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JEE Advanced Mathematics Sequences and Series 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Mathematics Question (2021) — Solution

Question

Paragraph: Let M= \ (x, y) R R : x^ 2 +y^ 2 r^ 2 \ , where r>0 . Consider the geometric progression a_ n = 1 2^ n-1 , n=1,2,3, . Let S_ 0 =0 and, for n 1, let S_ n denote the sum of the first n terms of this progression. For n 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 . Question: Consider M with r= 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

Options

  1. A. k + 2 l = 22
  2. B. 2 k + l = 26
  3. C. 2 k + 3 l = 34
  4. D. 3 k + 2 l = 40

Answer

D. 3 k + 2 l = 40

Step-by-step solution

S n = 1 + 1 2 + 1 2 2 + … + 1 2 n - 1 = 2 1 - 1 2 n = 2 - 1 2 n - 1 Centre of C n is 2 - 1 2 n - 2 , 0 and radius of C n is 1 2 n - 1 when r = 1025   513 < 2 C n will lie inside  m when 2 - 1 2 n - 2 + 1 2 n - 1 < 1025   513 ⇒ 1 - 1 2 n < 1025 1026 ⇒ 2 n < 1026 ⇒ n ≤ 10 Hence number of circles ⇒ k = 10 Also ℓ = 5 3 k + 2 ℓ = 30 + 10 = 40

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