Question

The inequality of $\sqrt{x^{{\log }_2\sqrt{x}}}\ge 2$ is satisfied by

• A. only one value of $x$
• B. $x\in (0,\frac{1}{4}]$
• C. $x\in [4,\infty )$
• D. $x\in (1,2)$

Answer 1

Answer: (B), (C)

$x\in (0,\frac{1}{4}]$
$x\in [4,\infty )$

$\sqrt{x^{{\log }_2\sqrt{x}}}\ge 2$, where, $x>0$ ...(1)
$\Rightarrow x^{{\log }_2\sqrt{x}}\ge 4$
$\Rightarrow {\log }_{2}x{\log }_2\sqrt{x}\ge {\log }_{2}4$
$\Rightarrow {\log }_{2}x\left(\frac{1}{2}{\log }_{2}x\right)\ge 2[\because \log a^m=m\log a,{\log }_{a}a=1]$
$\Rightarrow {\left({\log }_{2}x\right)}^2\ge 4$
$\Rightarrow {\log }_{2}x\ge 2or{\log }_{2}x\le -2$

$\Rightarrow x\ge 4orx\le \frac{1}{4}$
$\Rightarrow x\in (-\infty ,\frac{1}{4}]\cup [4,+\infty )$ ...(2)
From (1) & (2)
$x\in (0,\frac{1}{4}]\cup [4,+\infty )$
Ans: B,C