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=e−x and y=e−xsin4x , for 0⩽x⩽5π4 .
(a) On the same set of axes draw, on graph paper, the graphs, for 0⩽x⩽5π4. 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=e−x 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=e−xsin4x meets the graph of y=e−x . Give your answers in terms of π.
(d) (i) Show that when the graph of y=e−xsin4x meets the graph of y=e−x , their gradients are equal.
(ii) Hence explain why these three meeting points are not local maxima of the graph y=e−xsin4x .
(e) (i) Determine the y-coordinates, y1 , y2 and y3, where y1>y2>y3 , of the local maxima of y=e−xsin4x for 0⩽x⩽5π4 . 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) e−xsin(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) e−x=e−xsin(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=e−xsin4xdydx=−e−xsin4x+4e−xcos4x M1A1y=e−xdydx=−e−x A1verifying equality of gradients at one point R1verifying at the other two R1(ii) since dydx≠0 at these points they cannot be local maxima R1[6 marks] (e) (i) maximum when y′=4e−xcos4x−e−xsin4x=0 M1x=arctan(4)4, arctan(4)+π4, arctan(4)+2π4, ...maxima occur atx=arctan(4)4, arctan(4)+2π4, arctan(4)+4π4 A1so y1=e−14(arctan(4))sin(arctan(4)) (=0.696) A1y2=e−14(arctan(4)+2π)sin(arctan(4)+2π) A1(=e−14(arctan(4)+2π)sin(arctan(4))=0.145)y3=e−14(arctan(4)+4π)sin(arctan(4)+4π) A1(=e−14(arctan(4)+4π)sin(arctan(4))=0.0301) N3(ii) for finding and comparing y3y2 and y2y1 M1r=e−π2 A1Note: 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).