JEE Main 2020MathematicsQuadratic EquationMediumMCQ

JEE Main 2020 — Quadratic Equation Question with Solution

JEE Main 2020 (03 Sep Shift 1)

Question

If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then α-1αβ-1βα+1ββ+1α is equal to :

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Show full solutionCorrect option: D
Correct answer
D949-p2

Step-by-step explanation

α, β are roots of x2+px+2=0

⇒α+β=-p, αβ=2

1α,1β are roots of  2x2+2qx+1=0 

∴ α, β are roots of x2+2qx+2=0

 ⇒α+β=-2q, αβ=2

∴p=2q  ...(i)

⇒α-1αβ-1βα+1ββ+1α=αβ-αβ-βα+1αβ·αβ+1αβ+2

⇒2-α+β2-2αβαβ+12·2+12+2

⇒52-p2-42·92=949-p2 or, 949-4q2

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About this question

This is a previous-year question from JEE Main 2020, covering the Quadratic Equation 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.