Question

If $f\left(x\right)=\left(\frac{\sin 3x}{\sin x}\right),x\ne n\pi$, then the range of values of f(x) for real values of x is

• A. $[-1,3)$
• B. $(-\infty ,-1]$
• C. $(3,+\infty )$
• D. $[-1,3]$

Answer 1

The correct option is D $[-1,3]$

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

$=\frac{3\sin x-4{\sin }^{3}x}{\sin x}$

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

$=3-4{\sin }^{2}x$
Substitute $3-4{\sin }^{2}x=y$
$\Rightarrow {\sin }^{2}x=\frac{3-y}{4}$
Since, $0\le {\sin }^{2}x\le 1$
$\Rightarrow 0\le \frac{3-y}{4}\le 1$
$\Rightarrow 0\le 3-y\le 4$
$\Rightarrow -3\le -y\le 1$
$\Rightarrow -1\le y\le 3$
$\Rightarrow y\in [-1,3]$