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Home » QnA » Number of elements where n belongs to N and Sn Intersection Tn is phi

Number of elements where n belongs to N and Sn Intersection Tn is phi

For n \in N, Let {S_n} = \left\{ {z \in C:|z - 3 + 2i| = \frac{n}{4}} \right\} and {T_n} = \left\{ {z \in C:|z - 2 + 3i| = \frac{1}{n}} \right\}

Then the number of elements in the set \left\{ {n \in N:{S_n} \cap {T_n} = \phi } \right\} is

(A) 0
(B) 2
(C) 3
(D) 4

Solution:

Tip for solving this question:

Firstly Identify the given sets

According to the set type, apply intersection properties to find no. of elements.

Step 1 of 2:

Let z = x + iy

Given, {S_n} = \left\{ {z \in C:|z - 3 + 2i| = \frac{n}{4}} \right\}

\begin{gathered} |z - 3 + 2i| = \frac{n}{4} \hfill \\ |x + iy - 3 + 2i| = \frac{n}{4} \hfill \\ |(x - 3) + (y + 2)i| = \frac{n}{4} \hfill \\ {S_n}:{(x - 3)^2} + {(y + 2)^2} = {\left( {\frac{n}{4}} \right)^2} \hfill \\  \end{gathered}

Represents a circle with centre {C_1}(3, - 2) \text{ and radius } {r_1} = \frac{n}{4}

Also, {T_n} = \left\{ {z \in C:|z - 2 + 3i| = \frac{1}{n}} \right\}

\begin{gathered} |z - 2 + 3i| = \frac{1}{n} \hfill \\ |x + iy - 2 + 3i| = \frac{1}{n} \hfill \\ |(x - 2) + (y + 3)i| = \frac{1}{n} \hfill \\ {T_n}:{(x - 2)^2} + {(y + 3)^2} = {\left( {\frac{1}{n}} \right)^2} \hfill \\  \end{gathered}

Represents a circle with centre {C_2}(2, - 3) \text{ and radius } {r_2} = \frac{1}{n}

Step 2 of 2:

We have \left\{ {n \in N:{S_n} \cap {T_n} = \phi } \right\}

\text{ i.e.} {S_n} \cap {T_n} = \phi

So, distance between centre {C_1}(3, - 2) \text{ and } {C_2}(2, - 3)>{r_1} + {r_2}

\Rightarrow \sqrt {{{(2 - 3)}^2} + {{( - 1 - (2))}^2}} > \frac{n}{4} + \frac{1}{n}

{\because \text{ Distance between } ({x_1},{x_2})\& ({y_1},{y_2}) = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} }

\begin{gathered} \Rightarrow \sqrt {1 + 1} > \frac{{{n^2} + 4}}{{4n}} \hfill \\ \Rightarrow \sqrt 2 > \frac{{{n^2} + 4}}{{4n}} \hfill \\ \Rightarrow 4n\sqrt 2 > {n^2} + 4 \hfill \\ \Rightarrow {n^2} - 4n\sqrt 2 + 4 < 0 \hfill \\ \Rightarrow {n^2} - 4n\sqrt 2 + 8 - 4 < 0 \hfill \\ \Rightarrow {(n - 2\sqrt 2 )^2} - {(2)^2} < 0 \hfill \\ \Rightarrow (n - 2\sqrt 2 + 2)(n - 2\sqrt 2 - 2) < 0 \hfill \\ \Rightarrow (n - 2\sqrt 2 - 2) < 0 \hfill \\ \Rightarrow n < 2 + 2\sqrt 2 \hfill \\ \end{gathered}

Here, n = 1, 2, 3, 4 only satisfies the above equation.

Therefore, No. of elements in sets = 4

Final Answer:

Hence, Option (D) is correct.

« α^2021 + β^2021 + γ^2021 + δ^2021 for x^4 + x^3 + x^2 + x + 1 = 0
Lim n →∞, sqrt(n2-n-1)+nα+β=0 then 8(α+β) »


Filed Under: QnA Tagged With: JEE Main 2022, July 25th Shift 1

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