Answer :

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.


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