Question

If $f\left(x\right)=\frac{sin3x}{sinx},$ where $x\ne n\pi ,$ then the range of values of $f\left(x\right)$ 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 $[-1,3)$

$f\left(x\right)=\frac{sin3x}{sinx}=\frac{3sinx-4si{n}^{3}x}{sinx}=3-4si{n}^{2}x$
now we know that for all $xϵR-n\pi$, $0<si{n}^{2}x\le 1$
Therefore range of $f\left(x\right)$ is, [$-1,3)$
Hence, option 'D' is correct.