Question

The graph of $y=\ln (x^2+1)$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$y=\ln (x^2+1),$ Domain$:x\in R$
$x=0\Rightarrow y=0$
$y^{\prime }=\frac{2x}{x^2+1}$
$y^{\prime }=0\Rightarrow x=0$
$y^{\prime }\left(0^{+}\right)>0$ and $y^{\prime }\left(0^{-}\right)<0$
$\Rightarrow x=0$ is a point of minimum.
$y\left(0\right)=0\to$Local minimum value
$y^{\prime \prime }=\frac{2(1-x^2)}{(x^2+1{)}^2}$
$y^{\prime \prime }=0$
$\Rightarrow x=\pm 1\to$Point of inflection
Sign scheme for $y^{\prime \prime }$

Thus, Graph of $y=\ln (x^2+1)$ is