JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 9 April 2024, Shift 2, Maths Q3

Question 3 of 90 in this shift, Maths question 3 of 30, Section A.

Let

\alpha, \beta;\ \alpha>\beta

, be the roots of the equation

x^2-\sqrt{2}x-\sqrt{3}=0

. Let

P_n=\alpha^n-\beta^n,\ n\in\mathbb{N}

. Then

\left(11\sqrt{3}-10\sqrt{2}\right)P_{10}+\left(11\sqrt{2}+10\right)P_{11}-11P_{12}

is equal to

(1) $11\sqrt{3}P_9$
(2) $11\sqrt{2}P_9$
(3) $10\sqrt{3}P_9$ Official answer
(4) $10\sqrt{2}P_9$

Official answer

Option 3

NTA final key.

Chapter: Complex Numbers and Quadratic Equations
Topic: Quadratic Roots
Idea tested: Newton's Sums

Same idea in other shifts

Asked 9× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.