Question

The range of the function $f\left(x\right)=\frac{1}{3\sin x+4\cos x-2}$ is

• A. $R-\left\{0\right\}$
• B. $[-\frac{1}{7},\frac{1}{3}]$
• C. $R-[-\frac{1}{7},\frac{1}{3}]$
• D. $R-(-\frac{1}{7},\frac{1}{3})$

Answer 1

The correct option is D $R-(-\frac{1}{7},\frac{1}{3})$

Let $g\left(x\right)=3\sin x+4\cos x-2$
We know that,
$-\sqrt{3^2+4^2}\le 3\sin x+4\cos x\le \sqrt{3^2+4^2}$
$\Rightarrow -5\le 3\sin x+4\cos x\le 5$
$\Rightarrow -7\le 3\sin x+4\cos x-2\le 3$
$\Rightarrow g\left(x\right)\in [-7,3]$

$\therefore \frac{1}{g\left(x\right)}\in (-\infty ,-\frac{1}{7}]$ for $g\left(x\right)\in [-7,0)$
and $\frac{1}{g\left(x\right)}\in [\frac{1}{3},\infty )$ for $g\left(x\right)\in (0,3]$

Therefore, the range of $f\left(x\right)=R-(-\frac{1}{7},\frac{1}{3})$