Let a ∈ {1,2,3,4} and B ∈ {2,4,8,12}, then the probability that f (x) = X3 + ax-b has zero in interval (1,2) is () A. 12B. 58C. 1116D. 34

If f (x) = X3 + ax-b has zeros in the interval [1,2], we only need to satisfy the condition that f (1) ≤ 0, f (2) ≥ 0, and then we can get the solution that B-A ≥ 1 and b-2a ≤ 8, ｜ a + 1 ≤ B ≤ 2A + 8

Given the quadratic function f (x) = ax & # 178; + 4x + 3a, and f (1) = 0. Try to judge the number of zeros of function f (x)

ax²+4x+3a=0
a+4+3a=0
a=-1
f(x)=-x²+4x-3
Discriminant = 16-4 * (- 1) * (- 3) > 0
There are two intersections between the image of quadratic function and x-axis
So f (x) has two zeros

In the function f (x) = ax & # 178; + BX + C (a ≠ 0), if a and C are different signs, how many zeros are there?

The equation B ^ 2-4ac: if it is greater than 0, there are two zeros; if it is less than 0, there is no zero; if it is equal to 0, there is one zero
In the title, a and C are different signs, and (- 4ac) is greater than zero,
And B ^ 2 is greater than or equal to 0,
To sum up, B ^ 2-4ac is greater than zero, so there are two zeros

If the function y = ax2-x-1 has only one zero point, the value range of a is obtained

∵ when a = 0, y = - X-1 has a zero point, x = - 1, a = 0; when a ≠ 0, the image of y = ax2-x-1 and X-axis have only one intersection, a = (- 1) 2 + 4A = 0, the solution is a = - 14, a = 0 or a = - 14,

Let a ∈ {1,2,3,4} and B ∈ {2,4,8,12}, then the probability that f (x) = X3 + ax-b has zero in interval (1,2) is () A. 12B. 58C. 1116D. 34