Condition (1):
Since, f(x) = (x - 1)(2x - 3) is a polynomial and we know every polynomial function is continuous for all x ϵ R.
⇒ f(x) = (x -1)(2x - 3) is continuous on [1,3].
Condition (2):
Here, f’(x) = (2x - 3)+ 2(x - 1) which exist in [1,3].
So, f(x) = (x -1)(2x - 3) is differentiable on (1,3).
Condition (3):
Here, f(1) = [1 - 1][2(1) - 3] = 0
And f(5) = [3 -1][2(3) - 3] = 6
i.e. f(1) ≠ f(3)
Condition (3) of Rolle’s theorem is not satisfied.
So, Rolle’s theorem is not applicable.