JEE Main 2024 · 27 January · morning · Single correct · +4 / −1

ⁿ⁻¹ Cᵣ=(k²-8) ⁿ Cᵣ₊₁ if and only if :

  1. (a) 2 √2<k<2 √3
  2. (b) 2 √2<k ≤ 3
  3. (c) 2 √3<k<3 √3
  4. (d) 2 √3<k ≤ 3 √2
Answer and solution

Answer (b) 2 √2<k ≤ 3

Solution

Maths sir · 4 steps

Write ⁿCᵣ₊₁ in terms of ⁿ⁻¹Cᵣ, and k²-8 turns into the fraction (r+1)/n.

  1. Useful identity. Pull one factor out of the factorials: ⁿCᵣ₊₁ = n/(r+1) ⁿ⁻¹Cᵣ. This links the two coefficients in the question directly.

    k² - 8 = ⁿ⁻¹Cᵣ/ⁿCᵣ₊₁ = (r+1)/n

  2. Range of the ratio. For ⁿ⁻¹Cᵣ to exist, 0 ≤ r ≤ n-1, so 1 ≤ r+1 ≤ n. That makes (r+1)/n always positive and at most 1. It equals 1 exactly when r = n-1.
  3. Bound k. So 0 < k² - 8 ≤ 1, which gives 8 < k² ≤ 9. Taking k positive, 2√2 < k ≤ 3.
  4. So. The condition holds if and only if 2√2 < k ≤ 3, which is option (b).

The idea to keep. Whenever two binomial coefficients with neighbouring n and r appear, take their ratio. It collapses to a simple fraction like (r+1)/n, and the limits on r give the range.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2024.