Given that the range of function y = log2 (AX & # 178; + X + 1) (x ∈ R) is (- ∞, 1), find the value of real number a

Given that the range of function y = log2 (AX & # 178; + X + 1) (x ∈ R) is (- ∞, 1), find the value of real number a

It is known that the range of the function y = log2 (AX & # 178; + X + 1) (x ∈ R) is (- ∞, 1),
log2(aX²+X+1)
Given the function f (x) = loga (2-ax), whether there is a real number a, so that the function f (x) is an increasing function of X on [0,1]. If there is, find the value range of A
Given the function f (x) = loga (2-ax), whether there is a real number a, so that the function f (x) is an increasing function of X on [0,1]. If there is, find the value range of A
(is an increasing function)
Process: from the meaning, f (1) - f (0) > 0, then log a ((2-A) / 2) > 0
Then we discuss it by category
（1）0
Given that the function f (x) = 4x & # 179; + ax & # 178; + BX + 5 passes through points (1, - 12) and F "(- 1) = 0, find the analytic expression and monotone interval of the function
If f (x) = 4x ^ 3 + ax ^ 2 + BX + 5, the substitution point (1, - 12) has: - 12 = 4 + A + B + 5
f'(x)=12x^2+2ax+b
F "(x) = 24x + 2a, F" (- 1) = 2a-24 = 0, so a = 12
According to ①: B = 12 + 9 + 12 = 30
f(x)=4x^3+12x^2+30x+5
If the point (Mn, M + n) is in the fourth quadrant, then in which quadrant is the point (m-1, n-2)?
(Mn, M + n) is in the fourth quadrant, which means Mn > 0,
m. N has the same sign; also m + n
Given the set a = {x | x ^ 2 + (a + 2) x + 1 = 0, X belongs to R}, if the intersection of a and positive real number is equal to an empty set, then the value range of A
X = [- B + √ (b ^ 2-4ac)] / 2A or x = [- B - √ (b ^ 2-4ac)] / 2A
Make x less than or equal to 0
The value of a is greater than or equal to 0
If I'm right
If you fill in the blanks or multiple-choice questions, you can bring a few numbers
The coordinate of point a is (m, n) when Mn
Second or fourth quadrant
Because mn0, n
If a = {x | A-1 ≤ x ≤ a + 2}, B = {x | 3 ＜ x ＜ 5}, then the value range of the real number a with a ⊇ B is______ .
∵ a = {x | A-1 ≤ x ≤ a + 2} B = {x | 3 ＜ x ＜ 5} and a ⊇ B ⊇ a + 2 ≥ 5A − 1 ≤ 3, the solution is: 3 ≤ a ≤ 4, so the answer is: [3, 4]
If M (M + n), Mn) is in the second quadrant, n (m, n) is in the second quadrant
Because point m is in the second quadrant, so
m+n0
Solution
M
The second quadrant, M + N0, m, N, the same sign, M
Set a = x greater than - 1 less than or equal to 1 b = A-X greater than or equal to 0 a intersection B = empty set, then the value range of real number a
Simplify the set B = {x ≤ a}, and then draw the range of set a on the number axis, that is to draw the image of - 1 ＜ x ≤ 1. Because a ∩ B = & # 8709;, that is to say, after drawing the image of X ≤ a on the number axis, there is no overlapping part between the two images, so a ≤ - 1 is solved
If √ m ＋ Mn / 1 is meaningful, what quadrant is the position of point (m, - n)?
From the first term, we can see that M > 0, from the second term, the denominator √ Mn is not 0, and because m > 0, n > 0
-n
Is it wrong to type √ Mn / 1? It should be √ 1 / Mn
If √ m ＋ 1 / Mn is meaningful
Then m ≥ 0, m ≠ 0, n ≠ 0, Mn > 0
So m > 0, n > 0
(m, - n) is in the fourth quadrant