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Date May 2009 Marks available 22 Reference code 09M.2.hl.TZ2.13
Level HL only Paper 2 Time zone TZ2
Command term Draw, Find, Hence, and Show that Question number 13 Adapted from N/A

Question

Consider the graphs y=ex and y=exsin4x , for 0x5π4 .

(a)     On the same set of axes draw, on graph paper, the graphs, for 0x5π4. Use a scale of 1 cm to π8 on your x-axis and 5 cm to 1 unit on your y-axis.

 

(b)     Show that the x-intercepts of the graph y=ex sin 4x are nπ4 , n=0, 1, 2, 3, 4, 5.

 

(c)     Find the x-coordinates of the points at which the graph of y=exsin4x meets the graph of y=ex . Give your answers in terms of π.

 

(d)     (i)     Show that when the graph of y=exsin4x meets the graph of y=ex , their gradients are equal.

  (ii)     Hence explain why these three meeting points are not local maxima of the graph y=exsin4x .

 

(e)     (i)     Determine the y-coordinates, y1 , y2 and y3, where y1>y2>y3 , of the local maxima of y=exsin4x for 0x5π4 . You do not need to show that they are maximum values, but the values should be simplified.

  (ii)     Show that y1 , y2 and y3 form a geometric sequence and determine the common ratio r .

Markscheme

(a)

     A3

Note: Award A1 for each correct shape,

   A1 for correct relative position.

[3 marks]

 

(b)     exsin(4x)=0     (M1)

sin(4x)=0     A1

4x=0, π, 2π, 3π, 4π, 5π     A1

x=0, π4, 2π4, 3π4, 4π4, \(\frac{5\pi }{4}\)     AG

[3 marks]

 

(c)     ex=exsin(4x)=0 or reference to graph

sin4x=1,      M1

sin4x=1, fracπ2, frac5π2, frac9π2     A1

x=π8, 5π8, 9π8     A1     N3
[3 marks]
 
(d)     (i)     y=exsin4x
dydx=exsin4x+4excos4x     M1A1
y=ex
dydx=ex     A1
verifying equality of gradients at one point     R1
verifying at the other two     R1
(ii)     since dydx0 at these points they cannot be local maxima     R1
[6 marks]
 
(e)     (i)     maximum when y=4excos4xexsin4x=0     M1
x=arctan(4)4, arctan(4)+π4, arctan(4)+2π4, ...
maxima occur at
x=arctan(4)4arctan(4)+2π4, arctan(4)+4π4     A1
so y1=e14(arctan(4))sin(arctan(4))   (=0.696)     A1
y2=e14(arctan(4)+2π)sin(arctan(4)+2π)     A1
(=e14(arctan(4)+2π)sin(arctan(4))=0.145)
y3=e14(arctan(4)+4π)sin(arctan(4)+4π)     A1
(=e14(arctan(4)+4π)sin(arctan(4))=0.0301)     N3
(ii)     for finding and comparing y3y2 and y2y1     M1
r=eπ2     A1
Note: Exact values must be used to gain the M1 and the A1.
[7 marks]
 
Total [22 marks]

Examiners report

Although the final question on the paper it had parts accessible even to the weakest candidates. The vast majority of candidates earned marks on part (a), although some graphs were rather scruffy. Many candidates also tackled parts (b), (c) and (d). In part (b), however, as the answer was given, it should have been clear that some working was required rather than reference to a graph, which often had no scale indicated. In part d(i), although the functions were usually differentiated correctly, it was often the case that only one point was checked for the equality of the gradients. In part e(i) many candidates who got this far were able to determine the y-coordinates of the local maxima numerically using a GDC, and that was given credit. Only the exact values, however, could be used in part e(ii).

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Local maximum and minimum values.
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