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JEE Advance - Mathematics (2021 - Paper 2 Online - No. 13)

Let $$M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\} $$, where r > 0. Consider the geometric progression $${a_n} = {1 \over {{2^{n - 1}}}}$$, n = 1, 2, 3, ...... . Let S0 = 0 and for n $$\ge$$ 1, let Sn denote the sum of the first n terms of this progression. For n $$\ge$$ 1, let Cn denote the circle with center (Sn$$-$$1, 0) and radius an, and Dn denote the circle with center (Sn$$-$$1, Sn$$-$$1) and radius an.
Let $$M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\} $$, where r > 0. Consider the geometric progression $${a_n} = {1 \over {{2^{n - 1}}}}$$, n = 1, 2, 3, ...... . Let S0 = 0 and for n $$\ge$$ 1, let Sn denote the sum of the first n terms of this progression. For n $$\ge$$ 1, let Cn denote the circle with center (Sn$$-$$1, 0) and radius an, and Dn denote the circle with center (Sn$$-$$1, Sn$$-$$1) and radius an.
Let $$M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\} $$, where r > 0. Consider the geometric progression $${a_n} = {1 \over {{2^{n - 1}}}}$$, n = 1, 2, 3, ...... . Let S0 = 0 and for n $$\ge$$ 1, let Sn denote the sum of the first n terms of this progression. For n $$\ge$$ 1, let Cn denote the circle with center (Sn$$-$$1, 0) and radius an, and Dn denote the circle with center (Sn$$-$$1, Sn$$-$$1) and radius an.
Consider M with $$r = {{1025} \over {513}}$$. Let k be the number of all those circles Cn that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then
k + 2l = 22
2k + l = 26
2k + 3l = 34
3k + 2l = 40

Explanation

$$\because$$ $${a_n} = {1 \over {{2^{n - 1}}}}$$ and $${S_n} = 2\left( {1 - {1 \over {{2^n}}}} \right)$$

For circle Cn to be inside M.

$${S_{n - 1}} + {a_n} < {{1025} \over {513}} \Rightarrow {S_n} < {{1025} \over {513}}$$

$$ \Rightarrow 2\left( {1 - {1 \over {{2^n}}}} \right) < {{1025} \over {513}} \Rightarrow 1 - {1 \over {{2^n}}} < {{1025} \over {1026}}$$

$$ \Rightarrow 1 - {1 \over {{2^n}}} < 1 - {1 \over {1026}} \Rightarrow {{ - 1} \over {{2^n}}} < {{ - 1} \over {1026}}$$

$$ \Rightarrow {1 \over {{2^n}}} > {1 \over {1026}} \Rightarrow {2^n} < 1026 \Rightarrow n \le 10$$

$$\therefore$$ Number of circles inside be 10 = k. Clearly, alternate circle do not intersect each other i.e. C1, C3, C5, C7, C9 do not intersect each other as well as C2, C4, C6, C8 and C10 do not intersect each other.

Hence, maximum 5 set of circles do not intersect each other.

$$\therefore$$ l = 5

So, 3k + 2l = 40

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