Question

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

Answer 1

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

$\therefore f\text{'}\left(x\right)=0$ at x = $\frac{\mathrm{\pi }}{3}$

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=a\cos x+\frac{1}{3}\times 3\cos 3x$

$\Rightarrow f\text{'}\left(x\right)=a\cos x+\cos 3x$

Now,

$f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=0$

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

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

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

$\Rightarrow a=2$

Thus, the value of a is 2.


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