JEE Main 2021MathematicsBinomial TheoremMediumMCQ

JEE Main 2021 — Binomial Theorem Question with Solution

JEE Main 2021 (26 Feb Shift 1)

Question

The maximum value of the term independent of t in the expansion of tx15+1-x110t10where x∈0,1 is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C2.10!335!2

Step-by-step explanation

Tr+1=Cr10tx1/510-r1-x110tr

=Cr10t10-2rx10-r51-xr10

For the term independent of t, 

⇒10-2r=0

⇒r=5

T6=fx=C510x1-x; for maximum

f'x=0⇒x=23&f''23<0

so fxmax.=C51023·13

=2.10!335!2

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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.