JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 5)
Let $$S_1=\{z \in \mathbf{C}:|z| \leq 5\}, S_2=\left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$$ and $$S_3=\{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}$$. Then the area of the region $$S_1 \cap S_2 \cap S_3$$ is :
$$\frac{125 \pi}{24}$$
$$\frac{125 \pi}{6}$$
$$\frac{125 \pi}{12}$$
$$\frac{125 \pi}{4}$$
Explanation
$$S_1=\{z \in C:|z| \leq 5\}$$
_5th_April_Evening_Shift_en_5_1.png)
$$S_2=\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0$$
Take $$z=x+i y$$
$$=\frac{x+i y+1-\sqrt{3} i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}$$
$$\begin{aligned} = & \frac{x+i y+1-\sqrt{3} i+\sqrt{3} i x-\sqrt{3} y+\sqrt{3} i+3}{1+3} \\ & =y+\sqrt{3} x \geq 0 \\ S_3= & x \geq 0 \end{aligned}$$
_5th_April_Evening_Shift_en_5_2.png)
$$\begin{aligned} & y+\sqrt{3} x=0 \\ & y=-\sqrt{3} x \\ & \text { Slope }=-\sqrt{3} \\ & 90+\theta=120^{\circ} \\ & \theta=30^{\circ} \\ & 90-\theta=60^{\circ} \end{aligned}$$
In first quadrant we have angle $$\frac{\pi}{2}$$
so, total angle $$90^{\circ}+60^{\circ}=150^{\circ}$$
So, area $$\frac{\pi r^2}{2 \pi} \times \frac{5 \pi}{6}=\frac{125 \pi}{12}$$
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