Explanation

(b)
Given $y = x^{5/2}$
$\therefore\ \ \frac{dy}{dx} = \frac{5}{2}x^{3/2},\ \frac{d^{2}y}{dx^{2}} = \frac{15}{4}x^{1/2}$
At $x = 0,\ \frac{dy}{dx} = 0,\frac{d^{2}y}{dx^{2}} = 0$
and $\frac{d^{3}y}{dx^{3}}$ is not defined
When $x = 0,\ y = 0$
$\therefore\ (0,\ 0)$ is a point of inflexion