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Friday, 9 October

JEE Main 2024 · MathsMultiple choiceSingle correctHardConceptual

JEE Main 6 April 2024, Shift 2, Maths Q3

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

If z1,z2z_1, z_2 are two distinct complex number such that ∣z1−2z212−z1z2ˉ∣=2\left|\frac{z_1-2z_2}{\frac{1}{2}-z_1\bar{z_2}}\right|=2, then
  1. (1)both z1z_1 and z2z_2 lie on the same circle.
  2. (2)z1z_1 lies on a circle of radius 12\frac{1}{2} and z2z_2 lies on a circle of radius 1.
  3. (3)either z1z_1 lies on a circle of radius 12\frac{1}{2} or z2z_2 lies on a circle of radius 1.
  4. (4)either z1z_1 lies on a circle of radius 1 or z2z_2 lies on a circle of radius 12\frac{1}{2}.Official answer

Official answer

Option 4

NTA final key.

Chapter
Complex Numbers and Quadratic Equations
Topic
Complex Numbers - Polar Form
Idea tested
Modulus and Argument

Same idea in other shifts

Asked 5× in all
  1. 29 Jan 2024, Shift 1 · Q3If z=12−2iz=\frac{1}{2}-2i is such that ∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i), i=−1i=\sqrt{-1} and α,β∈R\alpha,\beta\in\mathbb{R}, then α+β\alpha+\beta is…EasySingle correct
  2. 31 Jan 2024, Shift 1 · Q22If α\alpha denotes the number of solutions of ∣1−i∣x=2x|1-i|^x=2^x and β=(∣z∣arg⁡(z))\beta=\left(\frac{|z|}{\arg(z)}\right), where…MediumNumerical value
  3. 9 Jan 2020, Shift 2 · Q53If zz be a complex number satisfying ∣Re(z)∣+∣Im(z)∣=4|\mathrm{Re}(z)|+|\mathrm{Im}(z)|=4, then ∣z∣|z| cannot be :MediumSingle correct
  4. 6 Sep 2020, Shift 2 · Q53Let z=x+iyz=x+iy be a non-zero complex number such that z2=i∣z∣2z^2=i|z|^2, where i=−1i=\sqrt{-1}, then zz lies on the :MediumSingle correct

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.