Let S_1=\ z ∈ C:|z| ≤ 5\ , S_2=\ z ∈ C: Im( z+1- 3 i 1- 3 i ) ≥ 0\ and S_3=\ z ∈ C: Re(z) ≥ 0\ . Then the area of the region S_1 S_2 S_3 is :

Correct answer: A. (125 π)/(12) S_1: x^2+y^2 ≤ 25...(1) aligned & S_2: lm of z+(1- 3 i) (1- 3 i) ≥ 0 & lm of ( x+i y 1- 3 i +1) ≥ 0 & lm of ( (x+i y)(1+ 3 i) 4 ) ≥ 0 & 3 x+y ≥ 0 ...(2) & S_3: x ≥ 0 ...(3) & Area =(5)/(12)(π(5)^2) aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Complex Number

Let S_1=\ z ∈ C:|z| ≤ 5\ , S_2=\ z ∈ C: Im( z+1- 3 i 1- 3 i ) ≥ 0\...

Let

S_1=\{z \in C:|z| \leq 5\}, S_2=\left\{z \in C: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}

and

S_3=\{z \in C: \operatorname{Re}(z) \geq 0\}

. Then the area of the region

S_1 \cap S_2 \cap S_3

is :

Options

A $\frac{125 \pi}{12}$
B $\frac{125 \pi}{4}$
C $\frac{125 \pi}{24}$
D $\frac{125 \pi}{6}$

Correct answer

A. $\frac{125 \pi}{12}$

Step-by-step solution

S_1: x^2+y^2 \leq 25...(1)

\begin{aligned} & S_2: \operatorname{lm} \text{\text{ of } } \frac{z+(1-\sqrt{3} i)}{(1-\sqrt{3} i)} \geq 0 \\ & \operatorname{lm} \text{ of }\left(\frac{x+i y}{1-\sqrt{3} i}+1\right) \geq 0 \\ & \operatorname{lm} \text{ of }\left(\frac{(x+i y)(1+\sqrt{3} i)}{4}\right) \geq 0 \\ & \Rightarrow \sqrt{3} x+y \geq 0 ...(2)\\ & S_3: x \geq 0 ...(3)\\ & \text{ Area }=\frac{5}{12}\left(\pi(5)^2\right)\end{aligned}

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