime}(x)=a \cos x+\frac{1}{3} \cos 3 x \cdot(3)\left\{\because \frac{d}{d x} \sin x=\cos x\right\} \\
& f^{\prime}(x)=a \cos x+\cos 3 x \\
&
\end{aligned}
$$
$\because$ At $x=\frac{\pi}{3}, f(x)$ is maximum.
$$
f^{\prime}(x)=0 \text {, at } x=\frac{\pi}{3}
$$
Hence, $a \cos x+\cos 3 x=0$
$$
\begin{gathered}
a \cos \frac{\pi}{3}+\cos \left(3 \cdot \frac{\pi}{3}\right)=0 \\
a \cdot \frac{1}{2}+(-1)=0 \\
\frac{a}{2}=1 \Rightarrow a=2
\end{gathered}
$$
$$
\begin{aligned}
& \therefore f^{\prime}(x)=a \cos x+\frac{1}{3} \cos 3 x \cdot(3)\left\{\because \frac{d}{d x} \sin x=\cos x\right\} \\
& f^{\prime}(x)=a \cos x+\cos 3 x \\
&
\end{aligned}
$$
$\because$ At $x=\frac{\pi}{3}, f(x)$ is maximum.
$$
f^{\prime}(x)=0 \text {, at } x=\frac{\pi}{3}
$$
Hence, $a \cos x+\cos 3 x=0$
$$
\begin{gathered}
a \cos \frac{\pi}{3}+\cos \left(3 \cdot \frac{\pi}{3}\right)=0 \\
a \cdot \frac{1}{2}+(-1)=0 \\
\frac{a}{2}=1 \Rightarrow a=2
\end{gathered}
$$
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