Question

Let $f\left(x\right)=\frac{e^x}{1+x^2}$ and $g\left(x\right)=f^{\prime }\left(x\right)$ then

• A. $g\left(x\right)$ has four points of local extremum
• B. $g\left(x\right)$ has two points of local extremum
• C. $g\left(x\right)$ has a point of local minimum at $x=1$
• D. $g\left(x\right)$ has a point of local maximum at some $x\in (-1,0)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $g\left(x\right)$ has two points of local extremum
C $g\left(x\right)$ has a point of local minimum at $x=1$
D $g\left(x\right)$ has a point of local maximum at some $x\in (-1,0)$

$f^{\prime }\left(x\right)=g\left(x\right)=\frac{(x-1{)}^2{e}^x}{(1+x^2{)}^2}{g}^{\prime }\left(x\right)=\frac{(x-1)(x^3-3x^2+5x+1)e^x}{(1+x^2{)}^3}$
let $h\left(x\right)=x^3-3x^2+5x+1$
$h^{\prime }\left(x\right)=3x^2-6x+5,D<0,\forall x\in R$
So $h^{\prime }\left(x\right)>0,\forall x\in R$
So, $h\left(x\right)$ has only one real root.
Now
$g^{\prime }(-1)g^{\prime }\left(0\right)<0$ so root $\to (-1,0)$
$g\left(x\right)$ has two points of extrema
maxima at $x\in (-1,0)$
minima at $x=1$