If the function y = sin [2 (x - π / 3) + φ] is even, tangent 0 ＜φ＜π, then=

If the function y = sin [2 (x - π / 3) + φ] is even, tangent 0 ＜φ＜π, then=

Y = sin [2 (x - π / 3) + φ] is even function
Then f (- x) = f (x)
sin[2(-x-π/3)＋φ]=sin[2(x-π/3)＋φ]
sin[-2x-2π/3＋φ]=sin[2x-π/3＋φ]
sin[-2x-2π/3＋φ]-sin[2x-2π/3＋φ]
=2cos（-2x-2π/3＋φ+2x-π/3＋φ)/2*sin(
-2x-2π/3＋φ-2x+π/3-φ）/2
=0
That is (- 2x-2 π / 3 + φ + 2x-2 π / 3 + φ) / 2 = π / 2 + K π
Then φ - 2 π / 3 = π / 2 + K π
φ=π/6+kπ
0＜φ＜π
Then φ = π / 6

Function y = sin (2x + Q) (0

Two Pai
If it is an even function, it needs to be transformed into cos, because the range of P can get the value of P

If the function y = sin (x / 2 + ψ) is even, then the first value of ψ is?

Y = sin (x / 2 + ψ) is an even function, that is to say, try to make y = sin (x / 2 + ψ) = cos (x '), any (x') in COS (x '), so we can make (x') = x / 2, then sin (x / 2 + ψ) = cos (x / 2), then ψ = K (π / 2), (k is an integer), so we can take k = 1, then ψ = (π / 2), k = - 1

It is known that the value range of independent variable x in a function y = KX + B is - 2 less than or equal to x less than or equal to 6, and the value range of corresponding function is - 11 less than or equal to 6
When y is less than or equal to 6, find the analytic expression of this function

When k > 0, it increases monotonically;
f(-2)=-11;f(6)=6;
-2k+b=-11(1)
6k+b=6(2)
8k=17;
k=17/8;
b=6-51/4=-27/4;
∴y=17x/8-27/4;
When k < 0, it decreases monotonically;
f(-2)=6;f(6)=-11;
-2k+b=6(1)
6k+b=-11(2)
8k=-17;
k=-17/8;
b=6-17/4=7/4;
∴y=-17x/8+7/4;
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In a function, when the independent variable x = 3, y = 3; when x = 1, how to draw the function image of y = - 1?

A linear function is a straight line,
Two points make a straight line,
Draw two points and connect them into a straight line

Let f (x) = 1 / 2-1 / 2 x + 1
It is proved that f (x) is an odd function
Finding the range of F (x) on [1,2]

Use f (- x) = - f (x) to prove whether it is an odd function. You should know this theorem. Then find out the extreme point of the machine, make the value of the function equal to 0, and find out the corresponding value of X

2. If the real numbers x, y and Z satisfy √ x ＋ √ (Y-1) ＋ √ (Z-2) = 1 / 2 (x + y + Z), find the value of logz (x + y)

The answer is logz (x + y) = 1, let √ x = a, √ (Y-1) = B, √ (Z-2) = C, then x = A & sup2;, y = B & sup2; + 1, z = C & sup2; + 2, the original formula is reduced to a + B + C = 1 / 2 (A & sup2; + B & sup2; + C & sup2; + 3), that is a & sup2; + B & sup2; + C & sup2; - 2a-2b-2c + 3 = 0

Solving a mathematical problem of function
The vertex of the parabola y = (X-2) ^ 2-m ^ 2 (M > 0) is p. if the intersection points with the X axis are a, B from left to right, and the angle APB = 90 degrees, calculate the perimeter of the triangle APB

Because P (2, - m ^ 2), a (2-m, 0), B (2 + m, 0),
And the angle APB is 90 degrees,
So (2 + m) - (2-m) = 2 * | - m ^ 2 |,
Because m > 0,
So m = 1,
So AB = 2m = 2,
AP = BP = radical 2,
So perimeter = AB + AP + BP = (2 + 2 root 2)

It is known that the function f (x) defined on (- 1,1) satisfies f (- x) = - f (x), and f (1-A) + F (1-a2) < 0. If f (x) is a decreasing function on (- 1,1), the value range of a is obtained

From F (1-A) + F (1-a2) < 0, we get f (1-A) < f (1-a2), ∵ f (- x) = - f (x). The inequality is equivalent to f (1-A) < f (A2-1). Since f (x) is a decreasing function on (- 1,1), the range of a is 0 < a < 1

Help me solve a function math problem!
The title is: known parabola y = x.x + PX + Q and X axis only have one common point, the intersection coordinate is (- 2,0). Find P, Q. hope that the expert can write out the process for me! The x.x in the title is the square of X!

Using the formula of symmetry axis: x = - B / 2A
Then: - 2 = - P / 2 gives P = 4, then the original formula is y = x.x + 4x + Q
Then substitute (- 2,0) into: 0 = (- 2) × (- 2) + 4 × (- 2) + Q
The solution is q = 4
So p = 4, q = 4