If the straight line y = KX + 1 and the curve X = radical y ^ 2 + 1 have two different intersections, then the value range of K is

If the straight line y = KX + 1 and the curve X = radical y ^ 2 + 1 have two different intersections, then the value range of K is

Curve X = radical y ^ 2 + 1
It is the right branch of the hyperbola x 2 - y 2 = 1
Make up the equations
x²-（kx+1）²=1
（1-k²）x²-2kx-2=0
△=b²-4ac= 4k²+8（1-k²）＞0
It is found that K 2 is less than 2
-√2＜k＜√2

If the curve y= If there is always an intersection point between 1-x2 and the straight line y = x + B, then the value range of B is___ If there is an intersection point, then the value range of B is___ If there are two intersections, then the value range of B is___ .

When the straight line is tangent to the semicircle, the distance d from the center of the circle (0, 0) to the line y = x + B is d = | b| 1 + 1 = r = 1, and the solution is b = 2, B = - 2 (round off). When the straight line passes through (- 1,0), substitute (- 1,0) into the equation y = x + B to get b = 1

If there are two different intersections between the curve y = 1 + root sign 4-x ^ 2 and the straight line y = K (X-2) + 4, find the value range of real number K My question is very simple, that is, the simplification of that formula 4-x ^ 2 under y = 1 + radical Add square This is where I have a problem First, y ^ 2 = 1 + 4-x ^ 2 Then x ^ 2 + y ^ 2 = 5 But I made Y-1 = root 4-x ^ 2 Squared, x ^ 2 + (Y-1) ^ 2 = 4 Who told me what was wrong,

The first step y ^ 2 = 1 + 4-x ^ 2 is wrong
Y=1+(4-x^2)^(1/2)
Y^2=1+(4-x^2)+2*(4-x^2)^(1/2)
"But I made Y-1 = 4-x ^ 2 under the radical
Square, x ^ 2 + (Y-1) ^ 2 = 4 "is right
The best way to solve this problem is to eliminate y first, and then move the root sign to one side and square it. Then we can get the bivariate equation of X and find the range of K according to two roots

If the curve y = root (1-x2) always has an intersection point with the straight line k (X-2) - y = 0, find the value range of K

A straight line: kx-y-2k = 0 curve y = √ (1-x?), converted into x ^ + y ^ = 1, y ≥ 0, (that is, the part of the circle above the x-axis, including the x-axis.)

If the line y = K (X-2) has an intersection point with curve y = root 1-x2, find the value range of K Y = 1-x2 under radical (2 is square)

Let's simplify the following two steps: (2) let's find the solution of the following equation: (2) let's simplify the equation!

When there are two intersections between the curve y = 1-radical (4-x ^ 2) and the straight line y = K (x-4) + 3, find the value range of K If there are two intersections, can we consider that there are two identical intersections? Seek detailed analysis

Two intersections cannot be considered as having two identical intersections. Having two intersections means having two different intersections
When there are two intersections between curve y = 1-radical (4-x ^ 2) and straight line y = K (x-4) + 3, find the value range of K
[explanation]:
y=1-√(4-x^2) ,-2≤x≤2,y≤1 .
It can be seen that the image y = 1 - √ (4-x ^ 2) is a circle C; X ＾ 2 + (Y-1) ＾ 2 = 4 is the lower half part cut by the line L: y = 1. The intersection point of C and l is a (- 2,1). B (2,1)
The straight line y = K (x-4) + 3 crosses the fixed point m (4,3),
When the straight line passes through point m (4,3) and point a (2,1), the slope is the smallest, Kmin = (3-1) / (4-2) = 1,
When the line y = K (x-4) + 3 is tangent to a semicircle,
The distance from the center of a circle (0,1) to the straight line y = K (x-4) + 3 D = 2,
|2-4k | / √ (1 + K ＾ 2) = 2, k = 0 or 4 / 3 (when k = 0, the straight line is tangent to the upper half of the circle, and is omitted),
∴1≤k＜4/3.

If the line y = x + B and the curve X= If 1 − Y2 has a common point, then the value range of B is______ .

The line y = x + B is a straight line with slope 1 and intercept B;
Curve X=
The deformation of 1 − Y2 is x2 + y2 = 1 and X ≥ 0
It is obviously a right semicircle with a center of (0,0) and a radius of 1
According to the meaning of the title, the straight line y = x + B and the curve X=
1 − Y2 has and has a common point
If we make their figures, we can easily get the value range of B is: - 1 < B ≤ 1 or B=-
2．
So the answer is: - 1 < B ≤ 1 or B=-
2．

F (x) = 2Sin (x) - 3cos (x) x belongs to the range of (0, PAI) to find f (x) emergency

F (x) = 2sinx-3cosx = √ 13sin (x - φ), where Tan φ = 3 / 2, sin φ = 3 / √ 13, that is, φ = arcsin3 / √ 13 because of 0

What is the range of F (x) = 2Sin (2x - π / 3) + 1?

Minimum = 2 (- 1) + 1 = - 1
Maximum = 2 (1) + 1 = 3
The range is [- 1,3]

Find the function f (x) = 2Sin ^ 2x + 2sinx-1 / 2, X belongs to the range of [π / 6,5 π / 6]

f(x)=2sin^2x+2sinx-1/2=2(sinx+1/2)^2-1
X belongs to [π / 6,5 π / 6]
So 1 / 2 ≤ SiNx ≤ 1
And is 1 ≤ f (x) ≤ 7 / 2
That is, the range is [1,7 / 2]