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**Hint:**Let us consider a particle which is free to move on x-axis, then restoring force acts on the body i.e., $F =-kx^{n}$ . If n will be odd. Force should be along the positive x-axis for negative points on x-axis. Force should be along the negative x-axis for positive points on x-axis. Force should be zero for zero on x-axis. Then the particle will move to oscillate about a given point.

**Complete step-by-step solution:**

a) Given: $F = (x-1)$

When, $x = 1; F = 0$

When, $x > 1; F = +ve$

When, $x < 1; F = -ve$

This equation resembles with the equation of Simple Harmonic motion. Hence, the motion is Simple Harmonic.

b) Given: $F = - (x-1)^{2}$

When, $x = 1; F = 0$

When, $x > 1; F = -ve$

When, $x < 1; F = -ve$

In this case, motion is rectilinear motion. In rectilinear motion, particle move along a straight line.

c) Given: $F = - (x-1)^{3}$

When, $x = 1; F = 0$

When, $x > 1; F = -ve$

When, $x < 1; F = +ve$

In this case, motion is oscillatory about $x = 1$.

d) Given: $F = (x-1)^{3}$

When, $x = 1; F = 0$

When, $x > 1; F = +ve$

When, $x < 1; F = -ve$

So, this motion is not oscillatory.

Option (c) will be correct.

**Note:**Let us consider a particle which is free to move on x-axis, then restoring force acts on the body i.e., $F = -k x^{n}$ . If n will be even. Force should be along the negative x-axis for negative and positive points on the x-axis. Then the particle will not oscillate about a given point but will move rectilinearly.

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