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Questions › JNU › JNU SSS 2013 › Q77
1 mark, −0.25Mathematics › Limits, continuity and differentiability › Differentiability and its link to continuityEasyabout 1 min

JNU SSS 2013 Question 77: Limits, continuity and differentiability

Let ∣x∣|x|∣x∣ stands for the absolute value of xxx. Then the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣ is

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Let |x| stands for the absolute value of x. Then the function f(x) = |x| is (A) differentiable everywhere including the point ‘0’ (B) differentiable everywhere excluding the point ‘0’ (C) differentiable everywhere excluding the points ‘0’, ‘1’ and ‘–1’ (D) None of the above

Limits, continuity and differentiability has come up 122 times in 7 exams since 2006.

What this question tests

This question is on Differentiability and its link to continuity, part of Limits, continuity and differentiability in Mathematics.

When a derivative exists and why differentiability implies continuity but not conversely.

TestStop has 12 past-paper questions on this subtopic; the nearest in year are below.

  • IIT JAM 2021 · Q35Let f be defined on (-π/2, π/2) by f(x) = (cos (π/2 + |x|))/(sin (π/2 - |x|)). Then,
  • IIT JAM 2022 · Q34Let f be defined by f(x) = |x| + |cos (π/2 - x)| for x ∈ (-π/2, π/2). Then
Practise Limits, continuity and differentiabilityMore on Differentiability and its link to continuity

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Question from the JNU SSS 2013 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.