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Ganesh
:
I need to know how to solve this problem. If |a tanbx| = {2-(2/c) x}, has 3 values between 0<x<pi, find the range of values of " c".
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August 23, 2014 at 7:20am
Kho Yen Hong
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As from the graph, to let the function |a tan bx| intersect the line {2-(2/c)x} at exactly 3 points, [deqn]-\frac{2}{\frac{2\pi}{b}}\lt-\frac{2}{c}\lt-\frac{2}{\frac{\pi}{b}} \\ \frac{\pi}{b}\lt c \lt \frac{2\pi}{b}[/deqn]
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August 24, 2014 at 10:04am
Kho Yen Hong
:
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August 24, 2014 at 10:04am
Ganesh
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Thank you. However i need some more explanation. How did you decide that c for g(x) is pi/b and for h(x) c = 2pi/b to plot the graphs. I need some explanation on how you plotted these 2 graphs. Thanks again
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August 24, 2014 at 6:45pm
Ganesh
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Let me rephrase my doubt. I know that |a tanbx| =0 only if bx = 0 or when bx is a full multiple of pi. Eg., 1 pi, 2 pi, etc. Since c has 3 values am i right in saying that we are equating | atan bx| =0 is when bx=0, bx=pi and bx= 2pi or conversely x = 0, x=pi/b and x = 2pi/b and we substitute it in { 2-(2/c) x} and we get c lies between " pi/b" and "2pi/b". if on the other hand were to have 4 values we would have to go more step isn't it by putting bx=3pi. Then c will lie between "pi/b" and "3pi/b". Is my understanding correct?
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August 24, 2014 at 8:14pm
Ganesh
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Please read the above line "If on the other hand c were to have 4 values, we would......."
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August 24, 2014 at 8:17pm
Kho Yen Hong
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Let two function f(x)=|a tan bx| and g(x)= {2-(2/c)x}. If f(x)=g(x) have 3 solution, it implies that f(x) intersect g(x) at exactly 3 point. Observe the graph of the curve and the straight line, the straight line will always cut the curve at 0<x<pi/2b at 1 point since the straight line have a positive y-intercept and negative gradient. As the gradient become less and less negative until it intersect the x-axis at pi/b, the line and the curve will intersect at exactly 2 points. If the gradient go less and less negative with the x-intercept c lies between pi/b and 2pi/b, the line and the curve will intersect at exactly 3 points until the line intersect the x-axis at 2pi/b. I am sorry to tell you that your understanding is incorrect. If u observed the graph that i plotted, if there are 4 values the straight line suppose to have an equation same as h(x), and the only value for c is 2pi/b. If the problem further extend to 'there are 5 values', the c will lies between the range of [deqn]\frac{2\pi}{b}\lt c \lt \frac{3\pi}{b}[/deqn]
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August 24, 2014 at 10:09pm
Ganesh
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Thank you very much.
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August 25, 2014 at 2:50am
Ganesh
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Dear Hen Yong, can this be solved only graphically. Yes your graph and the logic behind it makes perfect sense. Can we solve this without a graph and instead use pure algebra and the principles of trignometry to solve this. If so can you also share it?
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August 25, 2014 at 5:03am
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