| Date | November 2012 | Marks available | 3 | Reference code | 12N.1.hl.TZ0.4 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The diagram shows the graph of the function defined by \(y = x{(\ln x)^2}{\text{ for }}x > 0\) .
The function has a local maximum at the point A and a local minimum at the point B.
Find the coordinates of the points A and B.
Given that the graph of the function has exactly one point of inflexion, find its coordinates.
Markscheme
\(f'(x) = {(\ln x)^2} + \frac{{2x\ln x}}{x}\left( { = {{(\ln x)}^2} + 2\ln x = \ln x(\ln x + 2)} \right)\) M1A1
\(f'(x) = 0{\text{ }}( \Rightarrow x = 1,{\text{ }}x = {e^{ - 2}})\) M1
Note: Award M1 for an attempt to solve \(f'(x) = 0\).
\(A({e^{ - 2}},\,4{e^{ - 2}})\) and B(1, 0) A1A1
Note: The final A1 is independent of prior working.
[5 marks]
\(f''(x) = \frac{2}{x}(\ln x + 1)\) A1
\(f''(x) = 0{\text{ }}\left( { \Rightarrow x = {e^{ - 1}}} \right)\) (M1)
inflexion point \(({e^{ - 1}},{\text{ }}{e^{ - 1}})\) A1
Note: M1 for attempt to solve \(f''(x) = 0\).
[3 marks]
Examiners report
This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.
This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.
