Question

Suppose  f(x) =x3 +ax2 + bx + c where a, b, c are chosen respectively by throwing a  die three times, then the probability that f(x) is an increasing function is

Answer 1

Answer:

f'(x)=3x2+2ax+b=0 (1)

f''(x)=6x+2a=0, x=-1/3a (2)

Substiture (2) into (1):

3(-1/3a)^2+2a(-1/3a)+b=0

1/3a^2-2/3a^2+b=0

-1/3a^2+b=0 (3)

An increasing function (assuming you're talking about strictly increasing) is only possible if a stationary point of inflection is present. Equation (3) shows solutions that satisfy f'(x)=0 and f''(x)=0. Therefore, any values of a's and b's that satisfy equation (3) is what we aim for.

Since a and b can only be counting numbers, by inspection only a=3, b=3 is possible. Therefore, the probability is 1/(6*6*6)=1/216

Hope this is correct and useful :)

Step-by-step explanation: