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JEE Main 2018

JEE Main 2018 Mathematics question on Binomial Theorem

The sum of the coefficients of all odd degree terms in the expansion of (x+x3−1)5+(x−x3−1)5\left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5, (x>1)(x > 1) is
  1. A.
    -1
  2. B.
    0
  3. C.
    1
  4. D.
    2
Show answer and solution

Correct option: D

Solution

Using the binomial theorem, (a+b)5 + (a-b)5 = 2[a5 + 10a³b² + 5ab⁴] where a = x and b = √(x³-1).

This gives 2[x5 + 10x³(x³-1) + 5x(x³-1)²] = 2[x5 + 10x6 - 10x³ + 5x(x6 - 2x³ + 1)] = 2[x5 + 10x6 - 10x³ + 5x7 - 10x4 + 5x].

Expanding: 2[5x7 + 10x6 + x5 - 10x4 - 10x³ + 5x] = 10x7 + 20x6 + 2x5 - 20x4 - 20x³ + 10x.

The odd-degree terms are those with x7, x5, x³, x: coefficients are 10, 2, -20, 10. Their sum = 10 + 2 - 20 + 10 = 2.

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