JEE Main 2021MathematicsBinomial TheoremMediumMCQ

JEE Main 2021 — Binomial Theorem Question with Solution

JEE Main 2021 (17 Mar Shift 1)

Question

If the fourth term in the expansion of x+xlog2x7 is 4480, then the value of x where x∈N is equal to:

Choose an option

Show full solutionCorrect option: A
Correct answer
A2

Step-by-step explanation

Given x+xlog2x7. 

The general term in the expansion of x+xlog2x7 is Cr7xrxlog2x7-r.

Then, T4=C37x4xlog2x3=4480

⇒7·6·53·2·1x4+log2x=4480

⇒x4+3log2x=128

⇒x4+3log2x=27

Now, from the option only at x=2, LHS=RHS.

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About this question

This is a previous-year question from JEE Main 2021, 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.