Question

The range of the function $f(x)=\frac{1}{{\sin }^{2}x-2\sin x}$ is

• A. $\left[-1,\frac{1}{3}\right]$
• B. $(-\infty ,-1]\cup \left(\frac{1}{3},\infty \right)$
• C. $(-\infty ,-1)\cup \left[\frac{1}{3},\infty \right)$
• D. $(-\infty ,-1]\cup \left[\frac{1}{3},\infty \right)$

Answer 1

Answer: (D)

$f(x)=\frac{1}{(\sin x-1{)}^2-1}$
$(\sin x-1{)}^2-1\in [-1,3]$
$\therefore R_f=(-\infty ,-1]\cup \left[\frac{1}{3},\infty \right)$