JEE Main 2024MathematicsApplication of DerivativesHardMCQ

JEE Main 2024 — Application of Derivatives Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

Let g:R→R be a non constant twice differentiable such that g'12=g'32. If a real valued function f is defined as f(x)=12[ g(x)+g(2-x)], then

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Show full solutionCorrect option: A
Correct answer
Af"(x)=0 for atleast two x in (0,2)

Step-by-step explanation

Given: fx=12gx+g2-x

⇒f'x=12g'x-g'2-x

⇒f'12=12g'12-g'32

⇒f'12=0

Also, f'32=12g'32-g'12

⇒f'32=0

Now, f'1=12g'1-g'1

⇒f'1=0

So, f'x has three roots and thus f"x will have atleast two roots in 0,2

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About this question

This is a previous-year question from JEE Main 2024, covering the Application of Derivatives 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.