Paragraph: Let S=S_{1} \cap S_{2} \cap S_{3}, where S_{1}=\{z \in \mathbb{C}:|z|0\right\} and S_{3}=\{z \in \mathbb{C}: \operatorname{Re} z>0\}. Question: \min _{z \in S}|1-3 i-\mathrm{z}|=

JEE Advanced · Mathematics · 9. Straight Lines

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:

$\min _{z \in S}|1-3 i-\mathrm{z}|=$

A $\frac{2 - \sqrt{3}}{2}$
B $\frac{2 + \sqrt{3}}{2}$
C $\frac{3 - \sqrt{3}}{2}$
D $\frac{3 + \sqrt{3}}{2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{3 - \sqrt{3}}{2}$

Step-by-step Solution

Detailed explanation

min_{z \in s} |1 - 3 i - z|

perpendicular length of point (1, -3) from line

\sqrt{3} x + y = 0

|\frac{\sqrt{3} - 3}{\sqrt{3 + 1}}| = |\frac{\sqrt{3} - 3}{2}| = \frac{3 - \sqrt{3}}{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let $k$ be a positive real number and let $\begin{aligned} A & =\left[\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1\end{array}\right] \text { and } \\ B & =\left[\begin{array}{ccc}0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0\end{array}\right]\end{aligned}$
If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then $[k]$ is equal to
[Note : adj $M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$ ].

JEE Advanced 2010 Easy

Let

E_{1} =

{

x \in ℝ : x \neq 1

and

\frac{x}{x - 1} > 0

} and

E_{2} =

{

x \in E_{1} : \sin^{- 1} (\log_{e} (\frac{x}{x - 1}))

is a real number} (Here the inverse trigonometric function

\sin^{- 1} (x)

assumes values in

[- \frac{π}{2} . \frac{π}{2}]

.)
Let

f : E_{1} \to ℝ

be the function defined by

f (x) = \log_{e} (\frac{x}{x - 1})

And

g : E_{2} \to R

be the function defined by

g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1}))

LIST-I	LIST-II
A. The range of $f$ is	P. $(- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty)$
B. The range of $g$ contains	Q. (0, 1)
C. The domain of $f$ contains	R. $[- \frac{1}{2}, \frac{1}{2}]$
D. The domain of $g$ is	S. $(- \infty, 0) \cup (0, \infty)$
	T. $(- \infty, \frac{e}{e - 1}]$
	U. $(- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}]$

The correct option is:

JEE Advanced 2018 Hard

Explore more questions on app

From JEE Advanced

	List- $I$		List- II
$(I)$	$2 h + k$	$(P)$	$6$
$(I I)$	$\frac{L e n g t h o f Z W}{L e n g t h o f X Y}$	$(Q)$	$\sqrt{6}$
$(I I I)$	$\frac{A r e a o f t r i a n g l e M Z N}{A r e a o f t r i a n g l e Z M W}$	$(R)$	$\frac{5}{4}$
$(I V)$	$α$	$(S)$	$\frac{21}{5}$
		$(T)$	$2 \sqrt{6}$
		$(U)$	$\frac{10}{3}$

Explore more questions on app