What is the relation between f(a) and f(h) according to another form of Rolle’s theorem?

The correct option is f(a) = f(a+h).

To understand why this is correct, let us recall the standard statement of Rolle's Theorem and see how it is adapted to an alternative form.

1. Classical Rolle's Theorem:
If a real-valued function $f$ is:
(i) continuous on a closed interval $[a,b]$,
(ii) differentiable on the open interval $(a,b)$, and
(iii) satisfies the condition that the function values at the endpoints are equal, i.e.,

$f\left(a\right)=f\left(b\right)$

2. Alternative Form of Rolle's Theorem:
We can express the interval $[a,b]$ in terms of a starting point $a$ and a positive width $h$. Let us define the upper bound of the interval as:

$b=a+h$

Under this formulation, the third condition of Rolle's Theorem, which requires the function values at the endpoints of the interval to be equal ($f\left(a\right)=f\left(b\right)$), directly translates to:

$f\left(a\right)=f(a+h)$

Consequently, if this relation holds along with continuity and differentiability on the interval, there exists a number $\theta$ (where $0<\theta <1$) such that the derivative vanishes at the point $a+\theta h$, meaning $f\prime (a+\theta h)=0$.