QJEEM24Ma8139: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8139 Solution: Let and . Let in , be maximum and minimum at and respectively.…

Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}_1\right)$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$

Circle : $(x+2)^2+(y-3)^2=1$

$$\begin{array}{rl}& {\mathrm{L}}_1:\mathrm{x}+\mathrm{y}-1=0\\ \\ & {\mathrm{L}}_2:\mathrm{x}-\mathrm{y}+4=0\end{array}$$

On solving circle $\& \mathrm{~L}_1$ we get

$${\mathrm{z}}_1:(-2-\frac{1}{\sqrt{2}},3+\frac{1}{\sqrt{2}})$$

On solving $\mathrm{L}_1$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$ we get $z_2:\left(\frac{-3}{2}, \frac{5}{2}\right)$

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