JEE Main 2022MathematicsBinomial TheoremMediumNumerical

JEE Main 2022 — Binomial Theorem Question with Solution

JEE Main 2022 (28 Jun Shift 1)

Question

The number of positive integers k such that the constant term in the binomial expansion of 2x3+3xk12,x≠0 is 28·l, where l is an odd integer, is ______.

Enter your answer

Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

For 2x3+3xk12

Tr+1=Cr122x312-r·3xkr=Cr12·212-r·3r·x36-3r-kr

For the term to be independent of x,

36-3r-kr=0 ⇒r=36k+3 

Here k can be 0,1,3,6,9 for r to be a whole number.

But for Cr12·212-r·3r to be in the form of 28·l

Only k=3, 6 satisfies the above relation.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Binomial Theorem chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.