The range of the function $f(x)=x^2+\frac{1}{x^2+1}$ is
The range of the function $f(x)=x^2+\frac{1}{x^2+1}$ is
- $[1, \infty)$
- $[2, \infty)$
- $\left[\frac{3}{2}, \infty\right)$
- $(0,1]$
Solution
Given function $f(x)=x^2+\frac{1}{x^2+1}=y$ (let)
$
\begin{aligned}
& \Rightarrow \quad x^4+x^2+1=y\left(x^2+1\right) \text { and } y>0 \\
& \Rightarrow x^4+(1-y) x^2+(1-y)=0
\end{aligned}
$
$\because x \in \mathbf{R}$, so discriminant $\geq 0$
$
\begin{array}{lr}
\Rightarrow & (1-y)^2-4(1-y) \geq 0 \\
\Rightarrow & (1-y)(1-y-4) \geq 0 \\
\Rightarrow & (1-y)(-3-y) \geq 0 \\
\Rightarrow & (y+3)(y-1) \geq 0 \\
\Rightarrow y \in(-\infty,-3] \cup[1, \infty)
\end{array}
$
but $y>0$, so $y \in[1, \infty)$
Therefore range of $f$ is $[1, \infty)$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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