ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 11 - 4.1 complex nubers

અહી \(S=\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z+i|=\) \(|z-1|\}\) હોય તો \(x\) ની બધીજ કિમંતોનો ગણ મેળવો કે જેથી કોઈક \(y \in R\) માટે  \(w=2 x+i y \in S\) મળે.

  1. A \(\left(-\sqrt{2}, \frac{1}{2 \sqrt{2}}\right]\)
  2. B \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]\)
  3. C \(\left(-\sqrt{2}, \frac{1}{2}\right]\)
  4. D \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]\)

Step-by-step Solution

Detailed explanation

\(|z-1+i| \geq|z| ;|z|<2 ;|z+i|=|z-1|\) Hence \(w=2 x+i y \in S\) \(2 x \leq \frac{1}{2} \therefore x \leq \frac{1}{4}\) Now \((2 x )^{2}+(2 x )^{2}<4\) \(x ^{2}<\frac{1}{2} \Rightarrow x \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app