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Saturday, 31 October

JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 January 2020, Shift 2, Maths Q52

Question 52 of 75 in this shift, Maths question 2 of 25, Section A.

Let α=−1+i32\alpha = \frac{-1 + i\sqrt{3}}{2}. If a=(1+α)∑k=0100α2ka = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k} and b=∑k=0100α3kb = \sum_{k=0}^{100} \alpha^{3k}, then a and b are the roots of the quadratic equation :
  1. (1)x2+101x+100=0x^2 + 101x + 100 = 0
  2. (2)x2+102x+101=0x^2 + 102x + 101 = 0
  3. (3)x2−101x+100=0x^2 - 101x + 100 = 0
  4. (4)x2−102x+101=0x^2 - 102x + 101 = 0Official answer

Official answer

Option 4

NTA final key (Jan 2020).

Chapter
Complex Numbers and Quadratic Equations
Topic
Roots of Unity
Idea tested
Periodicity of Omega Powers

Same idea in other shifts

Asked 3× in all
  1. 8 Apr 2025, Shift 2 · Q5Let α\alpha be a solution of x2+x+1=0x^2 + x + 1 = 0, and for some aa and bb in R\mathbb{R},…MediumSingle correct
  2. 27 Jan 2024, Shift 1 · Q21If α\alpha satisfies the equation x2+x+1=0x^2+x+1=0 and (1+α)7=A+Bα+Cα2(1+\alpha)^7=A+B\alpha+C\alpha^2, A,B,C≥0A, B, C\geq 0, then 5(3A−2B−C)5(3A-2B-C) is equal to…MediumNumerical value

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 108 of the 174 JEE Main shifts held from 2017 to 2026.