A particle free to move along the `x-`axis has potential energy given by `U(x) = k[1 - e^(-x^(2))]` for `-oo le x le + oo`, where `k` is a positive constant of appropriate dimensions. Then select the incorrect option

Options

A. at points away from the origin, the particle is in unstable equilibrium
B. for any finite non-zero value of x, there is a force directed away from the origin
C. if its total mechanical energy is k//2, it has its minimum kinetic energy at the origin
D. for small displacements from x = 0, the motion is simple harmonic

Explanation

To solve the problem, we need to analyze the potential energy function given by \( U(x) = k[1 - e^{-x^2}] \) and determine the characteristics of the motion of the particle described by this potential energy. We will evaluate the statements provided in the options to identify the incorrect one.

### Step 1: Analyze the Potential Energy Function
The potential energy function is given by:
\[ U(x) = k[1 - e^{-x^2}] \]
where \( k \) is a positive constant.

### Step 2: Determine the Force Acting on the Particle
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = k \cdot \frac{d}{dx}[1 - e^{-x^2}] = k \cdot (0 - (-2x)e^{-x^2}) = 2kx e^{-x^2} \]
Thus, the force is:
\[ F = -2kx e^{-x^2} \]

### Step 3: Analyze the Equilibrium Points
To find the equilibrium points, we set the force to zero:
\[ -2kx e^{-x^2} = 0 \]
This gives us \( x = 0 \) as the only equilibrium point because \( e^{-x^2} \) is never zero.

### Step 4: Stability of the Equilibrium Point
To determine the stability of the equilibrium point at \( x = 0 \), we analyze the second derivative of the potential energy:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}[2kx e^{-x^2}] = 2k(e^{-x^2} - 2x^2 e^{-x^2}) = 2ke^{-x^2}(1 - 2x^2) \]
At \( x = 0 \):
\[ \frac{d^2U}{dx^2} = 2k \cdot 1 = 2k > 0 \]
This indicates that the equilibrium at \( x = 0 \) is stable.

### Step 5: Behavior Away from the Origin
For \( |x| \) large (away from the origin), \( e^{-x^2} \) approaches zero, and hence:
\[ U(x) \approx k \]
The force \( F \) becomes:
\[ F \approx -2kx \quad \text{(as \( e^{-x^2} \) approaches 0)} \]
This indicates that the force acts away from the origin, which suggests that the particle is in unstable equilibrium at points away from the origin.

### Step 6: Kinetic Energy and Simple Harmonic Motion
If the total mechanical energy \( E = \frac{k}{2} \), we can analyze the motion around the equilibrium point. For small displacements from \( x = 0 \), the potential energy can be approximated as:
\[ U(x) \approx kx^2 \]
This is characteristic of simple harmonic motion, where the motion is described by:
\[ F = -kx \quad \text{(with \( k \) being the spring constant)} \]

### Conclusion
Based on the analysis, we can conclude the following:
- The equilibrium at \( x = 0 \) is stable.
- At points away from the origin, the particle is in unstable equilibrium.
- For small displacements, the motion is simple harmonic.

### Final Answer
The incorrect statement is that "At points away from the origin, the particle is in unstable equilibrium." This is true; hence, the incorrect option is **A, B, and C**.