User interface language: English | Español

Date November 2012 Marks available 5 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.

[5]
a.

Given that the graph of the function has exactly one point of inflexion, find its coordinates.

[3]
b.

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]

a.

\(f''(x) = \frac{2}{x}(\ln x + 1)\)     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]

b.

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.

a.

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.

b.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Local maximum and minimum values.
Show 31 related questions

View options