Question

If $F=2x^2-3x-2$, then select the correct statement

• A. $x=-\frac{1}{2}$ is the position of stable equilibrium
• B. $x=2$ is the position of stable equilibrium
• C. $x=-\frac{1}{2}$ is the position of unstable equilibrium
• D. $x=2$ is the position of neutral equilibrium

Answer 1

Answer: (A)

$F=2x^2-3x-2$
Putting $F=0$
$2x^2-3x-2=0$
$2x^2-4x+x-2=0$
$2x(x-2)+(x-2)=0$
$(x-2)(2x+1)=0$
$\Rightarrow x=2,x=\frac{-1}{2}$
$\frac{d^{2}v}{d{x}^2}=\frac{-dF}{dx}=-(4x-3)$
at $x=\frac{-1}{2}$
$\frac{d^{2}v}{d{x}^2}>0$
$\Rightarrow$ Stable equilibrium