JEE Main 2025 · 2 April · morning · Single correct · +4 / −1

The term independent of x in the expansion of ((x+1)/(x^(2 / 3)+1-x^(1 / 3))-(x-1)/(x-x^(1 / 2)))¹⁰, x>1, is :

  1. (a) 240
  2. (b) 120
  3. (c) 150
  4. (d) 210
Answer and solution

Answer (d) 210

Solution

Maths sir · 5 steps

Clean up both fractions first. The bracket shrinks to two simple powers of x.

  1. First fraction. Write x + 1 = (x^(1/3))³ + 1³. The sum of cubes factorises as (x^(1/3) + 1)(x^(2/3) - x^(1/3) + 1), and the second factor is the denominator. So the first fraction is x^(1/3) + 1.
  2. Second fraction. Write x - 1 = (√x - 1)(√x + 1) and x - √x = √x(√x - 1). Cancel √x - 1, which is not zero since x > 1. That leaves (√x + 1)/√x = 1 + x^(-1/2).
  3. Simplified expansion. The difference is (x^(1/3) + 1) - (1 + x^(-1/2)) = x^(1/3) - x^(-1/2). So you are expanding (x^(1/3) - x^(-1/2))¹⁰.

    Tᵣ₊₁ = C(10, r)(x^(1/3))¹⁰⁻ʳ(-x^(-1/2))ʳ = (-1)ʳC(10, r) x^((10-r)/3 - r/2)

  4. Power of x zero. Set (10-r)/3 - r/2 = 0. Multiplying by 6 gives 20 - 2r - 3r = 0, so r = 4.
  5. So. The term is (-1)⁴C(10, 4) = (10 × 9 × 8 × 7)/24 = 210. The answer is (d).

The idea to keep. When the bracket looks messy, factorise first: sums of cubes and differences of squares often collapse it to two powers of x. Then set the power in the general term to zero.

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 2025.