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JEE Advanced Mathematics Limits 2019 JEE Advanced 2019 (Paper 2)

JEE Advanced Mathematics Question (2019) — Solution

Question

For a ∈ R ,   a > 1 , let lim n → ∞ ⁡ 1 + 2 3 + … + n 3 n 7 / 3 1 a n + 1 2 + 1 a n + 2 2 + … + 1 a n + n 2 = 54 . Then the possible value(s) of a is/are:

Options

  1. A. 8
  2. B. - 9
  3. C. - 6
  4. D. 7

Answer

B. - 9

Step-by-step solution

Let S = lim n → ∞ ⁡ 1 + 2 1 3 + 3 1 3 + … + n 1 3 n 7 3 1 a n + 1 2 + … + 1 a n + n 2 ⇒ S = lim n → ∞ ⁡ ∑ r = 1 n r 1 3 n 7 3 ∑ r = 1 n 1 n a + r 2 S = lim n → ∞ ⁡ ∑ r = 1 n r n 1 3 1 n × n 1 3 n 7 3 ∑ r = 1 n 1 a + r n 2 1 n 2 S = lim n → ∞ ⁡ ∑ r = 1 n r n 1 3 . 1 n ∑ r = 1 n 1 a + r n 2 1 n S = ∫ 0 1 x 1 3 d x ∫ 0 1 1 a + x 2 . d x S = x 4 3 4 3 0 1 - 1 x + a 0 1 S = 3 4 1 a - 1 a + 1 = 54 1 a a + 1 = 1 72 a 2 + a - 72 = 0 a = 8 or a = - 9

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