The value of ' $a$ ' for which the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$ has an extremum value at $x=\frac{\pi}{3}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } \because f(x)=a \sin x+\frac{1}{3} \sin 3 x \ & f(x)=a \cos x+\frac{1}{3} 	imes 3 \cos 3 x \ & \Rightarrow f(x)=a \cos x+\cos 3 x \ & \because f(x) 	ext { has an extremum at } x=\frac{\pi}{3} \ & 	herefore a x \cos \left(\frac{\pi}{3}ight)+\cos (\pi)=0 \ & \frac{a}{2}-1=0 \Rightarrow \frac{a}{2}=1 \Rightarrow a=2\end{aligned}$

The value of ' $a$ ' for which the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$ has an extremum value at…

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