The function y = f (x) is an even function on R and an increasing function on (- ∞, 0]. If f (a) ≤ f (2), then the value range of real number a is () A. A ≤ 2B. A ≥ - 2C. - 2 ≤ a ≤ 2D. A ≤ - 2 or a ≥ 2

The function y = f (x) is an even function on R and an increasing function on (- ∞, 0]. If f (a) ≤ f (2), then the value range of real number a is () A. A ≤ 2B. A ≥ - 2C. - 2 ≤ a ≤ 2D. A ≤ - 2 or a ≥ 2

From the meaning of the problem, f (x) is a monotone decreasing function on (0, + ∞), so there is a ＜ 0A ≤ - 2 or a ＞ 0A ≥ 2. The solution is a ≤ - 2 or a ≥ 2, so D

If the even function f (x) is an increasing function on (- ∞, 0], then the value range of real number a satisfying f (1) ≤ f (a) is______ ．

∵ even function f (x) is an increasing function on (- ∞, 0], and ∵ f (x) is a decreasing function on [0, + ∞). When a ≥ 0, 0 ≤ a ≤ 1 can be obtained from F (1) ≤ f (a); when a < 0, the inequality f (1) ≤ f (a), that is, f (- 1) ≤ f (a), and - 1 ≤ a < 0 can be obtained

Through the intersection of two straight lines 3x + 4Y-2 = 0 and 2x + y + 2 = 0, and the intercept on the two axes is equal, the linear equation is

Let the equation be x + y = a
The intersection of two lines is (- 2,2)
A = - 2 + 2 = 0 can be obtained by substituting it into the equation
So the linear equation is x + y = 0

If cosx is 1 / 2 + SiNx = - 1 / 2, then the value of cosx is 1 / 2

Cosx 1 / 2 + SiNx = - 1 / 2
Multiply both sides by (1-sinx) / cosx to get:
(1-sin²x)/cos²x=-(1/2)(1-sinx)/cosx
(1-sinx)/cosx=-2
Take the reciprocal from both sides
cosx/(1-sinx)=-1/2
So: cosx / (sinx-1) = 1 / 2

The line L passing through point (2,1) intersects x-axis, Y-axis at a and B respectively, and the area of triangle AOB is 4. The equation for solving line L (there should be several cases, it is necessary to go through

1. Let (2,1) be a linear equation: Y-1 = K (X-2)
That is y = kx-2k + 1
So the intersection of X axis: ((2k-1) / K, 0)
The intersection of y-axis (0,1-2k)
So s (AOB) = 1 / 2 * 2k-1 / k * 1-2k = 4
So 4K ^ 2-4k + 1 = 8|k|
When k > 0, 4K ^ 2-12k + 1 = 0, k = (3 ± 2 √ 2) / 2
When k

Finding the domain of definition and value of the function y = 2x + 1 divided by 1-3x

y = -2/3 + 5 / (3 - 9x)
The domain is
[negative infinity, 1 / 3) and (1 / 3, positive infinity]
The range is
[negative infinity, - 2 / 3) and (- 2 / 3, positive infinity]

-A rectangle of 4cm in length and 3cm in width is enlarged by 2:1, and the area of the figure is______ cm2．

4 × 2 = 8 (CM), 3 × 2 = 6 (CM), 8 × 6 = 48 (square cm). Answer: the area of the figure is 48 square cm

It is known that a, B and C are three real numbers which are not all zero. To find the root of the quadratic equation with one variable x ^ + (a + B + C) x + (a ^ + B ^ + C ^) = 0

∵△=（a+b+c）2-4（a2+b2+c2）
=-3a2-3b2-3c2+2ab+2bc+2ac
=-（a-c）2-（b-c）2-（a-b）2-a2-b2-c2,
And a, B and C are three real numbers which are not all zero,
∴（a-c）2-（b-c）2-（a-b）2-≤0,a2-b2-c2,＜0,
∴△＜0,
The original equation has no real roots

If the absolute value of M is m, then what is m? If n equals negative n, then what is n equal to

If the absolute value of M is m, then M is a number greater than 0. If the absolute value of n is equal to negative n, then n is less than 0

It is known that the function f (x) = X2 (ax-3) defined on R, where a is a constant. (1) if a = 1, find the tangent equation of the image of F (x) at points (1, - 2); (2) if x = 1 is an extreme point of function f (x), find the value of real number a; (3) if f (x) is an increasing function in the interval (- 1,0), find the value range of real number a

(1) When a = 1, f (x) = x3-3x2, then f ′ (x) = 3x2-6x = 3x (X-2), then k = f ′ (1) = - 3, then the tangent equation is y + 2 = - 3 (x-1), that is, 3x + Y-1 = 0; (2) f (x) = ax3-3x2, we get that f ′ (x) = 3ax2-6x = 3x (AX-2), ∵ x = 1 is an extreme point of F (x), ∵ f ′ (1) = 0, that is, 3 (A-2) = 0, ∵ a = 2; (3) when a = 0, f (x) = - 3x2 is in the interval (- 1) When a ≠ 0, f ′ (x) = 3ax (x-2a), Let f ′ (x) = 0, then X1 = 0, X2 = 2A, when a > 0, for any x ∈ (- 1,0), f ′ (x) > 0, then a > 0; when a < 0, when x ∈ (2a,0), f ′ (x) > 0, then 2A ≤ - 1, ■ - 2 ≤ a < 0, then a ≥ - 2 meets the requirements