If the intersection of the line L1: y = KX and L2: y = - 2x + 4 is in the first quadrant, the value range of the real number k is obtained

If the intersection of the line L1: y = KX and L2: y = - 2x + 4 is in the first quadrant, the value range of the real number k is obtained

Simultaneous equations y = KX, y = - 2x + 4
The solution is: x = 4 / (K + 2), y = 4K / (K + 2)
K / (K + 2) > 0
So k > 0
Draw a sketch to see if the two intersect in the first quadrant, you can only see k > 0
y=kx，:y=-2x+4
By solving the equations, x = 4 / (K + 2), y = 4K / (K + 2)
Because of the intersection coordinates of L1 and L2 (4 / (K + 2), 4K / (K + 2))
If the intersection point is in the first quadrant, then K + 2 > 0.k / (K + 2) > 0, the solution is k > 0
The value range of real number k is k > 0
Method 1, calculation as above
Method 2, image method
Draw L2 on the group coordinate system, and you can see it at a glance,
As long as k > 0, it's the simplest and fastest
The range of function y = 2Sin & # 178; X + 3cosx-3
Y = 2Sin & # 178; X + 3cosx-3 = 2 (1-cos & # 178; x) + 3cosx-3 = - 2cos & # 178; X + 3cosx-1 = - 2 (COS & # 178; x-3cosx / 2 + 9 / 16) - 1 + 9 / 8 = - 2 (cosx-3 / 4) & # 178; + 1 / 8 when cosx = 3 / 4, there is a maximum value of 1 / 8; when cosx = - 1, there is a minimum value of - 2 (- 1-3 / 4) & # 178; + 1 / 8 = - 6. The range is [-...]
y=2(1-cos²x)+3cosx-3
=-2cos²x+3cosx-1
=-2(cosx-3/4)²+1/8
-1
In the known quadrant of the line L1 + 2 x-k + 2 X-Y, the value of the intersection point of the line is obtained
kx-y+k+2=0
2x+y-4=0
By solving the above equations, it can be concluded that
X = (2-k) / (2 + k), y = (6K + 4) / (2 + k) (because K ≠ - 2, when k = - 2, the two lines are parallel and there is no intersection)
Because the intersection is in the first quadrant
(2-k)/(2+k)>0
(6k+4)/(2+k)>0
By solving the above inequalities, - 2 / 3 can be obtained
The definition field of "root sign (2Sin square x + 3cosx-3)"?
√（2sin^2x+3cos x-3)
2(1-cos^2x)+3cosx-3>=0
-2cos^2x+3cosx-1>=0
t=cosx
-2t^2+3t-1>=0
1/2
If the intersection of the line y = KX + 2K + 1 and the line y = − 12x + 2 is in the first quadrant, then the value range of K is ()
A. −12＜k＜12B. −16＜k＜12C. k＞12D. k＞−12
The intersection of two straight lines is: y = KX + 2K + 1y = − 12x + 2, the solution of the equations is: x = 2 − 4k2k + 1y = 6K + 12K + 1, ∵ the intersection of the straight line y = KX + 2K + 1 and the straight line y = − 12x + 2 is in the first quadrant, ∵ 2 − 4k2k + 1 ＞ 06k + 12K + 1 ＞ 0, the solution of the inequalities is: − 16 ＜ K ＜ 12, so choose B
Let f (T) = 2T & # 178; - 4 λ| t | - 1 (λ∈ R) (1) when λ = 1 / 2, find the maximum and minimum value of y = (SiNx) in X ∈ [- π / 6,2 / π]
(2) If the equation f (SiNx) = 0 of X has two different real roots on [- π / 2, π / 2], the range of real number λ is obtained
The 2 / π above should be π / 2, right?
T = SiNx in the range of X ∈ [- π / 6,2 / π] is [- 1 / 2,1], then | t | is in the range of [0,1]. Because λ = 1 / 2, the minimum | t | is taken as 1 / 2, and | t | is taken as 0 or 1 at the maximum. The maximum and minimum values are - 1, - 3 / 2
SiNx on [- π / 2, π / 2] is in the range of [- 1,1]. According to the symmetry of F (T) function about y axis, f (T) has a real root on [0,1]. Taking the point (1,0) into the equation, we get λ = 1 / 4. So when λ is less than or equal to 1 / 4, it is consistent with the problem
The two intersection points of hyperbola x ^ 2-2y ^ 2 + kx-4k = 0 and straight line y = KX + 1 are symmetric about the Y axis, and the coordinates of the two intersection points are obtained
1. Take y = KX + 1 into hyperbolic equation to get: x ^ 2-2 (KX + 1) ^ 2 + kx-4k = 0; 2. Sort out: (1-2k ^ 2) x ^ 2-3kx-4k-2 = 0; 3. Because the two intersection points are half symmetric on Y axis, then X1 + x2 = 0; 4. That is 3K / (1-2k ^ 2) = 0, the solution is k = 0; 5. Substitute k = 0 into 2 to get: x ^ 2-2 = 0, the solution is X1 = 2 ^ 1 / 2, X2 =
When x ∈ [π 6,7 π 6], the range of function y = 3-sinx-2cos2x is___ .
∵ y = 3-sinx-2cos2x = 2sin2x-sinx + 1 = 2 (sinx-14) 2 + 78, ∵ x ∈ [π 6, 7 π 6], when - 12 ≤ SiNx ≤ 1, ∵ when SiNx = 14, Ymin = 78; when SiNx = - 12, ymax = 2; ∵ the value range of function y = 3-sinx-2cos2x is [78, 2]. So the answer is: [78, 2]
It is known that: hyperbola C1: Y1 = TX (t is constant, t ≠ 0) passes through point m (1, 2, 2); its hyperbola about Y-axis symmetry is C2, and the intersection points of line L1: y = KX + B (k, B are constant, K ≠ 0) and hyperbola C2 are a (1, m), B (n, - 1); (1) find the analytic formula of hyperbola C2; (2) find the coordinates of two points a and B and the analytic formula of line L1; (3) if line L1 is translated The intersection points of the line L2 and the hyperbola C2 are respectively marked as C and D (A and D, B and C are on the same branch of the hyperbola C2), and the quadrilateral ABCD is just a rectangle. Please write the analytical formula of the line CD directly
(1) As shown in the figure, ∵ point m (- 2,2) is m ′ (2,2) with respect to the y-axis symmetry. The analytic formula of hyperbola C2 is y = 4x; (2) ∵ two points a (1, m), B (n, - 1) are on hyperbola C2, ∵ M = 4, n = - 4, ∵ two points a and B coordinates are a (1,4), B (- 4, - 1), ∵ a (1,4), B (- 4, - 1) two points are on line L1: y = KX + B, ∵ K + B = 4 − 4K + B = - 1 K = 1b = 3, the analytic formula of line L1 is y = x + 3; (3) the analytic formula of line CD is y = x-3
Let f (x) = [(x + 1) &# 178; + SiNx] / X & # 178; + 1 find the maximum and minimum of the function
f`(x)=((2(x+1)+cosx)x^2-((x+1)^2+sinx)2x)/(x^4)+1=(2x+2+cosx-(x+1)^2-sinx+x^2)/x^2=(1+cosx-sinx)/x^2.
F '(x) = 0, when x = 2K π + π, the maximum value of F (x) is 1 + 2 / π + 1 / π ^ 2
When x = 2K π + 3 π / 2, the minimum value of F (x) is 2 + 4 / 3 π
It is suggested to use derivative to solve the problem. (the process is a bit complicated) hope to adopt!!
The problem is to find the maximum and minimum!
f(x)=1+(2x+sinx)/(x^2+1)
Let g (x) = (2x + SiNx) / (x ^ 2 + 1), G (x) be an odd function
f(x)max+f(x)min=2