If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x²+α x+β>0, for all x ∈ R, is :

Solution

Maths sir · 3 steps

An upward parabola stays above zero only when it never touches the x-axis, so make the discriminant negative.

1. Condition on roots. The coefficient of x² is 1, so the parabola opens upward. It is above zero for every real x only if it has no real root, that is, D = α² - 4β < 0.

   α² < 4β β > α²/4

2. Count the pairs. Two throws give 6 × 6 = 36 equally likely pairs (α, β). For each α, count the values of β from 1 to 6 that are strictly greater than α²/4.

   α	Need β >	Allowed β	Count
   1	0.25	1 to 6	6
   2	1	2 to 6 (β = 1 fails)	5
   3	2.25	3 to 6	4
   4	4	5, 6 (β = 4 fails)	2
   5	6.25	none	0
   6	9	none	0

3. So. Favourable pairs: 6 + 5 + 4 + 2 = 17. The probability is 17/36, option (a).

The idea to keep. For ax² + bx + c > 0 for all real x, you need a > 0 and D < 0. Keep the inequality strict: D = 0 means the curve touches zero.

Answer from the NTA final answer key. Solution by PrepWiser.