Question

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

• A. [-1, 3]
• B. (-, -1)
• C. (3, )
• D. [-1, 3)

Answer 1

The correct option is D [-1, 3)

3 - 4 sin${}^2$x = y
$\therefore$${\sin }^{2}x=\frac{3-y}{4}$. But 0 $<si{n}^{2}x\le 1(\because sinx=0\Rightarrow x=n\pi )$
$\therefore$$0<\frac{3-y}{4}\le 1$ or 0 $<$ 3 - y $\le$ 4.