The correct option is D x=1 is a point of minima and x=−1 is a point of maxima of f.
Given limx→0(2+f(x)x3)=4
⇒limx→0f(x)x3=2
limx→0f(x)x3, Limit exists and it is finite.
∴f(x)=ax5+bx4+cx3
⇒limx→0(ax2+bx+c)=2
⇒c=2
Also, f′(x)=5ax4+4bx3+6x2
x=±1 are critical points.
⇒f′(1)=5a+4b+6=0
and f′(−1)=5a−4b+6=0
⇒b=0, a=−65
∴f(x)=−65x5+2x3
⇒f is odd.
f′(x)=−6x4+6x2
f′′(x)=−24x3+12x
f′′(1)<0 and f′′(−1)>0
So, at x=−1, there is local minima and at x=1, there is local maxima.
Also, f(1)−4f(−1)=4