JEE Main 2023 · 6 April · evening · Single correct · +4 / −1

If the coefficient of x⁷ in ( ax² + 1/2bx )¹¹ and x⁻⁷ in ( ax - 1/3bx² )¹¹ are equal, then :

  1. (a) 243ab = 64
  2. (b) 32ab = 729
  3. (c) 64ab = 243
  4. (d) 729ab = 32
Answer and solution

Answer (d) 729ab = 32

Solution

Maths sir · 4 steps

Write the general term of each expansion, match the power of x, then equate the two coefficients.

  1. First expansion. Tᵣ₊₁ = C(11, r)(ax²)¹¹⁻ʳ(1/2bx)ʳ = C(11, r)a¹¹⁻ʳ/(2b)ʳ x²²⁻³ʳ. For x⁷ we need 22-3r=7, so r=5. The coefficient is C(11, 5)a⁶/32b⁵.
  2. Second expansion. Tᵣ₊₁ = C(11, r)(ax)¹¹⁻ʳ(-1/3bx²)ʳ = C(11, r)((-1)ʳ a¹¹⁻ʳ)/(3b)ʳ x¹¹⁻³ʳ. For x⁻⁷ we need 11-3r=-7, so r=6. The sign is (-1)⁶=+1, and the coefficient is C(11, 6)a⁵/729b⁶.
  3. Equate them. Since C(11, 5)=C(11, 6)=462, the binomial coefficients cancel.

    a⁶/32b⁵ = a⁵/729b⁶ ⇒ a/32 = 1/729b

  4. So. Cross-multiplying gives 729ab = 32, which is option (d).

The idea to keep. For any power of x, write Tᵣ₊₁, collect the x-exponent as a function of r and solve for r. Check the sign (-1)ʳ, and remember that C(n, r)=C(n, n-r) often cancels.

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.