Let $$ \mathrm{M}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq \mathrm{r}^{2}\right\}, $$ where $\mathrm{r}>0$. Consider the geometric progression $\mathrm{a}_{\mathrm{n}}=\frac{1}{2^{\mathrm{n}-1}}, \mathrm{n}=1,2,3, \ldots$. Let $\mathrm{S}_{0}=0$ and, for $\mathrm{n} \geq 1$, let $\mathrm{S}_{\mathrm{n}}$ denote the sum of the first n terms of this progression. For $\mathrm{n} \geq 1$, let $\mathrm{C}_{\mathrm{n}}$ denote the circle with center $\left(\mathrm{S}_{\mathrm{n}-1}, 0\right)$ and radius $\mathrm{a}_{\mathrm{n}}$, and $\mathrm{D}_{\mathrm{n}}$ denote the circle with center $\left(\mathrm{S}_{\mathrm{n}-1}, \mathrm{~S}_{\mathrm{n}-1}\right)$ and radius $\mathrm{a}_{\mathrm{n}}$. Consider M with $\mathrm{r}=\frac{1025}{513}$. Let k be the number of all those circles $\mathrm{C}_{\mathrm{n}}$ that are inside M. Let $l$ be the maximum possible number of circles among these k circles such that no two circles intersect. Then

(C) $2 \mathrm{k}+3 \mathrm{l}=34$

Sequences and Series — JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 Question with Solution

Let

M = {(x, y) \in R \times R : x^{2} + y^{2} \leq 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 \geq 1

, let

S_{n}

denote the sum of the first n terms of this progression. For

n \geq 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}

.

Consider M with

r = \frac{1025}{513}

. Let k be the number of all those circles

C_{n}

that are inside M. Let

l

be the maximum possible number of circles among these k circles such that no two circles intersect. Then

Answer: (D)

S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 ^{n - 1}} = 2 (1 - \frac{1}{2 ^{n}}) = 2 - \frac{1}{2 ^{n - 1}}

Circle

C_{n}

has center

(S_{n - 1}, 0) = (2 - \frac{1}{2 ^{n - 2}}, 0)

and radius

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

.

For

C_{n}

to be inside M with

r = \frac{1025}{513} < 2

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

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

\frac{1}{2 ^{n}} > 2 - \frac{1025}{513} = \frac{1}{513}

2^{n} < 513

Since

2^{9} = 512 < 513 < 1024 = 2^{10}

, we have

n \leq 9

. But we need to check more carefully: the rightmost point of

C_{n}

is at

2 - \frac{1}{2 ^{n - 2}} + \frac{1}{2 ^{n - 1}} = 2 - \frac{1}{2 ^{n}}

. For this to be less than

\frac{1025}{513}

, we need

n = 10

. So

k = 10

.

For non-intersecting circles from the sequence

C_{1}, C_{2}, \dots, C_{10}

, we can select every other circle:

C_{1}, C_{3}, C_{5}, C_{7}, C_{9}

, giving

l = 5

3 k + 2 l = 3 (10) + 2 (5) = 30 + 10 = 40

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Single correct option