Let ℂ be the set of all complex numbers. Let S _ 1 = \ z ∈ ℂ: |z - 2| ≤ 1\ and S _ 2 = \ z ∈ ℂ: z(1 + i) + ¯ z (1 - i) ≥ 4\ . Then, the maximum value of |z - (5)/(2)| ^ 2 for z ∈ S _ 1 S _ 2 is equal to :

Correct answer: D. 5 + 2 2 4 Here, |z - 2| ≤ 1 Put z = x + iy (x - 2) ^ 2 + y ^ 2 ≤ 1 Also, z(1 + i) + ¯ z (1 - i) ≥ 4 Gives x - y ≥ 2 Let point on circle be A(2 + cosθ, sinθ) θ ∈ [- (3π)/(4), (π)/(4)] Let P be ((5)/(2), 0) So, |z - (5)/(2)| ^ 2 = (AP) ^ 2 = (2 + cosθ - (5)/(2)) ^ 2 + ^ 2 θ = ^ 2 θ - cosθ + (1)/(4) + ^ 2 θ = (5)/(4) - cosθ For (AP) ^ 2 to be maximum θ = - (3π)/(4) (AP) ^ 2 = (5)/(4) + 1 2 = 5 2 + 4 4 2 = 5 + 2 2 4

JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2021 · 27 July · Evening Shift · Mathematics · Complex Number

Let ℂ be the set of all complex numbers. Let S _ 1 = \ z ∈ ℂ: |z -...

Let

\mathbb{C}

be the set of all complex numbers. Let

{S}_{1} = \left\{z \in \mathbb{C}: \left|z - 2\right| \le 1\right\}

and

{S}_{2} = \left\{z \in \mathbb{C}: z\left(1 + i\right) + \overset{¯}{z}\left(1 - i\right) \ge 4\right\} .

Then, the maximum value of

{\left|z - \dfrac{5}{2}\right|}^{2}

for

z \in {S}_{1} \cap {S}_{2}

is equal to : $Let ℂ be the set of all complex numbers. Let S _ 1 = \ z ∈ ℂ: |z - 2| ≤ 1\ and S _ 2 = \ z ∈ ℂ: z(1 + i) + ¯ z (1 - i)$

Options

A $\dfrac{3 + 2\sqrt{2}}{4}$
B $\dfrac{5 + 2\sqrt{2}}{2}$
C $\dfrac{3 + 2\sqrt{2}}{2}$
D $\dfrac{5 + 2\sqrt{2}}{4}$

Correct answer

D. $\dfrac{5 + 2\sqrt{2}}{4}$

Step-by-step solution

Here, $\left|z - 2\right| \le 1$ Put $z = x + iy$

$Let ℂ be the set of all complex numbers. Let S _ 1 = \ z ∈ ℂ: |z - 2| ≤ 1\ and S _ 2 = \ z ∈ ℂ: z(1 + i) + ¯ z (1 - i)$

${\left(x - 2\right)}^{2} + {y}^{2} \le 1$

Also, $z\left(1 + i\right) + \overset{¯}{z}\left(1 - i\right) \ge 4$

Gives $x - y \ge 2$

Let point on circle be $A\left(2 + cos\theta, sin\theta\right)$

$\theta \in \left[- \dfrac{3\pi}{4}, \dfrac{\pi}{4}\right]$

Let $P$ be \left(\dfrac{5}{2}, 0\right) $So, $ {\left|z - \dfrac{5}{2}\right|}^{2} = {\left(AP\right)}^{2} = {\left(2 + cos\theta - \dfrac{5}{2}\right)}^{2} + \sin^{2}\theta $$ = \cos^{2}\theta - cos\theta + \dfrac{1}{4} + \sin^{2}\theta $$ = \dfrac{5}{4} - cos\theta $For$ {\left(AP\right)}^{2} to be maximum $\theta = - \dfrac{3\pi}{4}$

${\left(AP\right)}^{2} = \dfrac{5}{4} + \dfrac{1}{\sqrt{2}} = \dfrac{5\sqrt{2} + 4}{4\sqrt{2}} = \dfrac{5 + 2\sqrt{2}}{4}$

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