Verified Solution • 2025–26 Session moderate

QJEEM22Ma41d8: JEE Main 2022 Mathematics - Class 11 JEEM22Ma41d8 Solution: Let . Let attains minimum and maximum values, respectively, at…

Let S={z∈C:∣z−2∣≤1, z(1+i)+z‾(1−i)≤2}S = \{ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\} . Let ∣z−4i∣|z - 4i| attains minimum and maximum values, respectively, at z1 ∈\in S and z2 ∈\in S. If 5(∣z1∣2+∣z2∣2)=α+β55(|{z_1}{|^2} + |{z_2}{|^2}) = \alpha + \beta \sqrt 5 , where α\alpha and β\beta are integers, then the value of α\alpha + β\beta is equal to ___________.

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Complex Numbers Question 83 English Explanation
SS represents the shaded region shown in the diagram.

Clearly z1z_{1} will be the point of intersection of PAP A and given circle.

PA:2x+y=4P A: 2 x+y=4 and given circle has equation (x−2)2+y2=1(x-2)^{2}+y^{2}=1

On solving we get

z1=(2−15)+25i⇒∣z1∣2=5−45z_{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2}{\sqrt{5}} i \Rightarrow\left|z_{1}\right|^{2}=5-\frac{4}{\sqrt{5}}

z2z_{2} will be either BB or CC .

∵PB=17\because P B=\sqrt{17} and PC=13P C=\sqrt{13} hence z2=1z_{2}=1

So 5(∣z1∣2+∣z2∣2)=30−455\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=30-4 \sqrt{5}

Clearly α=30\alpha=30 and β=−4⇒α+β=26\beta=-4 \Rightarrow \alpha+\beta=26

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring