JEE Main 2022MathematicsBinomial TheoremEasyNumerical

JEE Main 2022 — Binomial Theorem Question with Solution

JEE Main 2022 (25 Jul Shift 1)

Question

If the maximum value of the term independent of t in the expansion of t2x15+1-x110t15,x≥0, is K, then 8 K is equal to _____ .

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Show full solutionCorrect answer: 6006
Correct answer
6006

Step-by-step explanation

Given,

t2x15+1-x110t15

Now rth term is given by,

Tr+1=Cr15t2x1515-r·1-xr10tr

For independent of t,

215-r-r=0

30-2r-r=0

⇒r=10

So, Maximum value of C1015x1-x will be at x=12as dydxx-x2=1-2x & 1-2x=0⇒x=12

=C101512×12=30034

So, K=30034 and 8K=6006

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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.