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QJEEM22Ma3f8e: JEE Main 2022 Mathematics - Class 11 JEEM22Ma3f8e Solution: Let and . Then, for and , the least value of is :

Let S1={z1∈C:∣z1−3∣=12}S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\} and S2={z2∈C:∣z2−∣z2+1∣∣=∣z2+∣z2−1∣∣}S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\} . Then, for z1∈S1z_{1} \in S_{1} and z2∈S2z_{2} \in S_{2} , the least value of ∣z2−z1∣\left|z_{2}-z_{1}\right| is :

Multiple Choice Options

  • A 0
  • B 12\frac{1}{2}
  • C 32\frac{3}{2}
  • D 52\frac{5}{2}

Step-by-Step Verified Solution

∵\because ∣Z2+∣Z2−1∣∣2=∣Z2−∣Z2+1∣∣2{\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}

⇒(Z2+∣Z2−1∣)(Z‾2+∣Z2−1∣)=(Z2−∣Z2+1∣)(Z‾2−∣Z2+1∣)\Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)

⇒Z2(∣Z2−1∣+∣Z2+1∣)+Z‾2(∣Z2−1∣+∣Z2+1∣)=∣Z2+1∣2−∣Z2−1∣2\Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}

⇒(Z2+Z‾2)(∣Z2+1∣+∣Z2−1∣)=2(Z2+Z‾2)\Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)

⇒\Rightarrow Either Z2+Z‾2=0{Z_2} + {\overline Z _2} = 0 or ∣Z2+1∣+∣Z2−1∣=2|{Z_2} + 1| + |{Z_2} - 1| = 2

So, Z2 lies on imaginary axis or on real axis within [−1,1][ - 1,1]

Also ∣Z1−3∣=12⇒Z1|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1} lies on the circle having center 3 and radius 12{1 \over 2} .

JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Complex Numbers Question 62 English Explanation

Clearly ∣Z1−Z2∣min⁡=32|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}

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