JEE Main 2022 · 28 June · evening · Single correct · +4 / −1

The term independent of x in the expansion of
(1 - x² + 3x³)( 5/2x³ - 1/5x² )¹¹, x ≠ 0 is :

  1. (a) 7/40
  2. (b) 33/200
  3. (c) 39/200
  4. (d) 11/50
Answer and solution

Answer (b) 33/200

Solution

Maths sir · 4 steps

Find which powers of x the binomial can give, then pair each term of 1 - x² + 3x³ with its partner.

  1. General term. In (5/2x³ - 1/5x²)¹¹, the power of x in Tᵣ₊₁ is 3(11-r) - 2r = 33 - 5r.

    Tᵣ₊₁ = C(11, r)(5/2)¹¹⁻ʳ(-1/5)ʳ x³³⁻⁵ʳ

  2. Match the powers. The 1 needs x⁰, so 33 - 5r = 0, which has no whole-number r. The 3x³ needs x⁻³, so 33 - 5r = -3, also not whole. Only the -x² works: it needs x⁻², so 33 - 5r = -2 and r = 7.
  3. Find the coefficient. At r = 7: C(11, 7)(5/2)⁴(-1/5)⁷ = 330 × 625/16 × (-1/78125) = -330/2000 = -33/200.
  4. So. Multiply by the -1 in front of x²: the constant term is (-1)×(-33/200) = 33/200, option (b).

The idea to keep. With a polynomial multiplying a binomial power, find the power of x in the general term first. Then each term xᵏ of the polynomial needs x⁻ᵏ from the binomial, and r must be a whole number.

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 2022.