Question

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

• A. $[1,\infty )$
• B. $[2,\infty )$
• C. $\left[\frac{3}{2},\infty \right]$
• D. None of these

Answer 1

The correct option is A

$[1,\infty )$

Explanation for the correct option:

Find the range of the given expression

Given, $f\left(x\right)=x^2+\frac{1}{[x^2+1]}$

$\Rightarrow f\left(x\right)=x^2+\frac{1}{1+x^2}+1-1$

$\Rightarrow f\left(x\right)=\left(x^2+1\right)+\frac{1}{(x^2+1)}-1$

Let the term $\left(x^2+1\right)+\frac{1}{\left(x^2+1\right)}$ be $g\left(x\right)$

For, $g\left(x\right)$

$$\begin{gathered} AM\ge GM \\ \Rightarrow \frac{\left(x^2+1\right)+{\displaystyle \frac{1}{\left(x^2+1\right)}}}{2}\ge \left(x^2+1\right)\times \frac{1}{\left(x^2+1\right)} \\ \Rightarrow \frac{\left(x^2+1\right)+{\displaystyle \frac{1}{\left(x^2+1\right)}}}{2}\ge 1 \\ \Rightarrow \left(x^2+1\right)+\frac{1}{\left(x^2+1\right)}\ge 2 \\ \Rightarrow g\left(x\right)\ge 2 \end{gathered}$$

But, $f\left(x\right)=g\left(x\right)-1$

So,

$\Rightarrow f\left(x\right)\ge 1$

therefore, the range is $[1,\infty )$

Hence the correct option is $A$.