$M=\left\{(x, y) \in R \times R: x^{2}+y^{2} \leq r^{2}\right\}$,
where \(r\)\(>\)\(0\). Consider the geometric progression $a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots $ 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 $\left(S_{n-1}, 0\right)$ and radius $a_{n}$, and $D_{n}$ denote the circle with center $\left(S_{n-1}, S_{n-1}\right)$ 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 $1$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect Then
- $k+2 l=22$
- $2 k +1=26$
- $2 k +3 l =34$
- $3 k +2 l =40$
The Correct Option is D
Solution and Explanation
\(a_n = \frac{1}{2^{n-1}}\)
And \(S_n = 2\left(1 - \frac{1}{2^n}\right)\)
For circles Cn to be inside M.
\(S_{n-1} + a_n < \frac{1025}{513}\)
\(⇒\) \(S_n < \frac{1025}{513}\)
\(⇒\) \(1 - \frac{1}{2^n} < \frac{1025}{1026}\)
\(⇒\) \(1 - \frac{1}{1026}\)
\(⇒\) \(2^n < 1026\)
\(⇒ n ≤ 10\)
∴ Number of circles inside be 10 = K
It's evident that alternate circles, namely C1, C3, C5, C7, and C9, do not intersect each other, and similarly, C2, C4, C6, C8, and C10 do not intersect each other. Thus, a maximum of 5 sets of circles do not intersect each other.
Hence, l=5.
Therefore, 3K+2l=40.
Thus, Option (D) is the correct answer.
Consider $M$ with $r=\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{198}}$. The number of all those circles $D_{n}$ that are inside $M$ is
- 198
- 199
- 200
- 201
The Correct Option is B
Solution and Explanation
M=199
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