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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 : 51 of 57

Marks: +1, -0

Consider

M

r=\frac{1025}{513}

k

be the number of all those circles

C_{n}

that are inside

M

l

be the maximum possible number of circles among these

k

circles such that no two circles intersect. Then

[JEE Adv 2021 P2]

$k+2l=22$
$2k+l=26$
$2k+3l=34$
$3k+2l=40$

Solution:

\because a_{n}=\frac{1}{2^{n-1}}

S_{n}=1+\frac{1}{2}+\frac{1}{2^{2}}+\dots+\frac{1}{2^{n-1}}=2\left(1-\frac{1}{2^{n}}\right)=2-\frac{1}{2^{n-1}}

For circles

C_{n}

to inside

M

S_{n-1}+a_{n} < \frac{1025}{513}

\Rightarrow 2-\frac{1}{2^{n-2}}+\frac{1}{2^{n-1}} < \frac{1025}{513}

\Rightarrow 1-\frac{1}{2^{n}} < \frac{1025}{1026}=1-\frac{1}{1026}

\Rightarrow 2^{n}<2026 \Rightarrow n \leq 10 \Rightarrow k=10

Also

l=5

3k+2l=30+10=40

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