By Rolle’s Theorem, for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then there exists some c in (a, b) such that f '(c) = 0.

If a function does not satisfy any of the above conditions, then Rolle’s Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9]

(c) f(a)= f(5) = [5] = 5

f(b) = f(9) = [9] = 9

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x ∈ [5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [-2, 2]

(b) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x = y.

So, f(x) is not differentiable in (-2, 2)

(c) f(a)= f(-2) = [-2] = -2

f(b) = f(2) = [2] = 2

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x ∈ [-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2]

(c) f(a) = f(1) = 1² – 1 = 1 – 1 = 0

f(b) = f(2) = 2² - 1 = 4 – 1 = 3

f(a) ≠ f(b)

Here, f(x) does not satisfy a condition of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = x² – 1 for x ∈ [1, 2].