JEE Main 2019MathematicsComplex NumberMediumMCQ

JEE Main 2019 — Complex Number Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

If α and β be the roots of the equation  x2-2x+2=0, then the least value of n for which αβn=1 is

Choose an option

Show full solutionCorrect option: B
Correct answer
B4

Step-by-step explanation

x2-2x+2=0

⇒ (x-1)2+1=0

⇒x=1±i , i=-1

Now let the roots α=i+i & β=1-i

αβn=1+i1-in=i1-i1-in

⇒αβn=(i)n

Using given information,

⇒(i)n=1

⇒ The minimum value of n∈N=4 because i4=1.

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About this question

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