JEE Main 2021 · 25 July · morning · Numerical answer · +4 / −0

The term independent of 'x' in the expansion of
( (x + 1)/(x^(2/3) - x^(1/3) + 1) - (x - 1)/(x - x^(1/2)) )¹⁰, where x ≠ 0, 1 is equal to ______________.

Answer and solution

Answer 210

Solution

Maths sir · 4 steps

Tidy each fraction first: the messy bracket shrinks to just x^(1/3) - x^(-1/2).

  1. First fraction. Put a = x^(1/3), so x + 1 = a³ + 1 = (a + 1)(a² - a + 1). The denominator is exactly a² - a + 1, so the fraction is a + 1 = x^(1/3) + 1.
  2. Second fraction. Put b = x^(1/2), so x - 1 = (b - 1)(b + 1) and x - x^(1/2) = b(b - 1). The fraction is (b + 1)/b = 1 + x^(-1/2).
  3. General term. The bracket becomes (x^(1/3) + 1) - (1 + x^(-1/2)) = x^(1/3) - x^(-1/2). So the general term is Tᵣ₊₁ = C(10, r)(-1)ʳ x^((10 - r)/3 - r/2).

    (10 - r)/3 - r/2 = 0 ⇒ 20 - 5r = 0 ⇒ r = 4

  4. So. The term independent of x is C(10, 4)(-1)⁴ = 210, so the answer is 210.

The idea to keep. When a binomial bracket looks frightening, factor first. Fractional powers usually hide a³ + 1 or b² - 1, and the expression collapses to two simple powers of x.

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 2021.