JEE Main 2019 · 9 Jan · Shift II

JEE Main 2019 Mathematics question on Complex Numbers and Quadratic Equations

If both the roots of the quadratic equation x2−mx+4=0x^2 - mx + 4 = 0 are real and distinct and they lie in the interval [1,5][1, 5] then mm lies in the interval
  1. A.
    (4, 5)
  2. B.
    (5, 6)
  3. C.
    (−5,−4)(-5, -4)
  4. D.
    (3, 4)
Show answer and solution

Correct option: A

Solution

For both roots to be real, distinct, and lying in [1,5], let f(x) = x² - mx + 4.

Conditions:
1. Discriminant > 0: m² - 16 > 0 → m > 4 or m < -4
2. f(1) ≥ 0: 1 - m + 4 ≥ 0 → m ≤ 5
3. f(5) ≥ 0: 25 - 5m + 4 ≥ 0 → m ≤ 29/5
4. Vertex x-coordinate in [1,5]: 1 ≤ m/2 ≤ 5 → 2 ≤ m ≤ 10

Combining all conditions (taking positive m): m > 4 AND m ≤ 5 AND m ≤ 29/5 AND 2 ≤ m ≤ 10 gives m ∈ (4, 5).

[JEE Main 2019 · 9 Jan · Shift II]

Practise Complex Numbers and Quadratic Equations the way it is asked

A timed set of real JEE Main Complex Numbers and Quadratic Equations questions, graded as the exam grades. No sign-up.

Start a Complex Numbers and Quadratic Equations test

More: all JEE Main Complex Numbers and Quadratic Equations questions · JEE Main 2019 · Mathematics chapter list