JEE Main 2024MathematicsComplex NumberMediumNumerical

JEE Main 2024 — Complex Number Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

Let α,β be the roots of the equation x2-6x+3=0 such that Im(α)>Im(β). Let a, b be integers not divisible by 3 and n be a natural number such that α99β+α98=3n(a+ib), i=-1. Then n+a+b is equal to ___________.

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Show full solutionCorrect answer: 49
Correct answer
49

Step-by-step explanation

Given,

x2-6x+3=0 has roots α & β

Now, solving x2-6x+3=0 we get,

x=6±i62=312±12i

Now, taking α=3eiπ4, β=3e-iπ4

Now, solving α99β+α98=α98αβ+1

=α98(α+β)β

=349eiπ49863e-iπ4

=349ei99π4×2

=349cos99π4+isin99π4×2

=349cos25π-π4+isin25π-π4×2

=349(-1+i)

=3n(a+ib)

So, on comparing we get,

n=49, a=-1, b=1

∴n+a+b=49-1+1=49

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About this question

This is a previous-year question from JEE Main 2024, covering the Complex Number 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.