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JEE Advance - Mathematics (2021 - Paper 2 Online - No. 14)

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 $$r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}$$. The number of all those circles Dn that are inside M is
198
199
200
201

Explanation

$$\because$$ $$r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}$$

Now, $$\sqrt 2 {s_{n - 1}} + {a_n} < \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2 $$

$$2\sqrt 2 \left( {1 - {1 \over {{2^{n - 1}}}}} \right) + {1 \over {{2^{n - 1}}}} < \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2 $$

$$\therefore$$ $$2\sqrt 2 - {{\sqrt 2 } \over {{2^{n - 2}}}} + {1 \over {{2^{n - 1}}}} < 2\sqrt 2 - {{\sqrt 2 } \over {{2^{198}}}}$$

$${1 \over {{2^{n - 2}}}}\left( {{1 \over 2} - \sqrt 2 } \right) < - {{\sqrt 2 } \over {{2^{198}}}} \Rightarrow {{2\sqrt 2 - 1} \over {2\,.\,{2^{n - 2}}}} > {{\sqrt 2 } \over {{2^{198}}}}$$

$$ \Rightarrow {2^{n - 2}} < \left( {2 - {1 \over {\sqrt 2 }}} \right)\,.\,{2^{197}}$$

$$\therefore$$ n $$\le$$ 199

So, number of circles = 199

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