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secondary 3 | A Maths
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how to solve this qn?
Date Posted: 4 years ago
Views: 444
Imagination is needed. Reasoning will be good here.
I will do this by 4 am if no one else has done them yet.
For part ii a we get c = 0 easily.
For part ii b I think it should be m > 0 but m < 8/3, since y = |3x - 8| has a vertex along the x-axis and the line y = mx passes the origin.
Horizontal line will only cut the graph at the vertex. As we incline the line it will cut the graph at two points. The limit is y = 8/3 x which is parallel to the upsloping portion of the graph ie no intersection with the upsloping part.
So, 0 < m < 8/3.
I will do this by 4 am if no one else has done them yet.
For part ii a we get c = 0 easily.
For part ii b I think it should be m > 0 but m < 8/3, since y = |3x - 8| has a vertex along the x-axis and the line y = mx passes the origin.
Horizontal line will only cut the graph at the vertex. As we incline the line it will cut the graph at two points. The limit is y = 8/3 x which is parallel to the upsloping portion of the graph ie no intersection with the upsloping part.
So, 0 < m < 8/3.
Sorry, 0 < m < 3. Misread the line.
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First part
Date Posted:
4 years ago
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Part iia
Date Posted:
4 years ago
You do not need to do any working for this one as long as you provide the correct solutions.
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Part iib, where I split the cases.
Date Posted:
4 years ago
Let me know if you need further explanations. By the way, you can also attempt to play around with the gradients in the Desmos online graphing calculator application to see how the lines change.
If the interval is followed, then 0 < m <= 7/5 since the corner point is (5, 7).
In the Long run without intervals, we get 0 < m < 3.
In the Long run without intervals, we get 0 < m < 3.
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On the application it looks like this.
Date Posted:
4 years ago
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