The probability of selecting integers a∈[- 5, 30] such that x² + 2(a + 4)x - 5a + 64 > 0, for all x∈R, is :

Solution

Maths sir · 4 steps

An upward parabola stays above the x-axis for every x only when it never touches it, so make D negative.

1. Condition needed. The coefficient of x² is 1, which is positive, so the parabola opens upward. It is positive for all real x exactly when it has no real roots, that is, when its discriminant satisfies D < 0.

   D = 4(a+4)² - 4(64 - 5a) < 0

2. Solve for a. Divide by 4: a² + 8a + 16 - 64 + 5a < 0, which is a² + 13a - 48 < 0. This factors as (a+16)(a-3) < 0, so -16 < a < 3.
3. Count integers. The interval [-5, 30] contains 30 - (-5) + 1 = 36 integers. Inside it, the integers with a < 3 are -5, -4, …, 2, which is 2 - (-5) + 1 = 8 values.
4. So. The probability is 8/36 = 2/9, which is option (b).

The idea to keep. For ax²+bx+c > 0 for all real x, you need a > 0 and D < 0, with strict inequality. Then count integers in an interval as last minus first plus one.

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