If the function f (x) = 2cos (2x + ϕ) is odd and is an increasing function on (0, π 4), then the real number ϕ may be () A. −π2B. 0C. π2D. π

If the function f (x) = 2cos (2x + ϕ) is odd and is an increasing function on (0, π 4), then the real number ϕ may be () A. −π2B. 0C. π2D. π

Because the function f (x) = 2cos (2x + ϕ) is an odd function, so ϕ = k π + π 2, the function is an increasing function on (0, π 4), so f (x) = 2sin2x, so ϕ = − π 2

How to get the image of function y = 2cos (- 0.5x + 0.25pai) from the image of function y = SiNx

There are many kinds of impassable process, I will say one ~ the other is just the exchange of steps
First, change y = SiNx to y = cos (- x): shift the image to the left by 0.5pai; then change y = cos (- x) to y = cos (- 0.5x): expand the abscissa to twice the original; then change y = cos (- 0.5x) to y = cos (- 0.5x + 0.25pai): shift the image to the right by 0.5pai; finally, change y = cos (- 0.5x) to y = 2cos (- 0.5x + 0.25pai): expand the ordinate to twice the original
That should be it^-^

Is there a real number a such that y = 2cos (2x + a) is an increasing function at (0, π / 4)?
I think it should be π / 2 or 3 π / 2, because according to the induction formula, the cosine function will become a sine function and - sin a, and whether it is + sin a or - sin a, the function is an increasing function in the first quadrant

It's too easy to solve with derivative, y '= - 4sin (2x + a) > = 0 in [0, π / 4]
Just make a

Let f (x) = sin2x + 2cos & # 178; X + 1
① Find the maximum value of F (x) and the corresponding value of X,
② Finding the minimum positive period of F (x)

f(x)=sin2x + 2cos²x+1
=sin2x+cos2x+2
=√2(√2/2sin2x+√2/2cos2x)+2
=√2cos(2x-π/4)+2
2 + √ 2 for 2x - π / 4 = 2K π
Then x = k π + π / 8
The minimum positive period is π

Given a > 0, f (x) = - A (2cos & # 178; X + √ 3 sin2x) + 3A + B, when x ∈ [0, π / 2], f (x) ≤ 1
(1) Find the value of constant a and B
(2) Let g (x) = f (x + π / 2) and LG [g (x)] > 0, find the monotone increasing interval of G (x)

f（x）=-a（2cos²x+√3 sin2x）+3a+b
=-a（cso2x+√3sin2x）+4a+b
=-a/2·sin(2x+π/6)+4a+b
x∈[0,π/2] ,
2x+π/6∈[π/6,7π/6]
sin(2x+π/6）∈[-1/2,1]
f(x)∈[15a/4+b,7a/2+b] -5≤f（x）≤1
So 15A / 4 + B = - 5, 7a / 2 + B = 1
a=-16,b=57

Let f (x) = | x + 2 | + | x-a | be symmetric with respect to the line x = 1, then the value of a is?
(- 2, 0), its symmetry point about x = 1 is (3, 0), are you kidding?

On the symmetry of line x = 1, then f (x) = f (2-x)
Substituting into the expression to solve the absolute value equation
a=4

Let f (x) = (2x + 3) / (x-1), the image of the function y = g (x) and the image of the function y = f ^ - 1 (x-1) be symmetric with respect to the line y = x, and find the value of G (3)

Starting from the function y = g (x)
Substitute with specific value
Let B = g (a), (where A.B is the specific value)
The image of the function y = g (x) and the image of y = f ^ - 1 (x-1) are symmetric with respect to the line y = X
We get a = f ^ - 1 (B-1)
The inverse function is symmetric with respect to the line y = X
B-1 = f (a)
Substituting f (x) = (2x + 3) / (x-1) and a = 3
B is solved

Given that the image of the function f (x) = 3x + B and the image of the function g (x) = 3 / X-1 are symmetric with respect to the line y = x, then B is equal to X

The symmetric functions of x = f (x) / 3-B / 3 with respect to y = x are reciprocal functions
So B / 3 = 1, B = 3

Given that the image of the function f (x) = 3x + B and the image of the function g (x) = x / 3-1 are symmetric with respect to the line y = x, then the value of B is only equal to
Need detailed process!

If two functions are symmetric with respect to y = x, they are inverse functions of each other
So, if x = f (x) / 3-B / 3, then B / 3 = 1, B = 3

Given that the image of the function f (x) = 3x + B and the image of the function g (x) = - 1 are symmetric with respect to the line y = x, what is B equal to?
Given that the image of the function f (x) = 3x + B and the image of the function g (x) = x / 3-1 are symmetric with respect to the line y = x, what is B equal to?

Firstly, f (x) = 3x + B and G (x) = - 1 are symmetric with respect to the line y = X. the intersection of G (x) = - 1 and y = x can be found at (- 1, - 1), and this point must be on f (x) = 3x + B, so B = 2 is obtained