It is known that the function f (x) = sin (Wx + &) (W > 0,0 <, & < < PAI) is an even function on R. its graph is symmetric about point m (3 Pai / 4,0) and monotone on [0, Pai / 2] Function, find the value of & and W

It is known that the function f (x) = sin (Wx + &) (W > 0,0 <, & < < PAI) is an even function on R. its graph is symmetric about point m (3 Pai / 4,0) and monotone on [0, Pai / 2] Function, find the value of & and W

&=π/2,w=2.
F (x) = sin (2x + π / 2) = cos2x, even function, symmetric about point m (3 π / 4,0), and monotone decreasing function on [0, π / 2]

It is known that the function f (x) = sin (ω x + φ) (ω > 0, 0 ≤ φ ≤ π) is an even function on R and its image is related to the point m (3 π) 4, 0) are symmetric and in the interval [0, π 2] Let's find the values of φ and ω

If f (x) is an even function, f (- x) = f (x),
That is sin (- ω x + φ) = sin (ω x + φ),
So - cos φ sin ω x = cos φ sin ω x,
It holds for any X and W > 0,
So cos φ = 0
Let 0 ≤ φ≤ π according to the problem, so the solution is φ = π
2，
The image of F (x) is symmetric about point M,
F (3 π) is obtained
4−x)＝−f(3π
4+x)，
Take x = 0 to get f (3 π)
4）=sin（3ωπ
4+π
2）=cos3ωπ
4，
∴f（3π
4）=sin（3ωπ
4+π
2）=cos3ωπ
4，
∴cos3ωπ
4=0，
And W > 0, we get 3 ω π
4=π
2+kπ，k=0，1，2，3，…
∴ω=2
3（2k+1），k=0，1，2，…
When k = 0, ω = 2
3，f（x）=sin（2
3x+π
2) At [0, π
2] The upper one is a minus function, which satisfies the meaning of the question;
When k = 1, ω = 2, f (x) = sin (2x + π)
2) = cos2x in [0, π]
2] The upper one is a minus function, which satisfies the meaning of the question;
When k = 2, ω = 10
3，f（x）=sin（10
3x+π
2) At [0, π
2] Is not a monotone function;
So, we get ω = 2
3 or 2

The function f (x) = sin (Wx + &) (W > 0,0 ≤π) is an even function on R. the image is symmetric about M (3 / 4 π, 0) and monotone in the interval [0, π / 2] Function, find the value of & and W

Since it is an even function on R, there is f (0) = 1
f(0)=sin(&)=1,&=π/2
In the interval [0, π / 2], it is monotone function, so t > = π, so 2 π / W > = π, W

Let f (x) be defined on (- L, l), and prove that f (x) + F (- x) is even function. I don't understand the next step in the solution, please parse it Let g (x) = f (x) + F (- x), - L

Multiply both sides by - 1 and change direction

Let even function f (x) satisfy f (x) = 2X-4 (x ≥ 0), then {x | f (X-2) > 0}=______ .

If the even function is full f (x) and f (x) = 2X-4 (x ≥ 0), then f (x) = f (| x |) = 2| x | - 4,
Then f (X-2) = f (| X-2 |) = 2 | X-2 | - 4, so that f (| X-2 |) > 0,
Only 2 | X-2 | 4 | 0, | X-2 | 2, the solution x | 4, or X ﹤ 0
So the answer is: {x| x < 0, or x > 4}

The image of the function y = f (x) defined on the interval [- π, 2 / 3 π] is symmetric with respect to the line x = - π / 6 The image of the function y = f (x) defined on the interval [- π, 2 / 3 π] is symmetric with respect to the line x = - π / 6 When x ∈ [- π / 6,2 / 3 π], The function f (x) = asin (ω x + φ), Find the expression of function y = f (x) in [- π, 2 / 3 π]

f(x)=Asin(ωx+φ),x∈[-π/6,2/3π]
It is known that the image of the function y = f (x) defined on the interval [- π, 2 / 3 π] is symmetric with respect to the straight line x = - π / 6,
If x ∈ [- π, - π / 6], the symmetric point of x = - π / 6 is set as x1, then (x1 + x) / 2 = - π / 6, that is, X1 = - X - π / 3, and x1 ∈ [- π / 6,2 / 3 π], so f (x) = asin (ω x + φ) is satisfied,
That is, f (x) = asin (ω (- X - π / 3) + φ), that is, f (x) = asin [- ω x + (φ - ω π / 3)],
So the function y = f (x) is a piecewise function on [- π, 2 / 3 π]:
f(x)=Asin[-ωx+(φ-ωπ/3)],x∈[-π,-π/6]；
f(x)=Asin(ωx＋φ),x∈[-π/6,2/3π].

Let the even function f (x) defined on [- 2,2] decrease monotonically on the interval [0,2], if f (1-m)

Function f (x) decreases monotonically on interval [0,2]
And f (x) is an even function, so the function f (x) increases monotonically on the interval [- 2,0]
Because f (x) is an even function, f (- x) = f (x) = f (| x|)
f(1-m)

If FX is an even function on the domain [- 2,2] and FX is an increasing function on the interval [0,2], then the value range of real number m of F (1-m) > F (1 + 2m) is satisfied A.（-1,0） B.【-1,0） C.【-1,0】 D.（-1,0】 High school mathematics can't do ah. I don't know what it means

Ideas:
Because the definition domain of its function has been given, the brackets around the inequality must be within this range
-2 < 1-m < 2 (1)
-2 < 1+2m < 2 (2)
Because this function is even, and 0-2 is an increasing function, so it is a minus function from - 2 to 0, and about the y-axis symmetry, you can draw a picture to experience, to ensure that f (1-m) > F (1 + 2m) holds
Only | 1 + 2m | > | 1-m | (3)
Three forms are combined to solve inequalities

It is known that even function f (x) defined on R is monotone decreasing function on interval [0, positive infinity). If f (1-x) < f (x), then the value range of real number x is

Because f (x) is a monotone decreasing function on the interval [0, positive infinity]
When x > 0
1-X>X
X

Let the even function y = f (x) defined on R be a decreasing function on the interval [0, + ∞). If the real number x satisfies f (x) > F (2x + 1), find the value range of X

F (x) > F (2x + 1), and f (x) is even function
So: F (| x |) > F (| 2x + 1|)
F (x) is a decreasing function on the interval [0, + ∞)
So: |x|0
x> - 1 / 3, or X