If the function

Question:

If the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$ has an extremum at $x=\frac{\pi}{3}$ then $a=$_____________

Solution:

It is given that, the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$ has an extremum at $x=\frac{\pi}{3}$.

$\therefore f^{\prime}(x)=0$ at $x=\frac{\pi}{3}$

$f(x)=a \sin x+\frac{1}{3} \sin 3 x$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=a \cos x+\frac{1}{3} \times 3 \cos 3 x$

$\Rightarrow f^{\prime}(x)=a \cos x+\cos 3 x$

Now,

$f^{\prime}\left(\frac{\pi}{3}\right)=0$

$\Rightarrow a \cos \left(\frac{\pi}{3}\right)+\cos 3\left(\frac{\pi}{3}\right)=0$

$\Rightarrow a \times \frac{1}{2}+\cos \pi=0$

$\Rightarrow \frac{a}{2}-1=0$

$\Rightarrow a=2$

Thus, the value of a is 2.

If the function $f(x)=a \sin x+\frac{1}{3} \sin 3 x$ has an extremum at $x=\frac{\pi}{3}$ then $a=$  ___2___.

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