Question

Let $[x]$ denote the largest integer $\le$ x. If the number of solutions of $\sin x \sqrt{4 \cos^2 x} = \frac{ 2 +x - [x]}{ 1 -x + [x]}$ is $k$, then for $x \in \left[ \frac{\pi}{4} , \frac{\pi}{3} \right]$, the value of $k^{\tan^2 x}$

• A. is equal to $1$
• B. lies in between $2^1$ and $2^3$
• C. is equal to zero
• D. lies in between $\frac{1}{2^3} $ and $\frac{1}{2}$

Answer 1

Answer: (C)

Given, $\sin\, x \sqrt{4 \cos ^{2} x}=\frac{2+x-[x]}{1-x+[x]}$
$\Rightarrow \sin\, x \sqrt{4 \cos ^{2} x}=\frac{2+\{x\}}{1-\{x\}} $
$[\because x-[x]=\{x\}]$
$\Rightarrow 2\, \sin \,x|\cos \,x|=\frac{2+\{x\}}{1-\{x\}}=y($ let $)$
$\because y=\frac{2+\{x\}}{1-\{x\}} > 0$
[because $\{x\} \in[0,1)$, so $2+\{x\}$ and $1-\{x\}$ are positive $]$
$\therefore |\cos\, x|=\cos \,x, \,\,\forall x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$
So, $2 \,\sin\, x \cos \,x=\frac{2+\{x\}}{1-\{x\}}$
$\because$ Maximum value of $2\, \sin \,x \,\cos\, x=\sin \,2 x$ is ' $1$'.
and minimum value of $\frac{2+\{x\}}{1-\{x\}}$ (at $\{x\}=0$ ) is '$2$ '
$\therefore $ For the given equation number of solution $k=0$
$\therefore \forall x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right], K^{\tan ^{2} x}=0$