Find the document: A.B.C is a three digit hundred, ten and one digit number, and a is less than or equal to B, less than or equal to C, then / A-B / + / B-C / + / C-A / can a. B. C is a three digit number of hundreds, tens and ones, and a is less than or equal to B, less than or equal to C, then / A-B / + / B-C / + / C-A / can take the maximum value

One
Let f (x) = {X & # 178;, X ≥ 0; (1 / 2) ^ X-1, X ≤ 0; if f (a) > 1, then the value range of a is?
Classification: F (a) > 1
（1）a≥0
That is a & # 178; > 1
A > 1 (rounding off)
（2）a1
∴ (1/2)^a>2=(1/2)^(-1)
∴ a
The maximum and minimum of function f (x) = sin2x / 3 + cos (2x / 3 - π / 6)
f(x)=sin(2x/3)+cos(2x/3)cox(π/6)+sin(2x/3)sin(π/6)
=Sin (2x / 3) + ((radical 3) / 2) cos (2x / 3) + (1 / 2) sin (2x / 3)
=((radical 3) / 2) cos (2x / 3) + (3 / 2) sin (2x / 3)
=(radical 3) ((1 / 2) cos (2x / 3) + ((radical 3) / 2) sin (2x / 3)
=(radical 3) (sin (π / 6) cos (2x / 3) + cos (π / 6) sin (2x / 3)
=(radical 3) sin (2x / 3 + π / 6)
Maximum root 3, minimum root 3
If f (x) = - X's square + 2mx and G (x) = m / (x + 1) are decreasing functions in the interval [1,2], the value range of real number m is obtained
G (x) is a function of inverse scale function after translation, and its center of symmetry is (- 1,0), so the interval [1,2] is on one of its branches
Ψ m ＞ 0 (Note 1)
And f (x) = - X & sup2; + 2mx = - (x-m) & sup2; + M & sup2;, is a quadratic function with an opening downward, and its axis of symmetry is x = M
∵ f (x) is a decreasing function on [1,2]
Ψ m ＜ 1 (Note 2)
0 ＜ m ＜ 1 (Note 2)
Note:
1. This conclusion can also be proved by the definition of monotonicity (let X &; < x &;...) The method of derivative function can also be used;
2. Or write m ≤ 1, depending on your teacher how to specify (monotone interval includes endpoints)
If 3sina + cosa = 0, then the value of 1 / cos ^ 2A + sin2a is
3sina+cosa=0
cosa=-3sina
cos²a=9sin²a
sin2a=2sinacosa=-6sin²a
The original formula = (Sin & # 178; a + cos & # 178; a) / (9sin & # 178; a-6sin & # 178; a) = 10sin & # 178; a / (3sin & # 178; a) = 10 / 3
3sina+cosa=0，tana=-1/3
1/(cos^2a+sin2a)=((sina)^2+(cosa)^2)/((cosa)^2-(sina)^2+2sinacosa)=((tana)^2+1)/(1- (tana)^2+2tana)=(1/9+1)/(1-1/9-2/3)=5
From the question, the tangent value is - 1 / 3, and then the cos ^ 2a, sin2a are tangent, there is a formula. The answer is 0.65
If f (x) = the square of x-2mx + 1 is a decreasing function from negative infinity to two and an increasing function from two to positive infinity, then what is f (1) equal to?
The derivative of F (x) is 2x-2m;
From negative infinity to 2, there are: 2x-2mm;
So m = 2; then f (1) = - 2
If 3sina + cosa = 0, then the value of 1 (COS square a sin2a) is?
3sina + cosa = 0, that is, Tana = - 1 / 3, sin2a = 2sinacosa = 2tana * (COSA) ^ 2
Original formula = 1 / (2tana * (COSA) ^ 4)
=((1 + (Tana) ^ 2) ^ 2) / (2tana) (where 1 / (COSA) ^ 2 = 1 + (Tana) ^ 2)
=-50/27
If the function f "X" is equal to the square of - x + 2aX and G is equal to x + 1 / A, both of them are decreasing functions in the interval [1.2], then the value range of A
Let X1 and X2 be on the interval [1,2], and X1 be greater than x2. From F (x) as the decreasing function, we can get that: F (x1) - f (x2) = (x2 ^ 2-x1 ^ 2) + 2A (x1-x2) is less than or equal to 0, that is, (x2-x1) (x1 + x2-2a) is less than or equal to 0, because x2-x1 is less than 0, so X1 + x2-2a is greater than or equal to 0, so a is less than or equal to (x1 + x2) / 2, so a is less than or equal to 1
If 3sina + cosa = 0, then the value of 1 / 2 of (COSA + sin2a) is 0
3sina + cosa = 0 cosa = - 3sina cosa = 1-sina Sina = 1 / 10 cosa + sin2a = 9sina + 2sinacosa = 9sina-6sina = 3sina, the formula = 10 / 3
If f (x) = x & # 178; + 2 (A-1) x + 2 is a decreasing function on (- ∞, 4], what is the value range of a
Give detailed reasons
For quadratic function of one variable, the coefficient of X & # is 1 > 0, so the curve changes from left to right with the opening upward
Because the curve is symmetric with respect to x = 1-A and is a decreasing function on (- ∞, 4]
(because the function is on the left side of the axis of symmetry, it is always a decreasing function. The upper limit of the interval is 4, which can be either the axis of symmetry or the point on the left side of the axis of symmetry.)
So, 1-A ≥ 4, the solution is a ≤ - 3
f(x)=x²+2（a-1）x+2
The axis of symmetry is x = 1-A
Because it is a decreasing function on (- ∞, 4]
For 1-A ≥ 4
So a ≤ - 3
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