Question

If $A=\frac{2}{3\sin x-4\cos x+7},$ then $A$ lies in the interval

• A. $[\frac{1}{6},1]$
• B. $(\frac{1}{6},1)$
• C. $[\frac{1}{12},\frac{1}{2}]$
• D. $(\frac{1}{12},\frac{1}{2})$

Answer 1

The correct option is A $[\frac{1}{6},1]$

$-\sqrt{3^2+(-4{)}^2}\le 3\sin x-4\cos x\le \sqrt{3^2+(-4{)}^2}$
$\Rightarrow -5\le 3\sin x-4\cos x\le 5$
$\Rightarrow 2\le 3\sin x-4\cos x+7\le 12$
$\Rightarrow \frac{1}{12}\le \frac{1}{3\sin x-4\cos x+7}\le \frac{1}{2}$
$\Rightarrow \frac{1}{6}\le \frac{2}{3\sin x-4\cos x+7}\le 1$
$\text{i.e.,}\frac{1}{6}\le A\le 1$