It is known that the coefficient of quadratic term of quadratic function f (x) is a, and the solution set of inequality f (x) > 0 is (1,2). If the maximum value of F (x) is less than 1, then the value of real number a is

It is known that the coefficient of quadratic term of quadratic function f (x) is a, and the solution set of inequality f (x) > 0 is (1,2). If the maximum value of F (x) is less than 1, then the value of real number a is

The solution set of F (x) > 0 is x ∈ (1,2), the solution set of F (x) = 0 is X1 = 1, X2 = 2, and a < 0  f (x) = a * (x - 1) * (x - 2). The remaining quadratic function f (x) is expressed by a, the maximum value of quadratic function f (x) can be expressed by the expression containing a, and then the right is less than 1

The quadratic function f (x) = ax & # 178; - BX + 1 is known. If a is a positive integer, B = a + 2, and the minimum value of function f (x) on [0,1] is - 1, the value of a is obtained
When the axis of symmetry is between 0 and 1, why can't a 2

A is a range, not a specific value
f（x）＝ax²－bx＋1=ax²-（a+2）x+1
The axis of symmetry is x = a + 2 / 2A
It can be discussed in three ways
1. When a + 2 / 2A ≤ 0, i.e. - 2 ≤ a < 0, f (x) is an increasing function on [0,1], and the minimum value is f (0) = 1
2. When 0 ＜ a + 2 / 2A ＜ 10, i.e. a ＞ 2 or a ＜ - 2, the minimum value of F (x) is: F (a + 2 / 2a) = - 1, the solution is a = 2, there is no solution
3. When a + 2 / 2A ≥ 1, that is, 0 < a ≤ 2, f (x) decreases monotonically on [0,1], and f (1) is the minimum,
f（1）=a-（a+2）+1=-1
To sum up, the minimum value of function f (x) on [0,1] is - 1, and the value range of a is 0 < a ≤ 2

It is known that the quadratic function f (x) = ax + BX satisfies the following conditions: 1) f (0) = f (1); 2) the minimum value of F (x) is - 1 / 8. Find the analytic expression of the function

Condition 1 can know that a + B = 0, condition 2 can know that the minimum value is taken on the symmetry line, and finally a = 0.5 is calculated

Quadratic function f (x) = AX2 + BX + C, (a is a positive integer), C ≥ 1, a + B + C ≥ 1, equation AX2 + BX + C = 0 has two unequal positive roots less than 1, then the minimum value of a is ()
A. 2B. 3C. 4D. 5

Let f (x) = a (X-P) (X-Q), where p, Q belong to (0, 1) and P is not equal to Q. from F (0) ≥ 1 and f (1) ≥ 1, we can get: Apq ≥ 1, a (1-p) (1-Q) ≥ 1, the multiplication of the two formulas has A2p (1-p) Q (1-Q) ≥ 1, that is, A2 ≥ 1p (1 − P) Q (1 − q), and from the basic inequality, we can get: P (1-p) Q (1-Q) ≤ 116 Therefore, the function f (x) = 5x2-5x + 1 satisfies all the conditions, so the minimum value of a is 5