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Anonymous users2024-02-071)) From the meaning of the title: -x+6=k x
can be squared as x-6x+k=0
To satisfy that there are two common points, then =36-4k>0
Solution: k<9
2) You might as well set the intersection of the curve and the straight line to the one with the larger abscissa as A, the one with the smaller abscissa as B, the intersection of the straight line and the Y axis as D, and the intersection of the straight line and the X axis as E. The equation of the curve and the straight line is combined to find the smaller x, that is, k x x+6, calculate the abscissa of a and b, and then substitute y=-x+6 to obtain the ordinate of a and b, and represent dob according to the abscissa ordinate of tan dob = point a
EOA.
aob=90+ dob+ eoa (denoted by k) and then k<9 can be found.
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Anonymous users2024-02-06-x+6=k x, you might as well let both solutions be greater than 0
can be squared as x-6x+k=0
To satisfy that there are two common points and both solutions are greater than 0, then =36-4k>0f(0) 0
0＜k<9
Let the intersection point be (x1,y1)(x2,y2), x1+x2=6, x1x2=k vector oa, and the angle of ob is aob
cos∠aob=(x1x2+y1y2)/√(x1²＋y1²)(x2²+y2²）=2k/√(36-2k)^2=k/（18-k）
0＜k<9
0＜k/（18-k）＜1
0＜cos∠aob＜1
0°＜∠aob＜90°
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Anonymous users2024-02-051) What condition is k, these two functions have two common points in the image in the same coordinate system.
When x is not equal to 0, there is -x+6=k x, giving x 2-6x+k=0
If there is a double solution, then (-6) 2-4*1*k>0, i.e. k<9
2) Let the two common points in the above question be a and b, and find the degree range of aob:
a) When k > 0: the function y=-x+6 and the function y=k x do not necessarily have an intersection point 0 degrees< aob<90 degrees.
When the images of the function y=-x+6 and the function y=k x (1,3 quadrants) are tangent, AOB has a limit value of 0. Hence AOB > 0 degrees.
When k approaches 0, the image of y=k x is very close to the x and y axes, and there is a maximum angle of about 90 degrees.
b) When k < 0: the function y=-x+6 and the function y=k x(2,4 quadrants) must have an intersection of 90 degrees < aob<180 degrees.
When k approaches 0, the image of y=k x is very close to the x and y axes, and there is a minimum angle AOB that is about 90 degrees.
When k approaches negative infinity, and the image of y=k x is far away from the x and y axes, then the intersection of the two functions is far away from the center of the circle (of course, on the same line y=-x+6), and the maximum angle aob is about 180 degrees.
Based on the results of the discussion, there are 0 degrees< AOB < 180 degrees and AOB is not equal to 90 degrees.
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