Mathematics · 2022
JEE Main · 28 July 2022, Shift 1 · Q15
Let S_1={z_1 ∈ C:|z_1-3|=1/2} and S_2={z_2 ∈ C:|z_2-|z_2+1||=|z_2+|z_2-1||}. Then, for z_1 ∈ S_1 and z_2 ∈ S_2, the least value of |z_2-z_1| is:
Let $\displaystyle \mathrm{S}_1=\left\{z_1 \in \mathbf{C}:\left|z_1-3\right|=\frac{1}{2}\right\}$ and $\displaystyle \mathrm{S}_2=\left\{z_2 \in \mathbf{C}:\left|z_2-\left|z_2+1\right|\right|=\left|z_2+\left|z_2-1\right|\right|\right\}$. Then, for $\displaystyle \mathrm{z}_1 \in \mathrm{~S}_1$ and $\displaystyle \mathrm{z}_2 \in \mathrm{~S}_2$, the least value of $\displaystyle \left|z_2-z_1\right|$ is :
Official answer
(3)
$\displaystyle \frac{3}{2}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.