It is known that the proposition p: equation x2 + MX + 1 = 0 has two unequal negative real roots. Proposition q: equation 4x2 + 4 (m-2) x + 1 = 0 has no real roots. If P or q are true, P and Q are false, then the range of real number m is () A. （1，2]∪[3，+∞） B. （1，2）∪（3，+∞） C. （1，2] D. [3，+∞）

If P is true, then
m2−4＞0
- M < 0, M > 2;
If q is true, then △ = [4 (m-2)] 2-16 < 0, the solution is: 1 < m < 3;
∵ P or q is true, P and Q are false,
/ / P and Q are true and false,
When p is true and Q is false, m ≥ 3 is obtained; when p is false and Q is true, the solution is 1 < m ≤ 2
In conclusion, 1 ＜ m ≤ 2 or m ≥ 3;
Therefore, a

It is known that P: equation x squared plus MX + 1 = 0 has two unequal negative roots: Q: equation 4x squared plus 4 (M minus 2) x plus 1 = 0 has no real heel, if "P or Q" is true, "P It is known that P: equation x squared plus MX plus 1 = 0 has two unequal negative roots: Q: equation 4x square plus 4 (M minus 2) x plus 1 = 0 has no real heel. If "P or Q" is true, "P and Q" is false, the more detailed the value range of M, the more urgent it is

When p is true, - M0, that is, M > 2;
When q is true, 16 (m-2) 2 - 16

For any real number x, y, a function f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0. It is proved that f (0) = 1

∵f(x+y)+f(x-y)=2f(x)f(y）
/ / 2F (0) = f (0) + F (0) = f (0 + 0) + F (0-0) = the square of 2F (0)
/ / F (0) = the square of F (0)
If the square of a number is equal to itself, the number must be 0 or 1
And ∵ f (0) ≠ 0
∴f（0)=1

Given that f (x) = radical (1-x ^ 2), find the tangent slope and equation of curve f (x) at x = 1 / 2

The derivative function equation = - X / root sign (1-x ^ 2) substituting x = 1 / 2 into k = negative 3 / 3 root sign 3, and then substituting x = 1 / 2 into the original equation, we get y = 2 / 2 root sign 3, and the point oblique type generation comes out

Given the root sign x of curve y = 5, find the equation of tangent line parallel to the curve y = 2X-4

Y '= 5 / 2 (x) ^ (- 1 / 2) is parallel to y = 2X-4, so y' = 2
In other words: 5 / 2 (x) ^ (- 1 / 2) = 2, x = 25 / 16
y=5(25/16)^(1/2)=25/4
Therefore, the tangent equation is obtained as follows:
y=2(x-25/16)+25/4

The distance between tangent at (0,1) of 2xcos3x curve y = E and straight line C is the root sign 5, and the equation of straight line C is obtained

2xcos3x of y = e
y'=2*e^2x*cos3x-3*e^2x*sin3x
k=y'=2
Tangent y = 2x + 1 at (0,1)
The distance of line C is root 5,
Let c y = 2x + C
Radical 5 = | C-1 | / √ 5 | C-1 | = 5, C = 6 or C = - 4
The straight line C Y = 2x + 6 or y = 2X-4

The distance from the center of x = 1 + cos α y = sin α to the line y = radical 3 / 3x is x=1+cosα The distance from the center of y = sin α to the line y = radical 3 / 3x is

The curve X = 1 + cos α y = sin α is a circle
（x-1)2+y2=1
The coordinates of the center of the circle are (1,0)
The formula of the distance from a point to a straight line
Distance d = 3 / 3 * 1-0 │ / √ [(√ 3 / 3) 2 + 1] = 1 / 2

The tangent equation of 3x-2 at point (1,1) is

Hello, use derivative function to find, the equation is 3x-2y-1 = 0

The distance between tangent of curve y = e ^ xcosx at (0,1) and straight line C is root 5, and the equation of straight line C is obtained

Derivation y '= e ^ x (cosx SiNx)
y'(0)=1
The straight line is y = x + 1
D = (ci-c2) / Radix 1 + K ^ 2 = Radix 5
The absolute value of C1-C2 is root 10
C2 = root 10 + 1 or 1 - root 10
Because straight lines are parallel to each other
So Y1 = x + 1 + radical 10
Y2 = x + 1 - radical 10

Point P is on the line 3x + Y-5 = 0, and the distance between point P and the line x-y-1 = 0 is 2, then the coordinates of point P are () A. （1，2） B. （2，1） C. (1,2) or (2, - 1) D. (2,1) or (- 2,1)

Let the coordinates of point p be (a, 5-3a),
From the meaning of the title: | a − (5 − 3a) − 1|
2=
2．
The solution is a = 1 or a = 2,
The coordinates of point P are (1,2) or (2, - 1)
Therefore, C