Question

The maximum value of $\cos x\left\{\sin x+\sqrt{{\sin }^{2}x+{\sin }^2\frac{\pi }{6}}\right\}$ is

• A. $\frac{\sqrt{5}}{3}$
• B. $\frac{\sqrt{5}}{2}$
• C. $\sqrt{\frac{5}{2}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (B)

Let $y=\cos x\left\{\sin x+\sqrt{{\sin }^{2}x+{\sin }^2\frac{\pi }{6}}\right\}$ is
$\Rightarrow y^2\left(1+{\tan }^{2}x\right)-2y\tan x=\frac{1}{4}$
$\Rightarrow y^2{\tan }^{2}x-2y\tan x+\left(y^2-\frac{1}{4}\right)\ge 0$
For real $\tan x$,
disc. $\ge 0$
$\Rightarrow 4y^2-4y^2\left(y^2-\frac{1}{4}\right)\ge 0$
$\Rightarrow 0\le y^2\le \frac{5}{4}$
$\therefore$ Maximum value of $y=\frac{\sqrt{5}}{2}$