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
- A \(40\)
- B \(75\)
- C \(80\)
- D \(85\)
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…
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