JEE Main 2024MathematicsDeterminantsEasyMCQ

JEE Main 2024 — Determinants Question with Solution

JEE Main 2024 (31 Jan Shift 1)

Question

If fx=x32x2+11+3x3x2+22xx3+6x3−x4x2−2 for all x∈ℝ, then 2f0+f'0 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
C42

Step-by-step explanation

Given: fx=x32x2+11+3x3x2+22xx3+6x3-x4x2-2

⇒f0=01120604-2

⇒f0=0+4+8

⇒f0=12

⇒f'x=x32x2+11+3x3x2+22xx3+63x2-102x+x32x2+11+3x6x23x2x3-x4x2-2+3x24x33x2+22xx3+6x3-x4x2-2

⇒f'0=011206-100+01102004-2+00320604-2

⇒f'0=0-0+6+0+0+0+0+38

⇒f'0=-6+24=18

⇒2f0+f'0=42

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About this question

This is a previous-year question from JEE Main 2024, covering the Determinants 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.