Let `f(x)=(e^(x))/(1+x^(2)) and g(x)=f'(x)` , then

Options

A. g(x) has two local maxima and two local minima points
B. g(x) has exactly one local maxima and one local minima point
C. x = 1 is a point of local maxima for g(x)
D. There is a point of local maxima for g(x) in the interval (-1,0)

Answer

B, D

Explanation

`f(x)=(e^(x))/(1+x^(2))`
`g(x)=f'(x)=(e^(x)(x-1)^(2))/((1+x^(2))^(2))`
`rArr" "g'(x)=((x-1)(x^(3)-3x^(2)+5x+1)e^(x))/((x^(2)+1)^(3))`
Now, `h(x)=x^(3)-3x^(2)+5x+1`
`rArr" "h'(x)=3x^(2)-6x+5` which is always positive.
Hence, `x^(3)-3x^(2)+5x+1=0` has only one real root which is not x = 1.
Also `h(0)=1,h(-1)=-8.` So h(x) = 0 has one negative root
`x=alpha in (-1,0).`
Now from the sign scheme of g'(x),
`x=alpha` is the point of maxima and x = 1 is the point of minima.