Answer:
Option A
Explanation:
Given , $y=x^{3}-3x^{2}+5$ .........(i)
On differentiating both sides w.r.t 'x' , we get
$\frac{dy}{dx}=3x^{2}-6x$ ........(ii)
For local maxima or local minima , put $\frac{dy}{dx}=0$
$\Rightarrow$ $3x^{2}-6x=0$
$\Rightarrow$ $3x(x-2)=0$
$\Rightarrow$ x=0 or x=2
Now, differentiating Eq.(ii) w.r.t .'x' we get
$\frac{d^{2}y}{dx^{2}}=6x-6$
$\Rightarrow$ $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0}=-6<0$
$\therefore$ x=0 is a point of local maxima
and $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=2}=6\times2-6=12-6=6>0$
$\therefore$ x=2 is a point of local minima
