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. $[\frac{3}{2},+\infty )$
• D. None of these

Answer 1

Answer: (A)

$[1,+\infty )$

Form the given $f\left(x\right)$ we know that $f\left(x\right)>0$
Now, the minimum value of $f\left(x\right)$ is 1 for $f\left(0\right)$.
$f\left(0\right)=0+\frac{1}{0+1}=1$
Differentiating with respect to $x$, we get
$f^{\prime }\left(x\right)=2x-\frac{2x}{(1+x^2{)}^2}$
$=2x(1-\frac{1}{x^4+2x^2+1})$
$=2x\left(\frac{x^4+2x^2}{(x^2+1{)}^2}\right)$
$=2x^3\left(\frac{x^2+2}{(x^2+1{)}^2}\right)$
Hence $f^{\prime }\left(x\right)>0$ for $x>0$ and $f^{\prime }\left(x\right)<0$ for $x<0$
Hence $f\left(x\right)$ has minimum value at $x=0$
Therefore the range of the given function $f\left(x\right)$ is
$[1,\infty )$