Given the function f (x) = LNX, G (x) = (1 / 2) ax ^ 2 + 2x (a ≠ 0), if f '(x) > G' (x) is constant at (0, + ∞), the value range of a is obtained

Given the function f (x) = LNX, G (x) = (1 / 2) ax ^ 2 + 2x (a ≠ 0), if f '(x) > G' (x) is constant at (0, + ∞), the value range of a is obtained

1 / x > ax + 2 because (0, + ∞), so ax * x + 2x-10) so the opening is downward A0, so the vertex - 1 / a can be obtained at (0, + ∞)

If the image of the function f (x) = ax + LNX is tangent to the line 2x-y-1 = 0, the value of a is obtained

2x-y-1=0
y=2x-1
The slope of the line k = 2
F (x) = ax + LNX is tangent to the straight line, let the tangent point be (x, y)
Then ax + LNX = 2x-1, LNX + 1 = (2-A) X. (1)
f(x)=ax+lnx
f'(x)=a+1/x=2,1/x=2-a,x=1/(2-a)
Because x > 0,2-a > 0, a

The function y = LG (AX + 1) is meaningful in (negative infinity, 1), and the value range of a is obtained

On (- ∞, 1), f (x) = ax + 1 > 0 holds;
If a = 0,1 > 0;
If a > 0, it is an increasing function, so it only holds for x > - 1 / A, not constant;
If a

If the definition of the function y = LG (X & # 178; - ax + 1) is r, then the value range of a is

The definition field is r, that is, X & # 178; - ax + 1 > 0
That is, the parabola y = x & # 178; - ax + 1 is always above the x-axis, that is, there is no intersection with the x-axis
So: △ = A & # 178; - 4

If the image of function y = 5 ^ (x + 1) + m does not pass through the second quadrant, then the value range of M is

Just move y = 5 ^ (x + 1) to - Y direction by 5, so M & lt; = - 5

Given that the first-order function y = (K-2) x + K does not pass through the third quadrant, then the value range of K is ()
A. k≠2B. k＞2C. 0＜k＜2D. 0≤k＜2

If the image of the linear function y = (K-2) x + K does not pass through the third quadrant, then it passes through the second and fourth quadrants or the first, second and fourth quadrants, and only passes through the second and fourth quadrants, then k = 0. If K < 0, the line must pass through the second and fourth quadrants, so K-2 < 0, that is, K < 2. If K-2 = 0, that is, k = 2, then it passes through the first and second quadrants, that is, the line intersects the positive half axis of Y axis, so k > 0, At this time, the line is only the third quadrant, so 0 ≤ K ≤ 2

If the image of the first function y = MX + n - 2 does not pass through the third quadrant, then the value range of M and N is given

Given that the function y = sinwx is an increasing function on [- 1,2], then the value range of real number W is?

Given that the function y = sinwx is an increasing function on [- 1,2], then the value range of real number W is?
Analytic: ∵ function y = sinwx is an increasing function on [- 1,2]
Y = sinwx is an odd function, which is centrosymmetric about the origin
∴T/2=2-(-2)=4==>T=8
To ensure that the function y = sinwx is an increasing function on [- 1,2], as long as t > = 8
w

The known function y = asin (ω x + Φ) + C (a > 0, C > 0, | Φ|

2A=2-(-4) => A=3
A+C=2 => C=-1
π/ω = 8-2 =6,=> ω= π/6
ωx+Φ =π/2 => Φ= π/6

Given the function f (x) = | x-a | + 1 / X (x > 0), if f (x) > = 1 / 2 is constant, then the value range of A

The absolute value of F (x) = x-a and the image of F (x) = 1 / X are drawn respectively for easy understanding. Because f (x) = │ x-a │ is a decreasing function (x ＞ 0) in the interval of (0, │ a │), f (x) = 1 / X (x ＞ 0) is also a decreasing function, so when x = a, there are two minimum values, so we substitute x = a into the original formula to get f (x) = 1 / x = 1 / 2 and x = a = 2