Question

The range of $f\left(x\right)=\frac{x^2}{x^4+1}$ is

• A. $(0,\frac{1}{2})$
• B. $[0,\frac{1}{2}]$
• C. $[0,\infty )$
• D. $[0,2]$

Answer 1

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

Domain of $f$ is $R$
Clearly, $f\left(x\right)\ge 0\forall x\in R$
We know that $(x^2-1{)}^2\ge 0$
$\Rightarrow x^4-2x^2+1\ge 0\Rightarrow x^4+1\ge 2x^2\Rightarrow \frac{x^2}{x^4+1}\le \frac{1}{2}$
$\therefore \frac{x^2}{x^4+1}\in [0,\frac{1}{2}]$

Alternate solution :
$f\left(x\right)=\frac{x^2}{x^4+1}$
$\Rightarrow f\left(x\right)=\frac{1}{x^2+\frac{1}{x^2}}$ for $x\ne 0$
We know that, $x^2+\frac{1}{x^2}\ge 2$
For $x=0,f\left(x\right)=0$
By observation, $f\left(x\right)\ge 0\forall x\in R$
Therefore, $f\left(x\right)\in [0,\frac{1}{2}]$