Challenging Maths Question

[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

[~ A.F ~]

m=e-1

c= (0,1)

[dlehddud]

m=e-1

c= (0,1)

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]

c must be 0,1

y= x(e-1) +1

x=0, y=1

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

[astarmathsandphysics]

I dont know if it is the right answer but is the right form.

[dlehddud]

I dont know if it is the right answer but is the right form.

fine.. answer is
now this is the challenging part XP why is the answer this? because it definitely is the answer.

[dlehddud]

And just to make sure, c should NOT be a co-ordinate, it is lower case 'c', which is part of the equation , so it should be the y intercept of the graph - a value of y.