Question

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

Answer 1

Given :$\sqrt{x^{{\log }_2\sqrt{x}}}\ge 2\Rightarrow x^{{\log }_2\sqrt{x}}\ge 4Taking\log bothsideswithbase2,weget\Rightarrow \left({\log }_2\sqrt{x}\right)\left({\log }_{2}x\right)\ge {\log }_{2}4\Rightarrow \frac{1}{2}\left({\log }_{2}x\right)\left({\log }_{2}x\right)\ge 2\Rightarrow {\left({\log }_{2}x\right)}^2\ge 4\Rightarrow \left(\left({\log }_{2}x\right)-2\right]\left(\left({\log }_{2}x\right)+2\right]\ge 0\Rightarrow {\log }_{2}x\le -2or{\log }_{2}x\ge 2\Rightarrow x\le 2^{-2}orx\ge 2^2\Rightarrow x\le \frac{1}{4}orx\ge 4\Rightarrow x\in (0,\frac{1}{4}]\cup [4,\infty )$Hence, The option B and C are correct answer