Verified Solution • 2025–26 Session moderate

QJEEM21Ma7012: JEE Main 2021 Mathematics - Class 11 JEEM21Ma7012 Solution: Let C be the set of all complex numbers. Let S 1 = {z C :

Let C be the set of all complex numbers. Let

S1 = {z ∈\in C : |z −- 2| ≤\le 1} and

S2 = {z ∈\in C : z(1 + i) + z‾\overline z (1 −- i) ≥\ge 4}.

Then, the maximum value of ∣z−52∣2{\left| {z - {5 \over 2}} \right|^2} for z ∈\in S1 ∩\cap S2 is equal to :

Multiple Choice Options

  • A 3+224{{3 + 2\sqrt 2 } \over 4}
  • B 5+222{{5 + 2\sqrt 2 } \over 2}
  • C 3+222{{3 + 2\sqrt 2 } \over 2}
  • D 5+224{{5 + 2\sqrt 2 } \over 4}

Step-by-Step Verified Solution

|t −- 2| ≤\le 1

Put t = x + iy

JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Complex Numbers Question 96 English Explanation

(x −- 2)2 + y2 ≤\le 1

Also, t(1 + i) + t‾\overline t (1 −- i) ≥\ge 4

x −- y ≥\ge 2

Let point on circle be A(2 + cos θ\theta , sin θ\theta )

θ∈[−3π4,π4]\theta \in \left[ { - {{3\pi } \over 4},{\pi \over 4}} \right]

(AP)2=(2+cos⁡θ−52)2+sin⁡θ{(AP)^2} = {\left( {2 + \cos \theta - {5 \over 2}} \right)^2} + \sin \theta

=cos⁡2θ−cos⁡θ+14+sin⁡2θ= {\cos ^2}\theta - \cos \theta + {1 \over 4} + {\sin ^2}\theta

=54−cos⁡θ= {5 \over 4} - \cos \theta

For (AP)2 maximum θ=−3π4\theta = - {{3\pi } \over 4}

(AP)2=54+12=52+442{(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2 + 4} \over {4\sqrt 2 }}

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