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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Complex Numbers and Quadratic Equations

The number of integral values of mm for which the quadratic expression, (1+2m)x2−2(1+3m)x+4(1+m)(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m), x∈Rx \in R, is always positive, is
  1. A.
    6
  2. B.
    8
  3. C.
    7
  4. D.
    3
Show answer and solution

Correct option: C

Solution

For the quadratic to be always positive, we need two conditions:

1) Leading coefficient > 0: 1 + 2m > 0 → m > -1/2

2) Discriminant < 0: D = 4(1+3m)² - 4·(1+2m)·4(1+m) < 0

Expanding: (1+3m)² - 4(1+2m)(1+m) < 0
= 1 + 6m + 9m² - 4(1 + 3m + 2m²) < 0
= 1 + 6m + 9m² - 4 - 12m - 8m² < 0
= m² - 6m - 3 < 0

Roots of m² - 6m - 3 = 0: m = (6 ± √48)/2 = 3 ± 2√3

So 3 - 2√3 < m < 3 + 2√3, i.e., approximately -0.464 < m < 6.464

Combined with m > -1/2 = -0.5, the binding constraint is: -0.464... < m < 6.464...

Since 3 - 2√3 ≈ -0.464 > -0.5, the intersection is (3-2√3, 3+2√3).

Integral values of m in this range: 0, 1, 2, 3, 4, 5, 6 → 7 values.

Answer: Option 3

[JEE Main 2019 · 12 Jan · Shift II]

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