It is known that f (x) = asin (ω x + π / 4) (where x ∈ R, a > 0, ω > 0) has a maximum value of 2 and a minimum period of 8 (1) Find the analytic formula of function f (x) (2) if the abscissa of two points P and Q on the image of function f (x) are 2,4,0, then find the value of cos ∠ p0q

It is known that f (x) = asin (ω x + π / 4) (where x ∈ R, a > 0, ω > 0) has a maximum value of 2 and a minimum period of 8 (1) Find the analytic formula of function f (x) (2) if the abscissa of two points P and Q on the image of function f (x) are 2,4,0, then find the value of cos ∠ p0q

It is known that f (x) = asin (ω x + π / 4) (where x ∈ R, a > 0, ω > 0) has a maximum value of 2 and a minimum period of 8
(1) Find the analytic formula of function f (x) (2) if the abscissa of two points P and Q on the image of function f (x) are 2,4,0, then find the value of cos ∠ p0q
(1) Analysis: the maximum value of ∵ function f (x) = asin (ω x + π / 4) (where x ∈ R, a ﹥ 0, ω ﹥ 0) has a maximum value of 2 and a minimum period of 8
∴ω=2π/8=π/4
∴f(x)=2sin(π/4x+π/4)
(2) Analysis: F (2) = 2Sin (π / 2 + π / 4) = √ 2 = = > P (2, √ 2)
f(4)=2sin(π+π/4)=-√2==>Q(4,-√2)
∴|OP|=√6,|OQ|=3√2,|PQ|=2√3
∴cos∠POQ=(OP^2+OQ^2-PQ^3)/(2OP*OQ)=(6+18-12)/(2*√6*3√2)=√3/3

If f (x) = asin (x + tt / 4) + 3sin (x-tt / 4) is an even function, then a=

When x = tt / 4, the expression is asin (tt / 2)
When x = - TT / 4, the expression is 3sin (- TT / 2) = - 3sin (tt / 2)
The two are equal
So the answer - 3

The known functions are (x) = 2x? - 3x + 1, G (x) = (x - π / 6), (a ≠ 0) (1) When 0 ≤ x ≤ π / 2, find the maximum value of y = f (SiNx) (2) If for any x1 ∈ [0,3], there is always x2 ∈ [0,3], so that f (x1) = g (x2) holds, find the value range of real number a (3) The equation f (SiNx) = a-SiNx has two solutions on [0,2 π]

f(x)=2x²-3x+1,g(x)=Asin(x-π/6)
(1) When 0 ≤ x ≤ π / 2, sin0 ≤ SiNx ≤ sin (π / 2), i.e. 0 ≤ SiNx ≤ 1
f(sinx)=2（sinx）²-3sinx+1=2（sinx-3/4）²-1/8
When SiNx = 0, the maximum value of y = f (SiNx) is 1
（2）
f(x)=2(x-3/4）²-1/8
X = [0,3] = > F (x) min = - 1 / 8 (x = 3 / 4), Max 10 (x = 3)
X = [0,3] = > G (x) min = - | a | / 2 (x = 0) max | a | (x = π / 6 + π / 2)
|A|>=10
-|A/ 2A > = 10 or a

Known function f (x)= −1 3x+1 6，x∈[0，1 2] 2x3 x+1，x∈(1 2,1], function g (x) = asin (π) 6x) - 2A + 2 (a > 0), if there is x1, X2 ∈ [0, 1], such that f (x1) = g (x2) holds, then the value range of real number a is () A. [-2 3，1] B. [1 2，4 3] C. [4 3，3 2] D. [1 3，2]

When x ∈ [0, 1
2] When y = 1
6-1
3x, the range is [0, 1
6]；
x∈（1
2,1], y = 2x3
x+1，y′=4x3+6x2
(x + 1) 2 > 0 always holds, so it is an increasing function with a range of (1)
6，1]．
Then when x ∈ [0,1], the value range of F (x) is [0,1],
When x ∈ [0,1], G (x) = asin (π)
6x）-2a+2（a＞0），
Is an increasing function, and the range is [2-2a, 2-3a]
2]，
∵ the existence of x1, X2 ∈ [0, 1] makes f (x1) = g (x2) hold,
∴[0，1]∩[2-2a，2-3a
2]≠∅，
If [0, 1] ∩ [2-2a, 2-3a
2]=∅，
2-2a > 1 or 2-3a
2 < 0, i.e. a < 1
2, or a > 4
3．
The value range of a is [1]
2，4
3]．
Therefore, B

Function f (x) = 3x? - 2x + 1 (x ≥ 0), find the function f (- x)

F (- x) = - 3x ^ 2 + 2x + 1 can't be simplified later

It is proved that the function f (x) = - 3x 2 + 2x is a decreasing function in (1 / 3, + ∞)

prove
f（x）=-3x²+2x
=-3（x²-2/3x+1/9-1/9）
=-3（x²-2/3x+1/9）+1/3
=-3（x-1/3）²+1/3
The opening of quadratic function is downward, and the axis of symmetry is x = 1 / 3
On the right side of the axis of symmetry is the subtraction function
That is, the function f (x) = - 3x 2 + 2x is a decreasing function in (1 / 3, + ∞)

If f (x) = asin (x + π / 4) + 3sin (x - π / 4) is even function, then the value of real number a is? f(x)=asin(x+π/4)+3sin(x-π/4)=asin(x+π/4)-3cos(x+π/4) F (x) = √ (a ^ 2 + 9) cos (x + π / 4 - φ). Where Tan φ = - A / 3 ∵ f (x) is an even function ᙽ π / 4 - φ = k π (K ∈ z) φ = - K π + π / 4, Tan φ = 1, so a = - 3 The answer is right. I hope you can help me to see if there is something wrong with the process? For example, should f (x) = √ (a ^ 2 + 9) cos (x + π / 4 - φ) be f (x) = √ (a ^ 2 + 9) sin (x + π / 4 - φ)? And how can tan φ = - A / 3 come out? Why is f (x) an even function then π / 4 - φ = k π? If I change it to f (x) = √ (a ^ 2 + 9) sin (x - π / 4 + φ), what should I do?

Landlord,
sin(a-b)=sinacosb-cosasinb
cos(a-b)=cosacosb+sinasinb
It is easy to know from the above process that sin φ = a, cos φ = - 3, so tan φ = - A / 3
In addition, the following methods can be adopted:
f(x)=asin(x+π/4)+3sin(x-π/4)=acos(x-π/4)+3sin(x-π/4)
=(a ^ 2 + 9) sin (x - π / 4 + φ), where Tan φ = A / 3
∵ f (x) is an even function ᙽ φ - π / 4 = k π + π / 2 (K ∈ z)
∴φ=Kπ+3π/4,∴tanφ=tan3π/4=-1
∴a=-3

If f (x) = asin (x + π) 4)+bsin(x−π 4) If (AB ≠ 0) is an even function, then the ordered pair of real numbers (a, b) can be______ (Note: write a set of numbers that you think are correct.)

ab≠0，f(x)＝asin(x+π
4)+bsin(x−π
4)
=a (
Two
2sinx+
Two
2cosx)+b(
Two
2sinx−
Two
2cosx)
=
Two
2（a+b）sinx+
Two
2（a-b）cosx．
∵ f (x) is an even function,
ν as long as a + B = 0,
We can take a = 1, B = - 1

If f (x) = asin (x + pi / 4) + 3sin (x-pi / 4) is an even function, then a=

F (x) = asin (x + π / 4) + 3sin (x - π / 4) + 3sin (x - π / 4) is the even function of the even function, and f (- x) = f (x) is the asin (- x + π / 4) + 3sin (- X - π / 4) = asin (x + π / 4) + 3sin (x - π / 4) a [sin (- x + π / 4) - sin (x + π / 4)] = 3 [sin (x-π / 4) - sin (- x-π / 4)] 2acos π / 4sin (- x-π / 4)] 2acos π / 4sin (- x = x = x) = 2acos π / 4sin (- x) = 4sin (- x) = 2acos π = 4sin (- x = x = x = x = 6 cos (- π / 4) sinx-2acos

Given that even function f (x) satisfies f (x + 1) = f (x-1) and X ∈ [0,1], if f (x) = x, then f (x) = (1 / 10) x power solution on [0,4] A1 B2 C3 D4

D4
F (x + 1) = f (x-1), so t = 2
Combined with images