Answer :

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].


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