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JEE Advanced 2021 Paper 2

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Let

M=\{(x, y) \in \mathbb{R}\times \mathbb{R}: x^{2}+y^{2} \le r^{2}\}

where

r>0

.

Consider the geometric progression

a_{n}=\;\frac{1}{2^{n-1}},\ n=1,2,3,\dots

Let

S_{0}=0

and, for

n \ge 1

, let

S_{n}

denote the sum of the first

n

terms of this progression. For

n \ge 1

, let

C_{n}

denote the circle with center

(S_{n-1}, 0)

and radius

a_{n}

, and

D_{n}

denote the circle with center

(S_{n}

-1, S_{n-1}

) and radius

a_{n}

.

Section: Mathematics

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Question : 52 of 57

Marks: +1, -0

Consider

M

r=\frac{(2^{199}-1)\sqrt{2}}{2^{198}}.

The number of all those circles

D_{n}

that are inside

M

[JEE Adv 2021 P2]

198
199
200
201

Solution:

\because \;\; r=\frac{(2^{199}-1)\sqrt{2}}{2^{198}}

Now,

\sqrt{2} S_{n-1}+a_{n} < \frac{2^{199}-1}{2^{198}} \sqrt{2}

\Rightarrow \;\; 2\sqrt{2}\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2^{n-1}} < \frac{2^{199}-1}{2^{198}}

\Rightarrow \;\; \frac{2\sqrt{2}-1}{2\cdot 2^{n-2}} > \frac{\sqrt{2}}{2^{198}} \Rightarrow 2^{n-2} < \left(2-\frac{1}{\sqrt{2}}\right) 2^{197}

\therefore \;\; n \leq 199 \Rightarrow n=199

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