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JEE Main 2019 · 8 April · Shift I

JEE Main 2019 Mathematics question on Binomial Theorem

The sum of the coefficients of all even degree terms is xx in the expansion of (x+x3−1)6+(x−x3−1)6(x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6, (x>1)(x > 1) is equal to
  1. A.
    29
  2. B.
    32
  3. C.
    26
  4. D.
    24
Show answer and solution

Correct option: D

Solution

When you expand (x + √(x³-1))⁶ + (x - √(x³-1))⁶, the odd-power terms cancel and you get:

2[C(6,0)x⁶ + C(6,2)x⁴(x³-1) + C(6,4)x²(x³-1)² + C(6,6)(x³-1)³]

= 2[x⁶ + 15x⁴(x³-1) + 15x²(x⁶-2x³+1) + (x⁹-3x⁶+3x³-1)]

Expanding:
= 2[x⁶ + 15x⁷ - 15x⁴ + 15x⁸ - 30x⁵ + 15x² + x⁹ - 3x⁶ + 3x³ - 1]

= 2[x⁹ + 15x⁸ + 15x⁷ - 2x⁶ - 30x⁵ - 15x⁴ + 3x³ + 15x² - 1]

The even-degree terms (even powers of x) are: x⁸, x⁶, x⁴, x², and the constant.

Coefficients: 2(15) + 2(-2) + 2(-15) + 2(15) + 2(-1) = 30 - 4 - 30 + 30 - 2 = 24.

The answer is **

[JEE Main 2019 · 8 April · Shift I]

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