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AIEEE 2004

JEE Main 2004 Mathematics question on Complex Numbers and Quadratic Equations

For all 'xx', x2+2ax+(10−3a)>0x^2 + 2ax + (10 - 3a) > 0, then the interval in which 'aa' lies is
  1. A.
    a<−5a < -5
  2. B.
    −5<a<2-5 < a < 2
  3. C.
    a>5a > 5
  4. D.
    2<a<52 < a < 5
Show answer and solution

Correct option: B

Solution

For the quadratic x² + 2ax + (10 - 3a) to be positive for all x, its discriminant must be negative (since the coefficient of x² is positive).

Discriminant < 0: (2a)² - 4(10 - 3a) < 0 → 4a² - 40 + 12a < 0 → a² + 3a - 10 < 0 → (a + 5)(a - 2) < 0.

This inequality holds when -5 < a < 2.

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