Question

If $x\in [-4,-1]$,
then $\frac{1}{x^2}$ belongs to

• A. $(-\infty ,\frac{1}{16}]\cup [1,\infty )$
• B. $[1,16]$
• C. $[\frac{1}{16},1]$
• D. $(-\infty ,1]\cup [16,\infty )$

Answer 1

The correct option is C $[\frac{1}{16},1]$

Here, $x\in [-4,-1]$
$\Rightarrow x^2\in [1,16]$
Since, $\left(1\right)\left(16\right)>0$
So, $\frac{1}{x^2}\in [\frac{1}{16},1]$