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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Binomial Theorem

If the fourth term in the binomial expansion of (2x+xlog⁡8x)6\left(\frac{2}{x} + x^{\log_8 x}\right)^6 (x>0)(x > 0) is 20×8720 \times 8^7, then the value of xx is
  1. A.
    8−28^{-2}
  2. B.
    838^3
  3. C.
    88
  4. D.
    828^2
Show answer and solution

Correct option: D

Solution

The fourth term in the expansion is T₄ = C(6,3) · (2/x)³ · (x^(log₈x))³ = 20 · (8/x³) · x^(3log₈x).

Setting this equal to 20 · 8⁷:

(8/x³) · x^(3log₈x) = 8⁷

x^(3log₈x) / x³ = 8⁶

x^(3log₈x - 3) = 8⁶

Let t = log₈x, so x = 8t. Substituting:

8^(t(3t - 3)) = 8⁶

3t² - 3t = 6 → t² - t - 2 = 0 → (t-2)(t+1) = 0

So t = 2 or t = -1, giving x = 8² or x = 8⁻¹. Since x > 0 and checking against the options, x = 8².

[JEE Main 2019 · 9 Apr · Shift I]

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