F (x) = 2 radical 3sin (3wx + Pai / 3) (W greater than 0) (1) if f (x + Z other) is an even function with period 2, find w and Z other values (2) f (x) is in (0, Pai / 3) F (x) = 2 radical sign 3sin (3wx + Pai / 3) (W greater than 0) (1) if f (x + Z other) is an even function with period 2, find w and Z other values (2) f (x) is an increasing function on (0, Pai / 3), and find the maximum value of W

(1) Because the period is 2, so w = 1 / 3, because even function, so 3W (x + Z) + Pai / 3 + 3W (x-z) + Pai / 3 = pie, so z = Pie / 6
（2）2

It is known that the period of the function FX = asin (Wx + ψ) + n is π, f (π / 4) = √ 3 + 1, and the maximum value of FX is 3 Write the expression of FX Write the symmetry center and axis equation of function FX

By F (π / 4) = asin (2 * π / 4 + 4 + ψ) + n = asin (π / 2 + ψ) + n = asin (π / 2 + 2 + ψ) + n = ACOS ψ + n = ACOS ψ + n = 3 + 1, from the maximum value of FX is 3, a + n = 3 can get n = 1, a = 2, ψ = π / 6 = 6, so, f (x) = 2Sin (2x + π / 6) + 1 symmetry center is: 2x + π / 6 = k π, the solution is: X = - (π / 12) + (K π) / 2 (K π) / 2 (K π) / 2 + (K π) / 2 + (K π) / 2 + (K π / 12) + (K π /

Given that the minimum positive period of the function f (x) = asin ω x + bcos ω x (ω > 0, a, B are not all zero) is 2, and f (1 / 4) = root 3, the value range of the maximum value of F (x) is obtained

The minimum positive period of F (x) = asin ω x + bcos ω x is t = 2 π / w = 2, w = π,
f(x)=asinπx+bcosπx
f(1/4)=√2/2*(a+b)=√3,
That is, a + B = √ 6,
The maximum value of F (x) is √ (a 2 + B 2)
Because of a 2 + B 2 ≥ (a + b) 2 / 2 = 3,
So √ (a 2 + B 2) ≥ 3,
In other words, the maximum value of F (x) is in the range of [√ 3, + ∞)

The absolute value of equation x - 1 = the square of 1 - (Y-1) under the radical sign

Method 1: discuss the positive and negative of X. when x > 0, the absolute value can be directly removed; when x < 0, add negative sign, and then square the left and right sides of the equation

The equation | [(x + 2) ^ 2 + y ^ 2] - [(X-2) ^ 2 + y ^ 2] | = 2 is simplified to obtain that "]" is the root, ^ 2 is the square and | is the absolute value

x^-y^2/3=1

Let a be the sum of all absolute values of the equation x square minus root 2003 times x-520 = 0?

Is it the sum of the absolute values of all roots?
First, the discriminant of root is
2003+2080=4083
A=|x1|+|x2|=√4083
A^2=4083

What is the curve represented by the root sign of the equation [(x + 3) ^ 2 + (Y-1) ^ 2] = | X-Y + 3 | Demand solution process

In the equation, the root sign [(x + 3) ^ 2 + (Y-1) ^ 2] = (| X-Y + 3 | / √ 2) * √ 2. In the equation, the root sign [(x + 3) ^ 2 + (Y-1) ^ 2] represents the distance d1 from the point (x, y) to the fixed point (- 3,1); | X-Y + 3 | / √ 2 represents the distance d2 from the point (x, y) to the fixed line X-Y + 3 = 0; that is, D1 = √ 2d2, then D1 /

What is the curve represented by the equation | X-1 | = root sign (1 - (Y-1) ^ 2)? The answer is two semicircles. Why? Please explain in detail

|X-1 | = root sign (1 - (Y-1) ^ 2) (x-1) ^ 2 = 1 - (Y-1) ^ 2 1 - (Y-1) ^ 2 ≥ 0 (x-1) ^ 2 + (Y-1) ^ 2 = 1 (Y-1) ^ 2 ≤ 1 (x-1) ^ 2 + (Y-1) ^ 2 = 1 0 ≤ y ≤ 2 means a circle with (1,1) as the center and 1 as the radius. If the answer is two semicircles, the answer is wrong

Equation (x + Y-1) The curve represented by x2 + Y2 − 4 = 0 is () A. B. C. D.

The original equation is equivalent to:
x+y−1＝0
X2 + Y2 ≥ 4, or x2 + y2 = 4; when x + Y-1 = 0, it is necessary to
If x2 + Y2 − 4 is meaningful, the equation holds, that is, X2 + Y2 ≥ 4. At this time, it represents the part of the straight line x-y-1 = 0 that is not in the circle x2 + y2 = 4, which is a very error prone link
Therefore, D is selected

In the following equation, where y is a function of X, there are several 3x-2y = 0 x ^ 2-y ^ 2 = 1, y = the absolute value of the radical x, y = x, x = the absolute value of Y

Whether y is a function of X depends on whether there is a unique y value corresponding to a given x value
The second expression is y? = x? - 1, y = ± √ (x? 2 - 1)
You can see that given an x value, there are two Y values corresponding to it
The fifth expression also has two Y values corresponding to X. for example, if x = 1, y can be ± 1
Therefore, only 1, 3, 4 are consistent
So there are three