NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice

The function f x = m a x 1 - x , 1 + x , 2 ∀ x ∈ R is

Options

Adiscontinuous at exactly two points
Bdifferentiable ∀ x ∈ R
Cdifferentiable ∀ x ∈ R - - 1 , 1
Dcontinuous ∀ x ∈ R - 0 , 1 , - 1

Correct answer

C. differentiable ∀ x ∈ R - - 1 , 1

Step-by-step solution

f x = 1 - x : x ≤ - 1 2 : - 1 < x ≤ 1 1 + x : x > 1 Continuity at x = - 1 f - 1 = 1 - - 1 = 2 f - 1 - = 1 - - 1 = 2 f - 1 + = 2 ∵ f - 1 = f - 1 - = f - 1 + ∴ continuous at x = - 1 f 1 = 2 ,   f 1 - = 2 f 1 + = 1 + 1 = 2 ∵ f 1 - = f 1 = f 1 + ∴ continuous at x = 1 For differentiability, f ′ x = − 1 x < − 1 0 − 1 < x < 1 1 x > 1 at x = - 1 , f ' - 1 - = - 1 ,   f - 1 + = 0 ∵ f ' − 1 − ≠ f ' − 1

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