JEE Main 2020MathematicsBinomial TheoremMediumMCQ

JEE Main 2020 — Binomial Theorem Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

Let α>0,β>0 be such that α3+β2=4. If the maximum value of the term independent of x in the binomial expansion of αx19+βx-1610 is 10k, then k is equal to

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Show full solutionCorrect option: A
Correct answer
A336

Step-by-step explanation

 Tr+1=Cr10αx1910-rβx-16r

Tr+1=Cr10α10-rβrx10-r9-r6

Term independent of x ⇒Power of x=0

10-r9-r6=0⇒  r=4

T5=C410α6β4

Now Let α3,β2 are two numbers. Clearly both are positive.

A.M.≥G.M.

⇒  α3+β22≥α3β21/2

⇒  α3β2≤4

⇒  α6β4≤16

⇒  T510C4≤16

⇒  T5≤16.10C4

⇒  T5max=16×10C4=10k

⇒10k=3360

⇒  k=336

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About this question

This is a previous-year question from JEE Main 2020, 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.