Date | May 2009 | Marks available | 4 | Reference code | 09M.2.hl.TZ2.4 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Solve | Question number | 4 | Adapted from | N/A |
Question
The graph of y=ln(x)y=ln(x) is transformed into the graph of y=ln(2x+1)y=ln(2x+1) .
Describe two transformations that are required to do this.
Solve ln(2x+1)>3cos(x)ln(2x+1)>3cos(x), x∈[0,10]x∈[0,10].
Markscheme
EITHER
translation of −12−12 parallel to the xx-axis
stretch of a scale factor of 1212 parallel to the xx-axis A1A1
OR
stretch of a scale factor of 1212 parallel to the xx-axis
translation of −1−1 parallel to the xx-axis A1A1
Note: Accept clear alternative terminologies for either transformation.
[2 marks]
EITHER
1.16<x<5.71∪6.75<x⩽10 A1A1A1A1
OR
]1.16, 5.71[ ∪ ]6.75, 10] A1A1A1A1
Note: Award A1 for 1 intersection value, A1 for the other 2, A1A1 for the intervals.
[6 marks]
Examiners report
This question was well done by many candidates. It would appear, however, that few candidates were aware of the standard terminology – Stretch and Translation - used to describe the relevant graph transformations. Most made good use of a GDC to find the critical points and to help in deciding on the correct intervals. A significant minority failed to note x=10 as an endpoint.
This question was well done by many candidates. It would appear, however, that few candidates were aware of the standard terminology – Stretch and Translation - used to describe the relevant graph transformations. Most made good use of a GDC to find the critical points and to help in deciding on the correct intervals. A significant minority failed to note x=10 as an endpoint.