Mathematics · 2022
JEE Main · 24 June 2022, Shift 2 · Q21
Let S={z ∈ C:|z-3| ≤ 1 and z(4+3 i)+z̄(4-3 i) ≤ 24}. If α+i β is the point in S which is closest to 4 i, then 25(α+β) is equal to \ \ \ \.
Let $\displaystyle S=\{z \in \mathbb{C}:|z-3| \leq 1$ and $\displaystyle z(4+3 i)+\bar{z}(4-3 i) \leq 24\}$. If $\displaystyle \alpha+i \beta$ is the point in $\displaystyle S$ which is closest to $\displaystyle 4 i$, then $\displaystyle 25(\alpha+\beta)$ is equal to \_\_\_\_.
Official answer
80
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.