Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z - 1|\ . Then the set of all values of x , for which w = 2x + iy ∈ S for some y ∈ ℝ , is

Correct answer: B. (- 1 2 , (1)/(4)] Given, |z - 1 + i| ≥ |z|, |z| Now solving |z - 1 + i| ≥ |z| put z = x + iy we get, x - y ≤ 1 so plotting we get, Now solving |z| Now solving |z + i| = |z - 1| we get x + y = 0 Now combine diagram of x + y = 0 and x - y ≤ 1 will be, Now combining all diagram we get, Hence w = 2x + iy ∈ S Now for w ∈ S , 2x ≤ (1)/(2) so x ≤ (1)/(4) Now 2x + y = 0 ⇒ y = - 2x putting the value in x ^ 2 + y ^ 2 ⇒ (2x) ^ 2 + (2x) ^ 2 ⇒ x ^ 2 So, x ∈ ( - 1 2 , (1)/(4)]

JEE Main 202229 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2022 · 29 July · Evening Shift · Mathematics · Complex Number

Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z -...

Let

S = \left\{z = x + iy: \left|z - 1 + i\right| \ge \left|z\right|, \left|z\right| < 2, \left|z + i\right| = \left|z - 1\right|\right\}

. Then the set of all values of

x

, for which

w = 2x + iy \in S

for some

y \in \mathbb{R}

, is $Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z - 1|\ . Then the set of all values of x , for which w =$ $Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z - 1|\ . Then the set of all values of x , for which w =$ $Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z - 1|\ . Then the set of all values of x , for which w =$ $Let S = \ z = x + iy: |z - 1 + i| ≥ |z|, |z| < 2, |z + i| = |z - 1|\ . Then the set of all values of x , for which w =$

Options

A $\left(- \sqrt{2}, \dfrac{1}{2\sqrt{2}}\right]$
B $\left(- \dfrac{1}{\sqrt{2}}, \dfrac{1}{4}\right]$
C $\left(- \sqrt{2}, \dfrac{1}{2}\right]$
D $\left(- \dfrac{1}{\sqrt{2}}, \dfrac{1}{2\sqrt{2}}\right]$

Correct answer

B. $\left(- \dfrac{1}{\sqrt{2}}, \dfrac{1}{4}\right]$

Step-by-step solution

Given,
$\left|z - 1 + i\right| \ge \left|z\right|, \left|z\right| < 2 \text{ and } \left|z + i\right| = \left|z - 1\right|$
Now solving \left|z - 1 + i\right| \ge \left|z\right| $put$ z = x + iy $we get,$ x - y \le 1 $so plotting we get,<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/924c118f-9058-4099-9f23-f896255d8fd6-Capture.PNG" style="width: 200px; height: 149px;" />Now solving$ \left|z\right| < 2 $we get,$ {x}^{2} + {y}^{2} < 4 $on plotting the diagram we get,<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b0a0c222-779d-4770-b490-97b4ca7fcb50-Capture1.PNG" style="width: 220px; height: 139px;" />Now solving$ \left|z + i\right| = \left|z - 1\right| $we get$ x + y = 0 $Now combine diagram of$ x + y = 0 \text{ and } x - y \le 1 $will be,<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/23f41058-595a-47a6-bc76-7b2f5099d853-Capture.PNG" style="width: 200px; height: 215px;" />Now combining all diagram we get,<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/1ba02c8c-baf0-4a3f-abc4-013fd97113fa-Capture1.PNG" style="width: 250px; height: 320px;" />Hence$ w = 2x + iy \in S $Now for$ w \in S $,$ 2x \le \dfrac{1}{2} $so$ x \le \dfrac{1}{4} $Now$ 2x + y = 0 ⇒ y = - 2x $putting the value in$ {x}^{2} + {y}^{2} < 4 $$ ⇒ {\left(2x\right)}^{2} + {\left(2x\right)}^{2} < 4 $$ ⇒ {x}^{2} < \dfrac{1}{2} ⇒ x \in \left(\dfrac{- 1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) $So,$ x \in \left(\dfrac{- 1}{\sqrt{2}}, \dfrac{1}{4}\right]

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