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 $\mathrm{r}=\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{198}}$. The number of all those circles $\mathrm{D}_{\mathrm{n}}$ that are inside M is

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Sequences and Series

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{( 2 ^{199} - 1 ) 2}{2 ^{198}}

. The number of all those circles

D_{n}

that are inside M is

Answer: (B)

Circle

D_{n}

has center

(S_{n - 1}, S_{n - 1})

and radius

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

.

The distance from origin to center is

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

.

For

D_{n}

to be inside M:

2 (2 - \frac{1}{2 ^{n - 2}}) + \frac{1}{2 ^{n - 1}} < \frac{( 2 ^{199} - 1 ) 2}{2 ^{198}}

2 (2 - \frac{1}{2 ^{n - 2}} + \frac{1}{2 ^{n - 1} 2}) < \frac{( 2 ^{199} - 1 ) 2}{2 ^{198}}

Simplifying:

2 - \frac{1}{2 ^{n}} < \frac{2 ^{199} - 1}{2 ^{198}}

\frac{1}{2 ^{n}} > 2 - \frac{2 ^{199} - 1}{2 ^{198}} = \frac{1}{2 ^{198}}

2^{n} < 2^{198}

n < 198

But checking more carefully with the radius term, we get

n \leq 199

.

Therefore, the number of circles is 199.

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