Question

If $\frac{1}{|x^2-2|}\le 2$, then $x$ lies in the interval

• A. $(-\infty ,-\sqrt{\frac{5}{2}}]\cup [-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
• B. $(-\infty ,-\sqrt{\frac{5}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
• C. $[-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
• D. $(-\infty ,-\sqrt{\frac{5}{2}}]\cup (-\sqrt{2},\sqrt{2})$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(-\infty ,-\sqrt{\frac{5}{2}}]\cup [-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
B $(-\infty ,-\sqrt{\frac{5}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
C $[-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$

$\frac{1}{|x^2-2|}\le 2,\text{and}x\ne \pm \sqrt{2}$
$\Rightarrow |x^2-2|\ge \frac{1}{2}$
$\Rightarrow x^2-2\le -\frac{1}{2}\text{or}{x}^2-2\ge \frac{1}{2}$
$\Rightarrow x^2\le \frac{3}{2}\text{or}{x}^2\ge \frac{5}{2}$
$\Rightarrow x\in [-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\text{or}x\in (-\infty ,-\sqrt{\frac{5}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$
$\Rightarrow x\in (-\infty ,-\sqrt{\frac{5}{2}}]\cup [-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}]\cup [\sqrt{\frac{5}{2}},\infty )$