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 $2\sin x\cos x=\sin 2x$ is ' $1$'.
and minimum value of $\frac{2+\{x\}}{1-\{x\}}$ (at $\{x\}=0$ ) is '$2$ '
$\therefore$ For the given equation number of solution $k=0$
$\therefore \forall x\in \left[\frac{\pi }{4},\frac{\pi }{3}\right],K^{{\tan }^{2}x}=0$