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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Complex Number

Let A=\ z ∈ ℂ:|z-2| ≤ 4\ and B=\ z ∈ ℂ:|z-2|+|z+2|=5\ . Then the...

Let A={z∈C:∣z−2∣⩽4} and B={z∈C:∣z−2∣+∣z+2∣=5}A=\{z \in \mathbb{C}:|z-2| \leqslant 4\} \text{ and }B=\{z \in \mathbb{C}:|z-2|+|z+2|=5\}. Then the max \left\{\left|z_{1}-z_{2}\right|: z_{1} \in \mathrm{~A}\right. and \left.z_{2} \in \mathrm{~B}\right\} is :Let A=\ z ∈ ℂ:|z-2| ≤ 4\ and B=\ z ∈ ℂ:|z-2|+|z+2|=5\ . Then the max \ |z_ 1 -z_ 2 |: z_ 1 ∈ ~A. and .z_ 2 ∈ ~B\ is :

Options

  1. A172\frac{17}{2}
  2. B8
  3. C9
  4. D152\frac{15}{2}

Correct answer

A. 172\frac{17}{2}

Step-by-step solution

∣z−2∣≤4⇒(x−2)2+y2≤16|z - 2| \leq 4 \Rightarrow (x - 2)^2 + y^2 \leq 16
∣z−2∣+∣z+2∣=5⇒x2a2+y2b2=1|z - 2| + |z + 2| = 5 \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1
⇒4x225+4y29=1\Rightarrow \dfrac{4x^2}{25} + \dfrac{4y^2}{9} = 1
Let A=\ z ∈ ℂ:|z-2| ≤ 4\ and B=\ z ∈ ℂ:|z-2|+|z+2|=5\ . Then the max \ |z_ 1 -z_ 2 |: z_ 1 ∈ ~A. and .z_ 2 ∈ ~B\ is :
Maximum value of ∣z1−z2∣=6+52=172|z_1 - z_2| = 6 + \dfrac{5}{2} = \dfrac{17}{2}
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