It is known that the minimum positive period of the function y = radical 3sinwxcoswx cos ^ 2wx + 3 / 2, (x ∈ R, w ∈ R) is π, and when x = π / 6, the function has a minimum value, (1) find the analytic formula of F (x), and (2) find the monotone increasing interval of F (x)

(1) (x) = (3 / 2) sin2 (3 / 2) sin2 ω x (Cos2 ω x + 1) / 2 + 3 / 2 = 3 / 2 = sin2 (2 ω x + 1) / 2 + 3 / 2 = sin (2 ω x-π / 6) + 1 because t = π, so t = 2 π / 2 ω = π = π, therefore, ω = 1, so f (x) = sin (2x - π / 6) + 1 (2) - π / 2 + 2K π ≤ 2x - π / 6 ≤ π / 2 + 2K π / 6 ≤ π / 2 + 2K π / 6 ≤ π / 2 + 2K π π / 2 + 2K π / π / π / 2 + 2K π / π / 2 + 2K 6 + K π ≤ x ≤ π / 3 + K π single

Given that the minimum positive period of the function f (x) = Radix 3sinwx coswx cos? Wx + 3 / 2 (W > 0, X ∈ R) is known to be TT (1). Find the analytic formula of F (x) (2) There is an intersection point between the image of y = 1 - ƒ (x) and the line y = a on [0, π]. The value range of real number a is calculated

(1) F (x) = √ 3 / 2sin2wx - (cos2wx + 1) / 2 + 3 / 2 = √ 3 / 2sin2wx-1 / 2coswx + 1 = sin (2wx - π / 6) + 1 since the minimum positive period of the function is π, i.e. 2 π / 2W = π, w = 1, f (x) = sin (2x - π / 6) + 1 (2) y = 1-f (x) = - sin (2x - π / 6) when x belongs to [0, π], - π / 6

Is f (x) = odd function and even function under root sign (1-x) + root sign (x-1)

The definition field is x = 1, not symmetric about the origin
So it's a non odd, non even function

Is f (x) = (x-1) radical 1 + X / 1-x an odd or even function? The process of problem solving is required

Define the domain first
F (x) = X-1 radical 1 + X / 1-x = radical (1 + x) (1-x)
F (- x) = - radical (1 + x) (1-x) = - f (x)
So it's even

Is f (x) = 1-x under radical + is X-1 odd or even?

F (x) = √ (1-x?) + √ (x ^ - 1) from the meaning of the title, we can get that 1-x ^ ≥ 0 and x ^ - 1 ≤ 0, ν 1 ≤ x ≤ 1, i.e. x = 1, so x = ± 1, the domain {- 1,1} is symmetric about the origin f (- x) = √ (1 - (- x) 2) + √ ((- x) - 1) = √ (1-x) + √ (x ^ -...)

Given that the quadratic function f (x) is an even function, then f (1 + radical 2) - f (one minus one half of the root sign) = online

(1 + radical 2) and (one minus one half of the root) are exactly opposite numbers (we can find it by rationalizing the denominator of the last number), while f (x) is an even function, so the calculated value is 0

Even function f (x) defined on R satisfies f (x + 1) = - f (x), and increases monotonically on [- 1, 0], a = f (3), B = F（ 2) If C = f (2), then the relationship between a, B, C is () A. a＞b＞c B. a＞c＞b C. b＞c＞a D. c＞b＞a

In this paper, we obtain that (f) is an even function (F, f) = 1, f (2) = 1, f (2) = 1, f (2) is a periodic function

F (x) = root 2 cos half x minus (A-1) sin half x is even function Find a positive period

f(x)=√2cos(x/2)-(a-1)sin(x/2)
f(-x)=√2cos(x/2)+(a-1)sin(x/2)
f(x)=f(-x)
a-1=0
A=1
f(x)=√2cos(x/2)
T=2π/ω=2π/(1/2)=4π

If the even function f (x) defined on R satisfies f (x + 1) = - f (x), then compare the size of radical two of F 3, F 2 and F

From F (x + 1) = - f (x), we know that f (3) = - f (2) = f (2)
f(2)=-f(1)=f(1)
It can be deduced that f (n) = f (n + k), N and K are integers
So f (3 = f (2)
The radical two of F cannot be compared

Let f (x) = radical 3sin (2x + FAI) - cos (2x + FAI) (0

（1）f(x)=√3sin(2x+φ)-cos(2x+φ）=2[√3/2*sin(2x+φ)-1/2*cos(2x+φ）]
=2Sin (2x + φ - π / 6) because it is an even function, the function f (x) takes the maximum or minimum value at x = 0
ν 2 * 0 + φ - π / 6 = π / 2 + K π φ = 2 π / 3 + K π, and ∵ 0

It is known that the minimum positive period of the function y = radical 3sinwxcoswx cos ^ 2wx + 3 / 2, (x ∈ R, w ∈ R) is π, and when x = π / 6, the function has a minimum value, (1) find the analytic formula of F (x), and (2) find the monotone increasing interval of F (x)