If there are two real numbers in the range of (a + a) - 1 + A + 1

If there are two real numbers in the range of (a + a) - 1 + A + 1

If there are two, then
△=(a-1)²-4a(a-1)>0
(a-1)(a-1-4a)>0
(a-1)(3a+1)
△=(a-1)²-4a(a-1)>0
(a-1)(-1-3a)>0
(a-1)(3a+1)
It is known that the equation LXL = ax + 1 about X has a negative root but no positive root, then the value range of real number a is
If we square both sides at the same time, we can get x ^ 2 = (AX + 1) ^ 2, we can get [(A-1) x + 1] [(a + 1) x + 1] = 0
When a = 1, x = - 1 / 2; when a = - 1, x = 1 / 2 does not hold
The other cases are x = - 1 / (A-1) or x = - 1 / (a + 1)
If the two roots are less than zero, we can get a > 1, a > - 1 and a > 1
It is concluded that a is greater than or equal to 1
There is a negative root and no positive root, so | x | = - X
-x=ax+1
x=-1/(1+a)0
a>-1
-x=ax+1 x=-1/(1+a)0 a>-1
Given the real number x, y satisfy x ^ 2 + y ^ 2 + 4x + 1 = 0, make ax + Y-3 > equal to 0, the value range of a is constant
X ^ 2 + y ^ 2 + 4x + 1 = (x + 2) ^ 2 + y ^ 2 = 3 can make the corresponding image as a circle with a radius of (- 2,0) and 3
Because x = 0 can be reduced to a
X ^ 2 + y ^ 2 + 4x + 1 = (x + 2) ^ 2 + y ^ 2 = 3 can make the corresponding image as a circle with a radius of (- 2,0) and 3
Because x = 0 can be reduced to a
For the real number x satisfying 1 ≤ x ≤ 2, what is the value range of the real number a that makes x ^ 2-ax ≤ 4x-a-3 constant?
The original inequality x ^ 2-ax ≤ 4x-a-3 is equivalent to x ^ 2 - (a + 4) x + A + 3 ≤ 0. Let f (x) = x ^ 2 - (a + 4) x + A + 3. To satisfy the original meaning, then Δ≥ 0, f (1) ≤ 0, f (2) ≤ 0, these three conditions Δ = (a + 4) ^ 2-4 (a + 3) = (a + 2) ^ 2 ≥ 0f (1) = 1 ^ 2 - (a + 4) * 1 + A + 3 = 0 ≥ 0f (2) = 2 ^ 2 - (a + 4) * 2 + A + 3 ≥ 0, a ≥ - 1
a≥-1
The known function y = {① X & # 178; - x-3, X ≥ 0, ② 2x + 1, X
f(2)=2^2-2-3=-1
f(-1)=2*(-1)+1=-1
So it's - 1
Finding monotone interval and range of function y = (1 / 3) ^ x ^ 2-2x + 2
This function can be regarded as a composite function of y = (1 / 3) ^ t and T = x & # 178; - 2x + 2
Exponential function y = (1 / 3) ^ t base 1 / 3
Let f (x) = AX2 + BX + C, and f (1) = - a △ 2. If a is greater than 0, we prove that the function has at least one zero point in the interval (0,2)
A>0
If f (1) = a + B + C = - A / 20, then f (0) and f (1) have different signs, and the function has at least one zero point in (0,1)
If f (0) = C0, then f (0) and f (2) have different signs, and the function has at least one zero point in (0,2)
So it is proved
The monotone increasing interval of function y = log3 ^ (x ^ 2-2x-3) is
Base 3 > 1
In the whole domain of definition, it is monotone increasing function, which requires true number
x^2-2x-3>0
(x-3)(x+1)>0
x> 3 or X
(3, positive infinity)
Let f (x) = ax & sup2; + BX + C, (a > 0), and f (1) = - A / 2. Find (1) and prove that f (x) has two zeros. (2) let X1 and X2 be the two zeros of F (x)
Find the value range of absolute value of difference between X1 and X2 (3) prove that the function f (x) has at least one zero point in (0,2)
1) F (1) = a + B + C = - A / 2, so B + C = - 3A / 2 discriminant, △ = B & sup2; - 4ac = B & sup2; - 4a (- 3A / 2-B) = B & sup2; + 6A & sup2; + 4AB = B & sup2; + 4AB + 4A & sup2; + 2A & sup2; = (B + 2a) & sup2; + 2A & sup2; because a ≠ 0, the equation has two unequal real roots 2 | x1-x2 | = √ ((x1 + x2)
The first question is to calculate delta > 0. The relationship of a, B and C can be obtained from F (1) = - A / 2. In Delta, a is replaced by B and C. The formula can be proved
In the second question, Weida's theorem, | x1-x2 | = radical [(x1 + x2) 2-4x1x2]
The third question is to prove f (0) * f (2)
Finding the monotone interval sum and value of function f (x) = log3 (x ^ 2-2x + 8)
x^2-2x+8>0
Because △ = 1
Monotone decreasing interval x = log3 (7)