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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2021 · Paper 2 · Mathematics · Sequences and Series

Paragraph: Let M=\ (x, y) ∈ ℝ × ℝ: x^ 2 +y^ 2 ≤ r^ 2 \ , where r>0...

Paragraph: Let M={(x,y)∈R×R:x2+y2≤r2}M=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^{2}+y^{2} \leq r^{2}\right\}, where r>0.r>0 . Consider the geometric progression an=12n−1,n=1,2,3,….a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots . Let S0=0S_{0}=0 and, for n≥1n \geq 1, let SnS_{n} denote the sum of the first nn terms of this progression. For n≥1n \geq 1, let CnC_{n} denote the circle with center \left(S_{n-1}, 0\right)andradiusand radiusa_{n},and, andD_{n}denotethecirclewithcenter(Sn−1,Sn−1)denote the circle with center \left(S_{n-1}, S_{n-1}\right) and radius ana_{n}.
Question: Consider MM with r=1025513r=\frac{1025}{513}. Let kk be the number of all those circles CnC_{n} that are inside MM. Let ll be the maximum possible number of circles among these kk circles such that no two circles intersect. Then

Options

  1. Ak+2l=22k + 2l = 22
  2. B2k+l=262k + l = 26
  3. C2k+3l=342k + 3l = 34
  4. D3k+2l=403k + 2l = 40

Correct answer

D. 3k+2l=403k + 2l = 40

Step-by-step solution

{S}_{n} = 1 + \dfrac{1}{2} + \dfrac{1}{{2}^{2}} + \ldots + \dfrac{1}{{2}^{n - 1}}</p><p></p><p>= 2\left(1 - \dfrac{1}{{2}^{n}}\right) = 2 - \dfrac{1}{{2}^{n - 1}}</p><p>Centreof</p><p>Centre of{C}_{n}is(2−12n−2,0)is \left(2 - \dfrac{1}{{2}^{n - 2}}, 0\right)

and radius of Cn{C}_{n} is 12n−1\dfrac{1}{{2}^{n - 1}}

when r=1025513<2r = \dfrac{1025}{513} < 2

Cn{C}_{n} will lie inside mm

when 2−12n−2+12n−1<10255132 - \dfrac{1}{{2}^{n - 2}} + \dfrac{1}{{2}^{n - 1}} < \dfrac{1025}{513}

⇒1−(12)n<10251026⇒ 1 - {\left(\dfrac{1}{2}\right)}^{n} < \dfrac{1025}{1026}

⇒2n<1026⇒n≤10⇒ {2}^{n} < 1026 ⇒ n \le 10

Hence number of circles

⇒k=10⇒ k = 10

Also ℓ=5ℓ = 5

3k+2ℓ=30+10=403k + 2ℓ = 30 + 10 = 40

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