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← JEE Advanced 2021 · Paper 2
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JEE Advanced 2021 · Paper 2, Mathematics question 14

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Let
M={(x,y)∈R×R:x2+y2≤r2},M = \{(x, y) \in \mathbb R \times \mathbb R : x^2 + y^2 \le r^2\},
where r>0r > 0. Consider the geometric progression an=12n−1a_n = \dfrac1{2^{n-1}}, n=1,2,3,…n = 1, 2, 3, \ldots. Let S0=0S_0 = 0 and, for n≥1n \ge 1, let SnS_n denote the sum of the first nn terms of this progression. For n≥1n \ge 1, let CnC_n denote the circle with center (Sn−1,0)(S_{n-1}, 0) and radius ana_n, and DnD_n denote the circle with center (Sn−1,Sn−1)(S_{n-1}, S_{n-1}) and radius ana_n.

Consider MM with r=(2199−1)22198r = \dfrac{(2^{199} - 1)\sqrt2}{2^{198}}. The number of all those circles DnD_n that are inside MM is

Choose one option
Answer and worked solution

Official answer: B

Condition for DnD_n. Its centre is at distance 2Sn−1\sqrt2S_{n-1} from the origin, so it lies inside M when
2Sn−1+an≤r=2(2−12198).\sqrt2S_{n-1}+a_n\le r=\sqrt2\left(2-\frac1{2^{198}}\right).

Simplify. With Sn−1=2−12n−2S_{n-1}=2-\frac1{2^{n-2}}, this becomes
−22n−2+12n−1≤−22198  ⟺  2⋅2 n−199≤22−1≈1.83.-\frac{\sqrt2}{2^{n-2}}+\frac1{2^{n-1}}\le-\frac{\sqrt2}{2^{198}}\iff\sqrt2\cdot2^{\,n-199}\le2\sqrt2-1\approx1.83.
That holds exactly when n≤199n\le199.

Answer. 199 circles. (B)