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Challenging Maths Question

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dlehddud:
Try it out:

i)Find the gradient of m of the line segment joining the points A and B with x-coordinates 0 and 1 respectively on the graph of y=.

With this value of m, the line y=mx+c is drawn to meet the y-axis at C and intersect the graph of y= twice between A and B, as shown in the diagram. A line parallel to the y-axis between the points of intersection of the straight line and the curve meets the straight line at P and the curve at Q.

ii)Find the value of c which makes the maximum value of lPQl equal to the value of the distance lACl

Relatively hard question.. even my teacher couldn't solve it.. but strangely the students could haha break a leg :D

~ A.F ~:
m=e-1

c= (0,1)

dlehddud:

--- Quote from: ~ A.F.N.M ~ on May 05, 2010, 09:59:00 am ---m=e-1

c= (0,1)

--- End quote ---

m is right, but c is wrong.
if c was (0,1), then AC would be zero (on the graph, the C is capital not lower case. The question asks for the lower case c, or the y-intercept of the equation y=mx+c

~ A.F ~:
c must be 0,1

y= x(e-1) +1

x=0, y=1

dlehddud:

--- Quote from: ~ A.F.N.M ~ on May 05, 2010, 10:18:07 am ---c must be 0,1

y= x(e-1) +1

x=0, y=1

--- End quote ---

yes, but y= x(e-1)+1 is not the equation of the form y=mx+c which will make lPQl equal lACl

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