If `F = 2x^(2) - 3x -2`, then choose correct option:-

Options

A. `x = -1/2` is position of stable equilibrium
B. `x = 2` is position of stable equilibrium
C. `x = -1/2` is position of unstable equilibrium
D. `x = 2` is position of neutral equilibrium

Explanation

To solve the problem, we need to determine the equilibrium points of the force function \( F = 2x^2 - 3x - 2 \) and classify them as stable or unstable equilibria.

### Step-by-Step Solution:

1. **Set the Force to Zero**:
To find the equilibrium points, we set the force function equal to zero:
\[ 2x^2 - 3x - 2 = 0 \]

2. **Factor the Quadratic Equation**:
We can factor the quadratic equation. First, we rearrange it:
\[ 2x^2 - 4x + x - 2 = 0 \]
Now, we can group the terms:
\[ (2x^2 - 4x) + (x - 2) = 0 \]
Factoring out common terms:
\[ 2x(x - 2) + 1(x - 2) = 0 \]
This gives us:
\[ (2x + 1)(x - 2) = 0 \]

3. **Find the Roots**:
Setting each factor to zero gives us the equilibrium points:
\[ 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

4. **Determine Stability**:
We need to check the stability of these points by finding the derivative of the force function \( F \):
\[ F' = \frac{dF}{dx} = 4x - 3 \]

5. **Evaluate the Derivative at Equilibrium Points**:
- For \( x = 2 \):
\[ F'(2) = 4(2) - 3 = 8 - 3 = 5 \quad (\text{positive}) \]
Since the derivative is positive, the force is in the same direction as the displacement, indicating that this point is **unstable**.

- For \( x = -\frac{1}{2} \):
\[ F'\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right) - 3 = -2 - 3 = -5 \quad (\text{negative}) \]
Since the derivative is negative, the force is in the opposite direction to the displacement, indicating that this point is **stable**.

6. **Conclusion**:
- The equilibrium point at \( x = 2 \) is unstable.
- The equilibrium point at \( x = -\frac{1}{2} \) is stable.

### Final Answer:
The correct option is that \( x = -\frac{1}{2} \) is stable.