JEE Main 2019MathematicsBinomial TheoremEasyMCQ

JEE Main 2019Binomial Theorem Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

The term independent of x in the expansion of 160-x881.2x2-3x26 is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D-36

Step-by-step explanation

To find the coefficient of term independent of x in given expression, first let's find constant term and coefficient of x-8 in the expansion of  2x2-3x26
Tr+1=  6Cr 2x26-r× -3x2r 
Tr+1=  6Cr 26-r× -3r ×  x12-4r
For constant term:  r=3 , hence constant term =  6C3 23×-33×x0
For coefficient of  x-8 : r=5 , hence coefficient of x-8=  6C5×21-35×x-8
Given expression can be written as
160-x881 6C3×23×-33×x0++   6C5×21×-35×x-8+
Hence, coefficient of term independent of x
=160× 6C3×23-33-181×  6C5×21×-35
= -72+36= -36

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About this question

This is a previous-year question from JEE Main 2019, 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.