JEE Main 2025 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 2 April 2025, Shift 1, Maths Q2

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

Let

z

be a complex number such that

|z|=1

. If

\frac{2+k^2z}{k+\bar{z}}=kz

k\in\mathbf{R}

, then the maximum distance of

k+ik^2

from the circle

|z-(1+2i)|=1

is :

(1) $\sqrt{3}+1$
(2)2
(3) $\sqrt{5}+1$ Official answer
(4)3

Official answer

Option 3

NTA final key.

Chapter: Complex Numbers and Quadratic Equations
Topic: Complex Numbers - Polar Form
Idea tested: Polar Triangle Inequality Extrema

Same idea in other shifts

Asked 4× in all

2 Apr 2026, Shift 2 · Q2Let the circles $C_1: |z| = r$ and $C_2: |z - 3 - 4i| = 5, z \in \mathbb{C}_e$ be such that $C_2$ lies within $C_1$ . If $z_1$ moves on…MediumSingle correct
9 Apr 2024, Shift 2 · Q2Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then, the maximum value of $|z-(6+8i)|$ is equal toMediumSingle correct
28 Jul 2022, Shift 1 · Q15Let $S_1=\left\{z_1\in\mathbf{C}:|z_1-3|=\frac{1}{2}\right\}$ and $S_2=\{z_2\in\mathbf{C}:|z_2-|z_2+1||=|z_2+|z_2-1||\}$ . Then, for…HardSingle 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.