If the function y = sin (2x + φ) (0 ≤ φ ≤ π) is an even function on R, then the value of φ is () A. 0B. π4C. π2D. π

If the function y = sin (2x + φ) (0 ≤ φ ≤ π) is an even function on R, then the value of φ is () A. 0B. π4C. π2D. π

The function y = sin (2x + φ) is an even function on R, that is, when x = 0, the function gets the maximum value, so f (0) = ± 1, that is, sin φ = ± 1, so φ = k π + 12 π (K ∈ z). If and only if k = 0, we get φ = 12 π, which conforms to 0 ≤ φ ≤ π, so we choose C

If the function y = sin (2x + φ) (0 ≤ φ ≤ π) is an even function on R, then the value of φ is______ ．

The function y = sin (2x + ϕ) is an even function on R, that is, when x = 0, the function gets the maximum value, so f (0) = ± 1, that is, sin ϕ = ± 1, so ϕ = k π + π 2 (K ∈ z). If and only if k = 0, we get φ = π 2, which conforms to 0 ≤ φ ≤ π, so the answer is: π 2

Function y = sin (2x + φ) (0

A: it can be solved by definition
If y (x) is an even function, then y (- x) = y (x)
y(-x)=sin(-2x+Φ)=y(x)=sin(2x+Φ)
-sin2xcosΦ+cos2xsinΦ=sin2xcosΦ+cos2xsinΦ
So: 2sin2xcos Φ = 0 holds for any x, then cos Φ = 0
Because 0

Some mathematical problems about function
1. The line y = MX + n intersects with y = 2x + 1 at (2, b), and intersects with y = - x + 2 at (a, 1) to find the value of M and n
2. When k is a value, the function y = 2-x, y = - X / 3 + 4, y = 4 / K x-3
2. When k is a value, the function y = 2-x, y = minus one third x plus four, y = four parts of K multiplied by x minus three has and only has one intersection.

1. The line y = MX + n intersects with y = 2x + 1 at (2, b), and intersects with y = - x + 2 at (a, 1) to find the value of M and n
Intersection with y = 2x + 1 at (2, b), substituting, B = 2 * 2 + 1 = 5, B = 5,
Intersection with y = - x + 2 at (a, 1), 1 = - A + 2, a = 1,
That is, the line y = MX + n goes through (2,5), (1,1)
5=M*2+N,
1=M*1+N
5-1=2M+N-1M-N,
4=M
1=4+N,N=-3,
2. When k is a value, the function y = 2-x, y = - X / 3 + 4, y = 4 / K x-3

Help to find several mathematical problems of function
Given that the domain of definition of function f (x) is [a, b], where 00 and a is not equal to 1), then f (x)=
If f (x-1 / x) = x ^ 2 + 1 / x ^ 2, then f (x)=
If f (x) = X-1 / x + 1, then f (x) + F (1 / x)=
Given f (√ x + 1) = x + 2 √ x, find f (x)=
If f (1-cosx) = sin ^ 2x, then f (x)=

brother, I can only say that you have more problems and no reward points. I guess you can't do it except those who are very idle.

Year X: 2000 2001 2002 2003
Number of people Y: 2520 2330 2140 1950
Is y a function of X

Year X: 2000 2001 2002 2003
Difference: 1
Number of people Y: 2520 2330 2140 1950
Phase difference: 190
So: y = 2520 - (x-2000) * 190

Ask for a math problem in the second semester of junior high school (piecewise function)
The admission fee of a park is 20 yuan / person for less than 20 people, and 10 yuan / person for more than 20 people
1. Write the analytic formula of the function between the ticket fee y (yuan) and the number of people x (people);
2. A class of 54 students are going to the park. How much does it cost to buy tickets?

Piecewise function

Ask a function math problem
For any quadratic function f (x) = x ＾ 2 + 2aX + C, and f (1) = - 1, if f (x) > 1 holds for any x ∈ [a, a + 2], the value range of real number a is obtained?
Thank you. Please speed up

f(x)=x＾2+2ax+c
You can see that the axis of symmetry of this function is x = - A
Therefore, it can be discussed in different situations
① When the axis of symmetry is in the interval:
That is: a ≤ - a ≤ a + 2
-1≤a≤0
Here: the minimum value of the function is f (- a)
-a²+c ＞-1
∵f（1） = -1
∴1-2a+c=-1
c = 2a-2
The original inequality is changed into a & sup2; - 2A + 1 < 0
Not to the point
② When the axis of symmetry is on the left side of the interval, i.e. - a < A
a > 0
The function is monotonically increasing, so a > 1
③ When the axis of symmetry is on the right side of the interval, i.e. a + 2 < - A, a < - 1
The minimum value of F (x) is f (a + 2)
therefore
f(a+2)= (a²+4+4a)+2a(a+2)+2(a-1)
The solution is as follows
A < - 3 or
A ＞ - 1 / 3 rounding off
in summary:
The value range of a is:
(-∞ ,-3) ∪ (1,+∞)

A mathematical problem about function
How to find the range of y = 2Sin & sup2; x-3cosx-1?

y=2sin²x-3cosx-1
=-2cos²x-3cosx+1
Let t = the range of cosx t [- 1,1]
Then y = - 2T & sup2; - 3T + 1 [- 1,1] becomes the problem of finding the range of quadratic function
The range of Y is [- 4,2]
If it's OK, I'll set the best task

On functions
For any real number, x > 0, f (x) > 1, and f (x + y) = f (x) * f (y), it is proved that

f(x+y)=f(x)*f(y),
x=y=0
f(0)=f(0)*f(0),
f（0）≠0
∴f(0)=1
X = - y, x < 0, then - x > 0
f(x-x)=f(x)*f(-x)=f(0)=1,
f(x)=1/f(-x)
-x＞0
∴f(-x)>1
0＜1/f(-x)＜1
x