Question

The range of $y=\sin 3x+\sin x$ is-

• A. $[\frac{-8}{3\sqrt{3}},\frac{8}{3\sqrt{3}},]$
• B. $[\frac{-4}{3\sqrt{3}},\frac{4}{3\sqrt{3}},]$
• C. $[\frac{-3}{\sqrt{2}},\frac{3}{\sqrt{2}},]$
• D. $[\frac{-3}{3\sqrt{2}},\frac{3}{2\sqrt{2}},]$

Answer 1

The correct option is A $[\frac{-8}{3\sqrt{3}},\frac{8}{3\sqrt{3}},]$

Given

$y=\sin 3x+\sin x$

or, $y=4\sin x-4{\sin }^{3}x$

or, $y=4(\sin x-{\sin }^{3}x)$

To find the range of $y$, we will try to find the maximum and minimum value of $y$.

Now,

$\frac{dy}{dx}=4(\cos x-3{\sin }^{2}x\cos x)$

and

$\frac{d^{2}y}{d{x}^2}=4(-\sin x-6\sin x{\cos }^{2}x+3{\sin }^{3}x)$.

To find the maximum and minimum value of $y$ we must have $\frac{dy}{dx}=0$.

Then

$4(\cos x-3{\sin }^{2}x\cos x)=0.$

or,$\cos x=0$ and ${\sin }^{2}x=\frac{1}{3}$

Now, $\sin x=-\frac{1}{\sqrt{3}}$ gives the minimum value of the $y$ as ${\frac{d^{2}y}{d{x}^2}|}_{\sin x=-\frac{1}{\sqrt{3}}}$<0 and $\sin x=\frac{1}{\sqrt{3}}$ gives the minimum value of the $y$ as ${\frac{d^{2}y}{d{x}^2}|}_{\sin x=\frac{1}{\sqrt{3}}}$>0.

The minimum value of $y$ is $-\frac{8}{3\sqrt{3}}$ and themaximum value of $y$ is $\frac{8}{3\sqrt{3}}$.

So the range of $y$ is Option A.