Question

The variable x satisfying the equation $\left|\sin x\cos x\right|+\sqrt{2+{\tan }^{2}x+{\cot }^{2}x}=\sqrt{3}$ belongs to the interval

• A. $[0,\frac{\pi }{3}]$
• B. $\frac{\pi }{3},\frac{\pi }{2}1$
• C. $\frac{3\pi }{4},\pi$
• D. non-existent

Answer 1

Answer: (D)

non-existent

$\sqrt{3}=\left|\sin x\cos x\right|+\sqrt{2+{\tan }^{2}x+{\cot }^{2}x}\Rightarrow \sqrt{3}=\left|\sin x\cos x\right|+\sqrt{{\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x}\Rightarrow \sqrt{3}=\left|\sin x\cos x\right|+\sqrt{{(\tan x+\cot x)}^2}$
$\Rightarrow \sqrt{3}=\left|\sin x\cos x\right|+\left|(\tan x+\cot x)\right|=\left|\sin x\cos x\right|+\left|\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right|$
$\Rightarrow \sqrt{3}=\left|\sin x\cos x\right|+\frac{1}{\left|\sin x\cos x\right|}\ge 2$ ...$\{\because AM\ge GM\}$
$\because \sqrt{3}\ge 2$ is not possible.
Therefore x have no solution.
Ans: D