Given:

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

At x = 1

LHL =

RHL =

∵ LHL = RHL = 2

and f(1) = 3 – x = 3 – 1 = 2

∴ f(x) is continuous at x = 1

Hence, condition 1 is satisfied.

Condition 2:

Now, we have to check f(x) is differentiable

On differentiating with respect to x, we get

Now, let us consider the differentiability of f(x) at x = 1

LHD ⇒ f(x) = 2x = 2(1) = 2

RHD ⇒ f(x) = -1 = -1

∵ LHD ≠ RHD

∴ f(x) is not differentiable at x = 1

Thus, Rolle’s theorem is not applicable to the given function.