JEE Main 2023 · 10 April · morning · Single correct · +4 / −1

If the coefficient of x⁷ in ( ax - 1/bx² )¹³ and the coefficient of x⁻⁵ in ( ax + 1/bx² )¹³ are equal, then a⁴b⁴ is equal to :

  1. (a) 22
  2. (b) 33
  3. (c) 44
  4. (d) 11
Answer and solution

Answer (a) 22

Solution

Maths sir · 4 steps

Write one general term, find the power of x, and the right r gives each coefficient directly.

  1. General term. In (ax - 1/bx²)¹³ the general term is C(13, r)(ax)¹³⁻ʳ(-1/bx²)ʳ. Collecting powers, this is C(13, r)(-1)ʳa¹³⁻ʳ/bʳ x¹³⁻³ʳ. The second expansion has the same term without the (-1)ʳ.
  2. Coefficient of x^7. Set 13 - 3r = 7, so r = 2. The sign is (-1)² = +1, and the coefficient is C(13, 2)a¹¹/b² = 78 a¹¹/b².
  3. Coefficient of x^-5. Set 13 - 3r = -5, so r = 6. The coefficient is C(13, 6)a⁷/b⁶ = 1716 a⁷/b⁶.

    78 a¹¹/b² = 1716 a⁷/b⁶ ⇒ a⁴b⁴ = 1716/78

  4. So. 1716/78 = 22, so a⁴b⁴ = 22, which is option (a).

The idea to keep. For a coefficient question, write the general term once, collect the power of x as a function of r, and solve for r. Do not forget the sign that comes with (-1)ʳ.

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.