Mathematics · 2026
JEE Main · 28 January 2026, Shift 2 · Q3
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:
Let
$$\begin{aligned}
& A=\{z \in \mathbb{C}:|z-2| \leqslant 4\} \text { and } \\
& B=\{z \in \mathbb{C}:|z-2|+|z+2|=5\} .
\end{aligned}
$$
Then the $\displaystyle \max \left\{\left|z_1-z_2\right|: z_1 \in \mathrm{~A}\right.$ and $\displaystyle \left.z_2 \in \mathrm{~B}\right\}$ is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{17}{2}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.