1. Given data

• Mass of particle: $m=1\,\text{kg}$
• Potential energy: $V(x)=\frac{x^4}{4}-\frac{x^2}{2}$
• Total mechanical energy: $E=2\,\text{J}$

We need the maximum speed.

2. Use conservation of mechanical energy

The total energy is $E=K+V(x)$ So kinetic energy is $K=E-V(x)$

Since speed is maximum when kinetic energy is maximum, and $E$ is constant, this happens when $V(x)$ is minimum.

3. Find the minimum of the potential energy

Given $V(x)=\frac{x^4}{4}-\frac{x^2}{2}$

Differentiate: $\frac{dV}{dx}=x^3-x=x(x^2-1)$

Critical points are: $x=0,\ \pm 1$

Now, $\frac{d^2V}{dx^2}=3x^2-1$

• At $x=0$: $\frac{d^2V}{dx^2}=-1<0$ so $x=0$ is a maximum.

• At $x=\pm1$: $\frac{d^2V}{dx^2}=2>0$ so $x=\pm1$ are minima.

Now evaluate the minimum value: $V(\pm1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\,\text{J}$

Thus, $V_{\min}=-\frac14\,\text{J}$

4. Find maximum kinetic energy

$K_{\max}=E-V_{\min}=2-\left(-\frac14\right)=\frac94\,\text{J}$

5. Convert to maximum speed

Using $K=\frac12 mv^2$ with $m=1$ kg, $\frac12 v_{\max}^2=\frac94$

So, $v_{\max}^2=\frac92$ $v_{\max}=\sqrt{\frac92}=\frac{3}{\sqrt2}\,\text{m/s}$

6. Check options

• A: $\dfrac{3}{\sqrt2}$ ✅
• B: $\sqrt2$ ❌
• C: $\dfrac{1}{\sqrt2}$ ❌
• D: $2$ ❌

Therefore, the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.