JEE Main 2022MathematicsBinomial TheoremMediumMCQ

JEE Main 2022 — Binomial Theorem Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

If the constant term in the expansion of 3x3-2x2+5x510 is 2k.l, where l is an odd integer, then the value of k is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D9

Step-by-step explanation

For 3x3-2x2+5x510the general term is given by

Tr+1=10!r1!r2!r3!3r1-2r25r3x3r1+2r2-5r3

For term independent of x the exponent 3r1+2r2-5r3=0     …1

Also we know r1+r2+r3=10    ⋯2

From 1 and 2, we get 

r1+210-r3-5r3=0

i.e. r1+20=7r3

So r1,r2,r3=1,6,3

Hence the constant term =10!1!6!3!31-2653

=29·32·54·71

⇒k=9

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About this question

This is a previous-year question from JEE Main 2022, covering the Binomial Theorem 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.