JEE Advanced 2013 · Paper 2 · Mathematics Q50

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MathematicsSingle correct+3 / -1Question 50 of 60

Passage

Paragraph for Questions 49 and 50

Let $S=S_1\cap S_2\cap S_3$ , where
$S_1=\{z\in\mathbb C:|z|<4\},\quad S_2=\left\{z\in\mathbb C:\operatorname{Im}\left[\frac{z-1+\sqrt3i}{1-\sqrt3i}\right]>0\right\}\quad\text{and}\quad S_3=\{z\in\mathbb C:\operatorname{Re}z>0\}.$

$\displaystyle\min_{z\in S}|1-3i-z|=$

\dfrac{2-\sqrt3}2

\dfrac{2+\sqrt3}2

\dfrac{3-\sqrt3}2

\dfrac{3+\sqrt3}2

Answer and worked solution

Official answer: C

The point $(1,-3)$ lies below the line $\sqrt3x+y=0$ , outside S.

Nearest point of S lies on that boundary line; the distance is
$\frac{|\sqrt3-3|}2=\frac{3-\sqrt3}2,$
and the foot of the perpendicular, about $(1.55,-2.68)$ , is inside the disc with $x>0$ , so it belongs to the closure of S.

(C)

Topics:Geometry in the complex plane Distance of a point from a line

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JEE Advanced 2013 · Paper 2, Mathematics question 50