Verified Solution • 2025–26 Session moderate

QJEEM22Maa321: JEE Main 2022 Mathematics - Class 11 JEEM22Maa321 Solution: For , let and . Then the number of elements in the set is :

For n∈N\mathrm{n} \in \mathbf{N} , let Sn={z∈C:∣z−3+2i∣=n4}\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\} and Tn={z∈C:∣z−2+3i∣=1n}\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\} . Then the number of elements in the set {n∈N:Sn∩Tn=ϕ}\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\} is :

Multiple Choice Options

  • A 0
  • B 2
  • C 3
  • D 4

Step-by-Step Verified Solution

Sn={z∈C:∣z−3+2i∣=n4}S_{n}=\left\{z \in C:|z-3+2 i|=\frac{n}{4}\right\} represents a circle with centre C1(3,−2)C_{1}(3,-2) and radius r1=n4r_{1}=\frac{n}{4}

Similarly TnT_{n} represents circle with centre C2(2,−3)C_{2}(2,-3) and radius r2=1nr_{2}=\frac{1}{n}

 As Sn∩Tn=ϕ\text { As } S_{n} \cap T_{n}=\phi

C1C2gt;r1+r2amp; OR amp;C1C2lt;∣r1−r2∣2gt;n4+1namp; OR amp;2lt;∣n4−1n∣n=1,2,3,4amp;amp;n may take infinite values \begin{array}{llr}C_{1} C_{2}>r_{1}+r_{2}& \text { OR } & C_{1} C_{2}<\left|r_{1}-r_{2}\right| \\\\\sqrt{2}>\frac{n}{4}+\frac{1}{n} & \text { OR } & \sqrt{2}<\left|\frac{n}{4}-\frac{1}{n}\right| \\\\n=1,2,3,4&& n \text { may take infinite values }\end{array}

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