JEE Main 2024MathematicsContinuity and DifferentiabilityEasyNumerical

JEE Main 2024 — Continuity and Differentiability Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

If the function fx=1|x|,|x|≥2ax2+2b,|x|<2 is differentiable on R, then 48(a+b) is equal to _______.

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Show full solutionCorrect answer: 15
Correct answer
15

Step-by-step explanation

Given,

fx=1x, x≥2ax2+2 b, -2<x<2-1x, x≤-2

Also given fx is differentiable on R

So, finding LHD and RHD at x=2 we get,

-1x2=2ax

⇒-14=4a

⇒a=-116

Now, for function to be continuous at x=2 we get,

12=4a+2b

⇒12=-416+2b

⇒b=38

Hence, 48a+b=4838-116

⇒48a+b=48516=15

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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.