Question

If $xϵR$ and $m=\frac{x^2}{(x^4-2x^2+4)}$, then m lies in the interval

• A. $[0,\frac{1}{4}]$
• B. $[0,\frac{1}{3}]$
• C. $[0,\frac{1}{2}]$
• D. $[0,\frac{1}{5}]$

Answer 1

The correct option is C $[0,\frac{1}{2}]$

$m=\frac{x^2}{(x^2-1{)}^2+3}=+ivei.e.\ge 0$
Again $m=\frac{1}{x^2+\frac{4}{x^2}-2}=\frac{1}{{(x-\frac{2}{x})}^2+2}\le \frac{1}{2}$
$mϵ[0,\frac{1}{2}]$