JEE Main 2018MathematicsBinomial TheoremHardMCQ

JEE Main 2018 — Binomial Theorem Question with Solution

JEE Main 2018 (16 Apr Online)

Question

The coefficient of x2 in the expansion of the product 2-x21+2x+3x26+1-4x26 is

Choose an option

Show full solutionCorrect option: D
Correct answer
D106

Step-by-step explanation

We have, (1+2x+3x2)6=∑r=06 6Cr2x+3x2r
=6C0+6C12x+3x2+6C22x+3x22+ …+6C62x+3x26
∴ Coefficient of x2=18+60=78
Again, coefficient of x2 in 1-4x26=-24
Constant term in 1+2x+3x26+1-4x26 is 1+1=2.
Thus, the required coefficient of x2=2 (coefficient of x2) in 1+2x+3x26+1-4x26-constant term in 1+2x+3x26+1-4x26
=278-24-2
=106.

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About this question

This is a previous-year question from JEE Main 2018, 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.