If the inequality x2 + AX-2 > 0 about X has a solution in the interval [1,5], then the value range of real number a is () A. (−235，+∞)B. (−235，1)C. （1，+∞）D. (−∞，−235)

If the inequality x2 + AX-2 > 0 about X has a solution in the interval [1,5], then the value range of real number a is () A. (−235，+∞)B. (−235，1)C. （1，+∞）D. (−∞，−235)

Let f (x) = x2 + AX-2, if the inequality x2 + AX-2 > 0 about X has no solution in the interval [1,5], then f (1) ≤ 0, f (5) ≤ 0, that is, a − 1 ≤ 052 + 5A − 2 ≤ 0, and the solution is a ≤− 235. Therefore, the range of a with solution in the interval [1,5] is (− 235, + ∞)
Given the set a = {x │ x ^ 2-5x + 4 ≤ 0}, B = {x │ x ^ 2-2ax + A + 2 ≤ 0}, and B is contained in a, find the value range of real number a
Please do not copy the answer, about the answer to this question in Baidu see a lot, basically wrong, please give a detailed problem-solving process, thank you! The answer on the first floor seems to be right...
The second floor equation factorization error...
Because the square of {a + B = - 4} is contained in the set a ≤ 4
That is, a = = {x | 1 ≤ x ≤ 4}
If B is an empty set, then the square of corresponding x-2ax + A + 2 = 0 has no real root, then (2a) ^ 2-4 (a + 2)
The answer is that x is greater than or equal to negative root two and less than three-thirds
A is (x-4) (x-1)
It is known that the equation MX square + 2x-1 = 0 about X has a real solution, and the value range of M is obtained
MX & sup2; + 2x-1 = 0 has real solution
So the discriminant △≥ 0
2²-4×m×（-1）≥0
4+4m≥0
4m≥-4
m≥-1
From 2 & sup2; - 4 × (- 1) ≥ 0
The results are as follows
m≥-1
But when m = - 1, there is a real solution, so the result is: M > - 1;
You can use that discriminant later. No mistakes...
M is greater than - 1
△=2^2-4m*(-1)
=4+4m>=0
m>=-1
This is discussed in terms of M = 0 and m not = 0
When m = 0, the equation is a linear equation, the image is a straight line, and there must be an intersection with the X axis, that is, there is a real solution
When m is not equal to 0, if the equation needs to have a real solution (the image needs to have an intersection with x), then Δ > 0 is OK, that is, M > - 1
B square - 4ac = 4-4m > = 0 then M
It is known that X1 and X2 are the two zeros of the function f (x) = ax ^ 2 + BX + C (a > 0),
The minimum value Yo of function f (x) and yo ∈ [x1, X2] find the number of zeros of y = f (f (x))
A 2 or 3 B 2 or 4 C 3 D 3 or 4
X1(X1+X2)/2）Y=0；
When xyo, then y has only one zero point, that is, when f (x) = X2 (x 1 + x2) / 2, f (x) > yo = x1, so y has only one zero point x, and X satisfies f (x) = x2;
Xyo = x1, then y has only one zero point x, and X satisfies f (x) = X2, but x also satisfies X
The monotone decreasing interval of function f (x) = | x2-1 | is______ .
The function f (x) = | x2-1 | = x2 − 1 & nbsp;, & nbsp; & nbsp; (x > 1 & nbsp;, & nbsp; or X & nbsp; < 1) 1 − X2, & nbsp; & nbsp; & nbsp; - 1 ≤ X & nbsp; ≤ 1, as shown in the figure: so the subtraction interval of function f (x) is (- ∞ - 1) and (0, 1), so the answer is (- ∞ - 1) and (...)
If the function f (x) = ax ^ 2-x-1 has only one zero point in the interval (0,1), find the value range of A
(1) When a = 0, f (x) = - X-1, the zero point is x = - 1, not in the interval (0,1)
(2) If a ≠ 0, f (x) has only one zero point in (0,1),
Then f (0) · f (1)
Classmate, this problem is simple enough, use your own brains
The monotone decreasing interval of function f (x) = x square - 4x is?
x≤2
Negative infinity to 2
Given that the function f (x) = 4x + m · 2x + 1 has and only has one zero point, the range of value of M is obtained and the zero point is obtained
Let 2x = t (T > 0), then T2 + MT + 1 = 0. When △ = 0, i.e. M2-4 = 0, M = - 2, t = 1. When m = 2, t = - 1 does not fit the problem, and is rounded off. When △ > 0, i.e. m > 2 or m < - 2, the equation T2 + MT + 1 = 0 should have a positive and a negative root, i.e. t1t2 < 0 In conclusion, when m = - 2, ƒ (x) has a unique zero, which is x = 0
What is the monotone decreasing interval of function f (x) = X3 square - 3x
F (x) '= 3x ^ 2-3 for the first derivative of a function
Let it be 0 to get 3x ^ 2-3 = 0 to get x = ± 1
So in (- 1, + 1) function monotone decreasing
Given that the function f (x) = 4 ^ x + m * 2 ^ x + 1 has zero point, find the value range of M
Senior math experts, please answer
t=2^x,t＞0
f(t)=t^2+mt+1
That is, the equation T ^ 2 + MT + 1 = 0 has roots greater than 0
t^2+mt+1=
f(t)=(t+m/2)^2+1-M^2/4
-m/2＞0
△≥0
The solution is m ≤ - 2