Does Rolle’s theorem applicable if f(a) is not equal to f(b)?

The correct option is No.

To understand why Rolle's theorem is not applicable when
$$f\left(a\right)\ne f\left(b\right)$$
let us review the three fundamental conditions that a function must satisfy for Rolle's theorem to hold true.

For a function $f$ defined on a closed interval
$$[a,b]$$
Rolle's theorem states that if:
1. The function $f$ is continuous on the closed interval $[a,b]$,
2. The function $f$ is differentiable on the open interval $(a,b)$, and
3. The values of the function at the endpoints are equal, meaning
$$f\left(a\right)=f\left(b\right)$$
then there exists at least one real number $c$ in the open interval $(a,b)$ such that the derivative is zero:
$$f^{\prime }\left(c\right)=0$$

The third condition, $f\left(a\right)=f\left(b\right)$, is absolutely essential. Geometrically, it ensures that the graph of the function starts and ends at the same height. This guarantees that if the function goes up or down, it must turn back to reach the initial height, resulting in a local extremum where the tangent line is horizontal (slope is zero).

If
$$f\left(a\right)\ne f\left(b\right)$$
this crucial guarantee is lost, and Rolle's theorem is not applicable. In such scenarios, the Mean Value Theorem (which generalizes Rolle's theorem for unequal endpoint values) is used instead.