JEE Main 2022MathematicsDeterminantsMediumMCQ

JEE Main 2022 — Determinants Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

Let fx=a-10axa-1ax2axa,a∈R. Then the sum of the squares of all the values of a for 2f'10-f'5+100=0 is

Choose an option

Show full solutionCorrect option: C
Correct answer
C125

Step-by-step explanation

Given, fx=a-10axa-1ax2axa

fx=a1-10xa-1x2axa

=a1a2+ax+1ax+x2

⇒fx=ax+a2

Now differentiating the function w.r.t x

We get, f'x=2ax+a

Given, 2f'10-f'5+100=0

⇒2×2a10+a-2a5+a+100=0

⇒40a+4a2-10a-2a2+100=0

2a2+30a+100=0

⇒a2+15a+50=0

a+10a+5=0

a=-10 or a=-5

Required =-102+-52=125

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About this question

This is a previous-year question from JEE Main 2022, 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.