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

Let \(S=\{z \in C :|z-3| \leq 1\) and \(z(4+3 i)+\bar{z}(4-3 i) \leq 24\}\). If \(\alpha+i \beta\) is the point in \(S\) which is closest to \(4 i\), then \(25(\alpha+\beta)\) is equal to

  1. A \(40\)
  2. B \(75\)
  3. C \(80\)
  4. D \(85\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(80\)

Step-by-step Solution

Detailed explanation

\(|z-3| \leq 1\) represent pt. i/s circle of radius \(1 \) centred at \((3,0)\) \(z(4+3 i)+\bar{z}(4-3 i) \leq 24\) \((x+i y)(4+3 i)+(x-i y)(4-3 i) \leq 24\) \(4 x+3 x i+4 i y-3 y+4 x-3 i x-4 i y-3 y \leq 24\) \(8 x-6 y \leq 24\) \(4 x-3 y \leq 12\) minimum of \((0,4)\) from…
From JEE Mains
Explore more questions on app