If there is a common point between the straight line y = 3 / 4x + B and the curve y = root sign (4x-x Square), then the value range of B is to explain how to draw the semicircle under y = root sign (4x-x Square)?

If there is a common point between the straight line y = 3 / 4x + B and the curve y = root sign (4x-x Square), then the value range of B is to explain how to draw the semicircle under y = root sign (4x-x Square)?

If y = √ (4x-x 2), the square of both sides can get y 2 = 4x-x 2 (Y ≥ 0). The formula of X is y = - (X-2) 2 + 4, i.e. (X-2) 2 + y 2 = 4 (Y ≥ 0). This is a semicircle with (2,0) as the center and 2 as the radius (above the X axis). The slope of the straight line y = 3 / 4x + B is 3 / 4

If the square of the line y = x + B and the square of 1-x under the root of curve y = have exactly one common point, then the value range of real number B is?

Because they have only one common point, the solution of the system of equations formed by them has only one solution. By substituting y = x + B into the square of 1-x under the root sign, x + B = the square of 1-x under the root sign, and then the square of both sides gets: x 2 + 2 BX + B 2 = 1-x 2 x 2 + 2 BX + B 2 - 1 = 0, which shows that the equation has two equal

X = x + B curve If 9 − Y2 has a common point, then the value range of B is______ .

According to the meaning of the question, the equation of curve C can be arranged as Y2 + x2 = 9 (x ≥ 0). In order to make the straight line L and the curve C have only one common point, there are two situations as follows: (1) the straight line is tangent to the semicircle, the distance from the origin to the straight line is 3, tangent to point a, d = | b| 2 = 3, because B < 0, we can get b = - 32, which is full of the meaning of the question; (2) the straight line passes through

If the line y = x + m and the curve y = root (1-x Square) have two common points, then the value range of M is

The straight line y = x + M represents the horizontal axis intercept-m, and the vertical axis intercept M. the curve y = root (1-x Square) represents the upper semicircle with radius 1. From the image, we can see that when the root 2 > m > = 1, the two curves have two intersections

F (x) = sin (x) - 3cos (x) x belongs to the range of (0, PAI) to find f (x) emergency

In fact, I remember this formula when I was in high school, but now I can't remember it clearly when I was in college. Here's an idea: a * sin (a) + b * cos (a) = sqrt (a ^ 2 + B ^ 2) [sin (a + C)] [where Tan (c) = B / a a a * sin (a) - B * cos (a) = sqrt (a ^ 2 + B ^ 2) [cos (A-C)], Tan (c) = A / b the upper C is the degree! Sqrt is the root sign. First, you dissolve f (x) = sin (x) - 3cos (x), so it's easy. I hope it's useful for you. As for calculation, I'll forget it

According to the fraction of sin + 4x (2X-4) of sin + 4x, we can find the following formula Value: 1, x = 4 / 2, x = 3 / 4 urgent PI is pi

1.x=π/4
Then f (x) = sin π / 2 + 2sin0 - 4cos π / 2 + 3cos π
=1+0-0-3=-2
2.x=3π/4
Then f (x) = sin π + 2Sin π / 2 - 4cos3 π / 2 + 3cos3 π / 2
=0+2-0+0=2

How fast is the minimum positive period sum range of F (x) = 2Sin (Pie / 6-x) + 1

The minimum positive period is 2 π / | - 1| = 2 π
The range is [- 1,3]

F (x) = 2-cos (2x - π / 3) - 2Sin? X, X ∈ {0, π / 2}, find the range of F (x)

f(x)=2-cos(2x-π/3)-2sin²x
=2-1/2cos2x-√3/2sin2x-1+cos2x
=1+1/2cos2x-√3/2sin2x
=1+cos(2x+π/3)
Because x ∈ {0, π / 2}, then 2x + π / 3 ∈ {π / 3,4 π / 3},
Therefore, f (x) ∈ {0,3 / 2} means that the range of F (x) is: {0,3 / 2}

What is the range of y = - 2Sin (2x - π / 3) x ∈ [0,5 / 3]

x∈[0,5/3],
Then 2x - π / 3 ∈ [- π / 3, (10 - π) / 3]
π / 2

F (x) = 2Sin (2x + π / 6), if x belongs to [0, π / 2], find the value range of function f (x)

x∈[0,π/2]
2x+π/6∈[π/6,7π/6]
When 2x + π / 6 = π / 2, f (x) has a maximum value of 2
When 2x + π / 6 = 7 π / 6, f (x) has a minimum value of - 1
So the range is [- 1,2]