Consider `f(x) =x^3+ax^2+bx+c` Parameters `a, b, c` are chosen as the face value of a fair dice by throwing it three times Then the probability that `f(x)` is an invertible function is (A) `5/36` (B) `8/36` (C) `4/9` (D) `1/3`

Options

A. `5//36`
B. `8//36`
C. `4//9`
D. `1//3`

Answer

C

Explanation

`f'(x) = 3x^(2) + 2ax + b`
y = f(x) is increasing
implies `f'(x) ge 0, AA x` and for f'(x) = 0 should not form an interval
`implies (2a)^(2) - 4 xx 3 xx b le 0 implies a^(2) - 3b le 0`
This is true for exactly 16 ordered pairs (a, b), `1 le a, b le 6`, namely (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), and (4, 6) Thus, the required probability is `16//36 = 4//9`