The potential energy of a conservative system is given by
$U=a{x}^2-bx$
Here $a$ and $b$ are positive constants. Then, choose the correct option.

The correct option is A The equilibrium position is $x_{eq}=\frac{b}{2a}$ and it is a point of stable equilibrium.
In a conservative force - field,

$F=-\frac{dU}{dx}$

$\therefore F=-\frac{d}{dx}(a{x}^2-bx)$

$\Rightarrow F=b-2ax$

For equilibrium, $F=0$

or, $b-2ax=0$

$\Rightarrow x=\frac{b}{2a}$

Also,

$\frac{d^{2}U}{d{x}^2}=2a=+ve$

So, $U$ is minimum.

So, $x=\frac{b}{2a}$ is the stable equilibrium position.

Why this Question ? Key concept- Stable equilibrium: When a particle is displaced slightly from a position and a force acting on it brings it back to the initial position, it is said to be stable equilibrium position. Necessary condition: $F=-\frac{dU}{dx}=0$ and $\frac{d^{2}U}{d{x}^2}=+ve$