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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Complex Numbers and Quadratic Equations

The number of real solutions of the equation 3(x2+1x2)−2(x+1x)+5=03\left(x^2 + \dfrac{1}{x^2}\right) - 2\left(x + \dfrac{1}{x}\right) + 5 = 0, is
  1. A.
    3
  2. B.
    4
  3. C.
    0
  4. D.
    2
Show answer and solution

Correct option: C

Solution

Let t = x + 1/x. Then x² + 1/x² = t² - 2.

The equation becomes 3(t² - 2) - 2t + 5 = 0 → 3t² - 2t - 1 = 0 → (3t + 1)(t - 1) = 0, giving t = -1/3 or t = 1.

For real x, x + 1/x ≥ 2 or x + 1/x ≤ -2 (since |x + 1/x| ≥ 2 for x ≠ 0). Neither t = -1/3 nor t = 1 satisfies |t| ≥ 2, so there are no real solutions.

[JEE Main 2023 · 24 Jan · Shift 2]

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