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Question: Answered & Verified by Expert
Let $[x]$ denote the largest integer $\leq x$. 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 $k^{\tan ^2 x}$
MathematicsFunctionsAP EAMCETAP EAMCET 2019 (21 Apr Shift 1)
Options:
  • 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 $\frac{1}{2}$
Solution:
1048 Upvotes Verified Answer
The correct answer is: is equal to zero
Given, $\sin x \sqrt{4 \cos ^2 x}=\frac{2+x-[x]}{1-x+[x]}$
$$
\begin{aligned}
& \Rightarrow \quad \sin x \sqrt{4 \cos ^2 x}=\frac{2+\{x\}}{1-\{x\}} \quad[\because x-[x]=\{x\}] \\
& \Rightarrow \quad 2 \sin x|\cos x|=\frac{2+\{x\}}{1-\{x\}}=y(\text { let }) \\
& \because \quad y=\frac{2+\{x\}}{1-\{x\}}>0
\end{aligned}
$$
[because $\{x\} \in[0,1)$, so $2+\{x\}$ and $1-\{x\}$ are positive $]$
$$
\therefore \quad|\cos x|=\cos x, \quad \forall x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]
$$

So, $\quad 2 \sin x \cos x=\frac{2+\{x\}}{1-\{x\}}$
$\because$ 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 $k=0$
$$
\therefore \forall x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right], K^{\tan ^2 x}=0
$$

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