A force F=kx (where k is a positive constant) is acting on a particle Work done:
`{:(,"Column-1",," Column-2"),("(A)","in displacing the body from x=2 to x=4",,"(P) Negative"),("(B)","In displacing the body from x=-4 to x=-2",,"(Q) Positive"),("(C)","In displacing the body from x=-2 to x=+2",,"(R) Zero"):}`

Answer

To solve the problem, we need to calculate the work done by the force \( F = kx \) during the specified displacements. The work done by a variable force can be calculated using the integral of the force over the displacement.

### Step-by-Step Solution:

1. **Understanding the Force**:
The force acting on the particle is given by \( F = kx \), where \( k \) is a positive constant.

2. **Work Done from \( x = 2 \) to \( x = 4 \)**:
\[ W_{2 \to 4} = \int_{2}^{4} F \, dx = \int_{2}^{4} kx \, dx \]
- Integrating:
\[ W_{2 \to 4} = k \left[ \frac{x^2}{2} \right]_{2}^{4} = k \left( \frac{4^2}{2} - \frac{2^2}{2} \right) = k \left( \frac{16}{2} - \frac{4}{2} \right) = k (8 - 2) = 6k \]
- Since \( k \) is positive, \( W_{2 \to 4} = 6k \) is **positive**.

3. **Work Done from \( x = -4 \) to \( x = -2 \)**:
\[ W_{-4 \to -2} = \int_{-4}^{-2} F \, dx = \int_{-4}^{-2} kx \, dx \]
- Integrating:
\[ W_{-4 \to -2} = k \left[ \frac{x^2}{2} \right]_{-4}^{-2} = k \left( \frac{(-2)^2}{2} - \frac{(-4)^2}{2} \right) = k \left( \frac{4}{2} - \frac{16}{2} \right) = k (2 - 8) = -6k \]
- Since \( k \) is positive, \( W_{-4 \to -2} = -6k \) is **negative**.

4. **Work Done from \( x = -2 \) to \( x = 2 \)**:
\[ W_{-2 \to 2} = \int_{-2}^{2} F \, dx = \int_{-2}^{2} kx \, dx \]
- Integrating:
\[ W_{-2 \to 2} = k \left[ \frac{x^2}{2} \right]_{-2}^{2} = k \left( \frac{(2)^2}{2} - \frac{(-2)^2}{2} \right) = k \left( \frac{4}{2} - \frac{4}{2} \right) = k (2 - 2) = 0 \]
- Thus, \( W_{-2 \to 2} = 0 \) is **zero**.

### Final Answers:
- (A) Work done from \( x = 2 \) to \( x = 4 \) is **(Q) Positive**.
- (B) Work done from \( x = -4 \) to \( x = -2 \) is **(P) Negative**.
- (C) Work done from \( x = -2 \) to \( x = 2 \) is **(R) Zero**.