JEE Main 2023 · 11 April · evening · Single correct · +4 / −1

If the 1011ᵗʰ term from the end in the binominal expansion of ((4 x)/5-5/(2 x))²⁰²² is 1024 times 1011ᵗʰR term from the beginning, then |x| is equal to

  1. (a) 5/16
  2. (b) 8
  3. (c) 12
  4. (d) 15
Answer and solution

Answer (a) 5/16

Solution

Maths sir · 4 steps

Turn the term from the end into a term from the beginning, then divide: the binomial coefficients cancel.

  1. Locate the term. With n = 2022 there are 2023 terms. The 1011ᵗʰ term from the end is term number 2023 - 1011 + 1 = 1013 from the beginning. Write a = 4x/5 and b = -5/2x.
  2. Write both terms. T₁₀₁₃ = C(2022, 1012) a¹⁰¹⁰ b¹⁰¹² and T₁₀₁₁ = C(2022, 1010) a¹⁰¹² b¹⁰¹⁰. Since C(2022, 1012) = C(2022, 1010), the coefficients cancel when you divide.

    T₁₀₁₃/T₁₀₁₁ = (b/a)² = (-25/8x²)² = 625/64x⁴ = 1024

  3. Solve for x. x⁴ = 625/(64 × 1024) = 625/65536 = 5⁴/16⁴, so |x| = 5/16.
  4. So. |x| = 5/16, which is option (a).

The idea to keep. The rᵗʰ term from the end of (a+b)ⁿ is the (n-r+2)ᵗʰ from the start. Terms equally far from both ends share a coefficient, so their ratio is just a power of b/a.

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 2023.