**Introduction:**
When a particle moves in a straight line and experiences a retardation (deceleration) proportional to its displacement, we can analyze its motion using the concept of simple harmonic motion (SHM). In SHM, the restoring force acting on the particle is directly proportional to its displacement from the equilibrium position. In this case, the restoring force is acting against the motion of the particle, causing it to decelerate.

**Deriving the Equation:**
Let's consider a particle of mass m moving along the x-axis. The equation of motion for the particle can be written as:

ma = -kx

Where:
- m is the mass of the particle,
- a is the acceleration of the particle,
- k is the proportionality constant, and
- x is the displacement of the particle.

The negative sign indicates that the acceleration is opposite in direction to the displacement. Rearranging the equation, we have:

a = -(k/m) * x

**Kinetic Energy and Displacement:**
The kinetic energy of a particle is given by the equation:

K.E. = (1/2) * m * v^2

Where:
- K.E. is the kinetic energy,
- m is the mass of the particle, and
- v is the velocity of the particle.

Since the particle is moving in a straight line, the velocity v can be related to the displacement x using the equation:

v^2 = u^2 + 2as

Where:
- u is the initial velocity of the particle, and
- s is the displacement of the particle.

Substituting the value of acceleration a from the equation of motion, we get:

v^2 = u^2 - 2(k/m) * x

**Loss of Kinetic Energy:**
The loss of kinetic energy for any displacement x can be calculated by subtracting the final kinetic energy (when displaced by x) from the initial kinetic energy (when at rest, x = 0). Let's denote the initial kinetic energy as K.E.0 and the final kinetic energy as K.E.x.

K.E.0 = (1/2) * m * u^2
K.E.x = (1/2) * m * v^2

Substituting the value of v^2 from the previous derivation, we have:

K.E.x = (1/2) * m * (u^2 - 2(k/m) * x)

The loss of kinetic energy, ΔK.E., can be calculated as:

ΔK.E. = K.E.0 - K.E.x
ΔK.E. = (1/2) * m * u^2 - (1/2) * m * (u^2 - 2(k/m) * x)
ΔK.E. = (1/2) * m * (2(k/m) * x)
ΔK.E. = kx

Hence, the loss of kinetic energy for any displacement x is proportional to x, which corresponds to option (c) in the given options.

Given that the retardation is proportional to the displacement x -a is directly proportional to x or a=-I'd dv/dt=-kx (or) dv/dx × dv/dt =-kx vdv/dx=-kx (or) vdv=-kxdx Integrating we have integration of vdv(range is v to u) =-k integration of xdx ( range is x to 0) 1/2 mu^2-1/2mv^2 =mkx^2 :K.E loss is proportional to x^2 ANSWER "A" IS CORRECT. Hope this helps!