JEE Main 2024MathematicsContinuity and DifferentiabilityHardMCQ

JEE Main 2024 — Continuity and Differentiability Question with Solution

JEE Main 2024 (27 Jan Shift 2)

Question

Consider the function f:(0,2)→R defined by f(x)=x2+2x and the function g(x) defined by gx=min{f(t)},0<t≤x and 0<x≤132+x,1<x<2. Then

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Correct answer
Ag is continuous but not differentiable at x=1

Step-by-step explanation

Given,

f:(0,2)→R;f(x)=x2+2x

⇒f'x=12-2x2

⇒f'x=x2-42x2

∴fx is decreasing in domain 0,2.

⇒gx=x2+2x   0<x≤132+x    1<x<2

⇒g1=12+21=52

⇒g1+=32+1=52

⇒gx is continuous in 0,2.

⇒g'x=f'x, 0<x≤11,      1<x<2

⇒g'1=f'1=-32

⇒g'1≠g'1+

So, gx is not differentiable at x=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.