Exam

JEE Advanced

Year

2013

Paper

Paper 2

Subject

Mathematics

Chapter

Straight Lines

JEE Advanced 2013 · Paper 2 · Mathematics · Straight Lines

Paragraph: Let S=S_ 1 S_ 2 S_ 3 , where S_ 1 =\ z ∈ ℂ:|z| 0\ and S_...

Paragraph: Let $S=S_{1} \cap S_{2} \cap S_{3}$, where $S_{1}=\{z \in \mathbb{C}:|z|<4\}, \quad S_{2}=\left\{z \in \mathbb{C}: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]>0\right\}$ and $S_{3}=\{z \in \mathbb{C}: \operatorname{Re} z>0\}$.
Question: Area of $S=$

Step-by-step solution

S₁ : x² + y² < 16
${\text{S}}_{2} \text{:} \dfrac{\text{z} - 1 + \sqrt{3 \text{i}}}{1 - \text{i} \sqrt{3}} = \dfrac{\left(\text{x} - 1\right) + \text{i} \left(\text{y} + \sqrt{3}\right)}{1 - \sqrt{3} \text{i}}$
$= \dfrac{\left\{\left(\text{x} - 1\right) + \text{i} \left(\text{y} + \sqrt{3}\right)\right\} \left\{1 + \sqrt{3 } \text{i}\right\}}{1 + 3}$


${\text{S}}_{2} \text{:} \dfrac{\left(\text{x} - 1\right) \sqrt{3} + \text{y} + \sqrt{3}}{9} > 0$
${\text{S}}_{2} \text{:} \sqrt{3} \text{x} + \text{y} > 0 \text{ and } {\text{S}}_{3} \text{:} \text{x} > 0$
$\text{A} = \dfrac{1}{2} {\text{r}}^{2} \theta = \dfrac{1}{2} \times 1 6 \times \dfrac{5 \pi}{6} = \dfrac{4 0 \pi}{6} = \dfrac{2 0 \pi}{3}$