Question

The expression $\frac{36}{4\sin (x+\frac{\pi }{3})-4\sqrt{3}\cos x+8}$ lies in the interval

• A. $[4,12]$
• B. $[-4,4]$
• C. $(4,12)$
• D. $[3,9]$

Answer 1

The correct option is D $[3,9]$

$\sin (x+\frac{\pi }{3})=\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x$
$4\sin (x+\frac{\pi }{3})-4\sqrt{3}\cos x+8=2\sin x-2\sqrt{3}\cos x+8$
Since,
$2\sin x-2\sqrt{3}\cos x\in [-\sqrt{2^2+(-2\sqrt{3}{)}^2},\sqrt{2^2+(-2\sqrt{3}{)}^2}]$
$\Rightarrow 2\sin x-2\sqrt{3}\cos x\in [-4,4]$
$\Rightarrow 2\sin x-2\sqrt{3}\cos x+8\in [4,12]$
$\Rightarrow \frac{36}{4\sin (x+\frac{\pi }{3})-4\sqrt{3}\cos x+8}\in [3,9]$