JEE Main 2020MathematicsComplex NumberHardMCQ

JEE Main 2020 — Complex Number Question with Solution

JEE Main 2020 (08 Jan Shift 2)

Question

Let α=-1+i32. If a=1+α∑k=0100α2k and b=∑k=0100α3k, then a and b, are the roots of the quadratic equation.

Choose an option

Show full solutionCorrect option: B
Correct answer
Bx2-102x+101=0

Step-by-step explanation

Given,

α=-1+i32

a=1+α∑k=0100α2k and

b=∑k=0100α3k

Now

α=ω  ; ω=-1+i32

Using this a and b can be written as 
a=1+ω1+ω2+ω4+…ω198+ω200
=1+ω1-ω21011-ω2=1+ω1-ω1-ω2=1

Similarly,

b=1+ω3+ω6+…+ω300=101
So, the required quadratic equation is 

x2-a+bx+ab=0

⇒x2-102x+101=0.

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About this question

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