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Question

# We can’t apply rolle’s theorem on f(x) = |x| on the interval [-2, 2] because -

A

x is not a function.

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B

f(x) = |x| is not continuous in the interval [-2, 2]

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C

f(x) = |x |is not differentiable in the interval (-2, 2)

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D

f(-2) f (2)

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Solution

## The correct option is C f(x) = |x |is not differentiable in the interval (-2, 2) Let’s consider each option individually. a. |x| is a function because the function is defined on [-2, 2] and no x value from this interval left in the domain and there is no such x in the domain which are giving two values as output. As all the conditions for being a function is satisfied we can say it is a function. b. f(x) = |x| is definitely continuous in the interval [-2, 2] . We can see that very easily by making graph that there is no point where function breaks its continuity. c. f(x) = x is not differentiable at x = 0. As L.H. D and R.H.D are not same. This is because the function definition on the right side of zero is y = x and on the left side it is y = -x. They give different slopes on differentiating making L.H.D and R.H.D not equal.So, we can’t apply Rolle’s theorem since f(x) is not differentiable in the given interval. d. f(-2) = f (2) AS we can see f(-2) = |-2| = 2 & f(2) = 2 So both are equal.

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