Verified Solution • 2025–26 Session moderate

QJEEM22Ma5e2c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma5e2c Solution: Let S = { z = x + i y :

Let S={z=x+iy:|z−1+i|≥|z|,|z|<2,|z+i|=|z−1|}. Then the set of all values of x, for which w=2x+iy∈S for some y∈R, is :

Multiple Choice Options

  • A (−2,122]
  • B (−12,14]
  • C (−2,12]
  • D (−12,122]

Step-by-Step Verified Solution

S:\{z=x+i y:|z-1+i| \geq|z|,|z|&lt;2,|z-i|=|z-1|\}

|z−1+i|≥|z|

JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 59 English Explanation 1
|z|&lt;2

JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 59 English Explanation 2
|z−i|=|z−1|

JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Complex Numbers Question 59 English Explanation 3
∵w∈S\because \quad w \in S and w=2x+iyw=2 x+i y

2 x&lt;\frac{1}{2} \quad\quad \therefore x&lt;\frac{1}{4}

(2 x)^{2}+(-2 x)^{2}&lt;4

4 x^{2}+4 x^{2}&lt;4

x^{2}&lt;\frac{1}{2} \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)

∴x∈(−12,14]\therefore \quad x \in\left(-\frac{1}{2}, \frac{1}{4}\right]

Related Learning Resources & Exam Preparation