If the inequality ax + (2a-1) y + 1 < 0 represents the lower region of the straight line ax + (2a-1) y + 1 = 0, then the value range of the real number a is______ .

If the inequality ax + (2a-1) y + 1 < 0 represents the lower region of the straight line ax + (2a-1) y + 1 = 0, then the value range of the real number a is______ .

Because the straight line ax + (2a-1) y + 1 = 0 passes through the fixed point (- 2,1), and obviously the point (- 2,0) is below the point (- 2,1), it should satisfy the inequality. Substituting the point (- 2,0) into the inequality, that is, - 2A + 1 < 0, the solution is a > 12, so the answer is: a > 12
If the system of inequalities 3 − 2x ≥ 0 x ≥ m has a solution, then the range of M is ()
A. m＜32B. m≤32C. m＞32D. m≥32
3 − 2x ≥ 0 (1) x ≥ m (2) from (1): X ≤ 32 (2): X ≥ m (2). Its solution set is m ≤ x ≤ 32  m ≤ 32
If there is no solution to the system of inequalities x1, find the range of A
3x-1/2>1
3x>3/2
x>1/2
Another x
If there is no solution, then A1
x>1/2
x1/2
Then 1 / 22 X1 X1 + 1 / 2
So if x > 1 / 2, we have to make x
If the system of inequalities x ＞ 2A − 1x ＜ a + 1 has no solution, then the value range of a is ()
A. a＜2B. a=2C. a＞2D. a≥2
It can be judged that 2a-1 ≥ a + 1, and the solution is a ≥ 2
(2011 Jiading District the first mock exam) two times function y=ax2+bx+c, a C < 0, the zero point of the function is ()
A. 1b. 2C. 0d. Not sure
∵ AC ＜ 0, ∵ △ = b2-4ac ＞ 0, ∵ the corresponding equation AX2 + BX + C = 0 has two unequal real roots, so the quadratic function and X-axis have two intersections
When the function f [x] = x & sup2; - X-2 is known, the value range of X is obtained when the zero point [3] f [x] of the range [2] f [x] of [1] f [x] is 0
[range]
f(x)=x²-x-2
=(x - 1/2)² - 9/4
≥-(9/4)
[zero point]
That is, f (x) = 0
x²-x-2=0
(x+1)(x-2)=0
X = - 1 or x = 2
[value range]
According to the meaning of the title:
x²-x-2
Let f (x) = 2, x > 0, f (x) = x ^ 2 + BX + C, X
f(-4)=f(0),
The axis of symmetry is - B / 2 = - 2, B = 4
f(-2)=-2,
(-2)^2+4(-2)+c=-2
c=-6
g(x)=f(x)-x=x^2+3x-6
Δ=9+24=33>0
The opening of the image is upward, and there is no intersection with the X axis
So the number of zeros is zero
If f (x) = - 1 / 2x ^ 2 + bln (x + 2) is a decreasing function on (- 1, + infinity), then the value range of B is?
F (x) = - 1 / 2x ^ 2 + bln (x + 2) is a decreasing function on (- 1, + infinity)
If f '(x) = - x + B / (x + 2) - 1, B / (x + 2)
If the two zeros of function f (x) = x2-ax + B are 2 and 3, then the zeros of function g (x) = bx2-ax-1 are ()
A. - 1 and 16b. 1 and - 16C. 12 and 13D. - 12 and - 13
The two zeros of ∵ function f (x) = x2-ax + B are 2 and 3, ∵ 2, 3 are the two roots of equation x2-ax + B = 0, then 2 + 3 = a = 5, 2 × 3 = B, that is, a = 5, B = 6, ∵ g (x) = bx2-ax-1 = 6x2-5x-1, from G (x) = 6x2-5x-1 = 0, x = 1 and - 16 are obtained, so the zeros of function are 1 and - 16, so B is selected
Given the function f (x) = a ^ x (a > 0, a ≠ 1), if f (2x ^ 2-3x + 1) > F (x ^ 2 + 2x-5), find the value range of X
If a > 1, f (x) is monotone increasing, there are: 2x ^ 2-3x + 1 > x ^ 2 + 2x-5,
x^2-5x+6>0
Get x > 3 or X
(1)0