JEE Main 2024MathematicsContinuity and DifferentiabilityMediumMCQ

JEE Main 2024 — Continuity and Differentiability Question with Solution

JEE Main 2024 (31 Jan Shift 2)

Question

Consider the function f:0,∞→R defined by fx=e−logex. If m and n be respectively the number of points at which f is not continuous and f is not differentiable, then m+n is

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Show full solutionCorrect option: C
Correct answer
C1

Step-by-step explanation

We know that, logx is continuous in 0,∞.

⇒fx=e-logx is continuous in 0,∞

Thus, the number of points where fx is discontinuous is 0.

⇒m=0

fx=elogx,0<x<1e-logx,x≥1

⇒fx=x,0<x<11x,x≥1

⇒f'x=1,0<x<1-1x2,x≥1

⇒f'1-=1 & f'1+=-1

So, the number of points where fx is non-differentiable is 1.

⇒n=1

⇒m+n=1

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About this question

This is a previous-year question from JEE Main 2024, covering the Continuity and Differentiability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.