Given that the maximum value of function f (x) = AX2 + (2a-1) x-3 in the interval [− 32, 2] is 1, find the value of real number a

Given that the maximum value of function f (x) = AX2 + (2a-1) x-3 in the interval [− 32, 2] is 1, find the value of real number a

When a = 0, f (x) = - x-3, f (x) can't get 1 on [− 32, 2], so a ≠ 0, then the equation of symmetry axis of F (x) = AX2 + (2a-1) x-3 (a ≠ 0) is x0 = 1 − 2a2a, ① Let f (− 32) = 1, then a = - 103, then x0 = - 2320 ∈ [− 32, 2], ∵ a < 0, ∵ f (x0) is the largest, so f (− 32) = 1 is not suitable; ② Let f (2) = 1, then a = 34, then x0 = - 13 ∈ [− 32, 2] because a = 34 (3) Let f (x0) = 1, get a = 12 (− 3 ± 22), a = 12 (− 3 − 22), a = 34 or a = 12 (− 3 − 22)
Let a ≠ 0, for the function f (x) = log3 (AX & # 178; - x + a), if the domain of definition is r, find the value range of real number a
ax^2-x+a>0
It holds for X to be any real number
A = 0 is not suitable
A0, and △ = 1-4a ^ 21 / 2
ax^2-x+a>0
The domain is r
△1/4
a1/2
The range of a is: (- ∞, - 1 / 2) U (1 / 2, + ∞)
Let a ≠ 0, for the function y (x) = log3 (AX & # 178; - x + a), if the domain of definition is r, find the value range of real number a
The true number of logarithmic function is greater than 0
So ax & # 178; - x + a > 0
The domain is r
So whatever the value of X is
The value of ax & # 178; - x + A is always greater than 0
So a > 0
Minimum value a * [1 / (2a)] ² - 1 / (2a) + a > 0
1/(4a)-1/(2a)+a>0
-1/(4a)+a>0
-1+4a²>0
a²>1/4
So a > 1 / 2 or A0
So a > 1 / 2
In conclusion, a > 1 / 2
It is known that the set a consists of elements 1 and a ^ 2. The set a can't get the value of is
It is known that a = {1, 2, 3} B = {1, 2} C = {(x, y) | x belongs to a, y belongs to B} the set C is represented by enumeration=___
1. {- 1,1}. Because a ^ 2 ≠ 1, a ≠ - 1,1
2.{(1,2),(1,2),(2,1),(2,2),(3,1),(3,2)}
{（1,1），（1,2），（2,1），（2,2），（3,1），（3,2）}
If P (m-1, N + 1) and Q (4 + N, 2m-1) are symmetric about the origin, what quadrant is x (m, n)
If two points are symmetrical to the origin, the midpoint is the origin
m-1+4+n=0/2=0
n+1+2m-1=0/2=0
So m + n = - 3
2m+n=0
So m = 3, n = - 6
Point x (m, n) is in the fourth quadrant
Given the set a = {the square of X - 4x + 2A + 6 = 0, X belongs to R}, if a is not an empty set, find the set of all the real numbers a that satisfy the condition
sudu!
Because a is not an empty set
B squared - 4ac ≥ 0 (here is 16-4 (2a + 6) ≥ 0)
16≥4（2a+6）
4≥2a+6
-2≥2a
-1≥a
All real numbers R with a ≤ - 1
When m is the value, the symmetric point of point P (- 3,2m-1) about the origin is in the fourth quadrant
The symmetric point of point P (- 3,2m-1) about the origin is (3,1-2m)
The coordinates of the points in the fourth quadrant should be greater than 0 in abscissa and less than 0 in ordinate
Let (3,1-2m) be the point in the fourth quadrant
Must be 1-2m1 / 2
We know that the set a = {x belongs to the square of R | X - 4ax + 2A + 6 = 0} if the intersection of a and negative real number is not empty
Because the function values are all nonnegative,
So the minimum value should be greater than or equal to 0
That is - 4A ^ 2 + 2A + 6 ≥ 0
The solution is - 1 ≤ a ≤ 1.5
Given (m-n) ^ 2 = 8, (M + n) ^ 2 = 2, find the value of Mn and m ^ 2 + n ^ 2
It is known that (m-n) &# 178; = 8, (M + n) &# 178; = 2
that
m²+n²-2mn=8①
m²+n²+2mn=2②
(1 + 2) / 2 is M & # 178; + n & # 178; = (8 + 2) / 2 = 5
(2 - 1) / 4 is Mn = (2-8) / 4 = - 3 / 2
If you don't understand, I wish you a happy study!
The former is minus two thirds, the latter is five
The known set a = {x | - 2
First of all, let's look at the set B. because B is not an empty set, B must have a root, so Δ > = 0, that is, Δ = 9A ^ 2-8a ^ 2 > = 0, a ^ 2 > = 0, so a takes any real number, B is not an empty set
B = {x | (x-a) (x-2a) = 0}, when a > 0,2a > A, so a > = 4, a ∩ B = empty set; if a > 0,2a > A, then a ∩ B = empty set; if a > 0,2a > A, then a ∩ B = empty set