| Date | May 2009 | Marks available | 19 | Reference code | 09M.2.hl.TZ1.12 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find, Show that, and Write down | Question number | 12 | Adapted from | N/A |
Question
Let \(f\) be a function defined by \(f(x) = x + 2\cos x\) , \(x \in \left[ {0,{\text{ }}2\pi } \right]\) . The diagram below shows a region \(S\) bound by the graph of \(f\) and the line \(y = x\) .
A and C are the points of intersection of the line \(y = x\) and the graph of \(f\) , and B is the minimum point of \(f\) .
(a) If A, B and C have x-coordinates \(a\frac{\pi }{2}\), \(b\frac{\pi }{6}\) and \(c\frac{\pi }{2}\), where \(a\) , \(b\), \(c \in \mathbb{N}\) , find the values of \(a\) , \(b\) and \(c\) .
(b) Find the range of \(f\) .
(c) Find the equation of the normal to the graph of f at the point C, giving your answer in the form \(y = px + q\) .
(d) The region \(S\) is rotated through \({2\pi }\) about the x-axis to generate a solid.
(i) Write down an integral that represents the volume \(V\) of this solid.
(ii) Show that \(V = 6{\pi ^2}\) .
Markscheme
(a) METHOD 1
using GDC
\(a = 1\), \(b = 5\), \(c = 3\) A1A2A1
METHOD 2
\(x = x + 2\cos x \Rightarrow \cos x = 0\)
\( \Rightarrow x = \frac{\pi }{2}\), \(\frac{{3\pi }}{2}\) ... M1
\(a = 1\), \(c = 3\) A1
\(1 - 2\sin x = 0\) M1
\( \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi }{6}\) or \(\frac{{5\pi }}{6}\)
\(b = 5\) A1
Note: Final M1A1 is independent of previous work.
[4 marks]
(b) \(f\left( {\frac{{5\pi }}{6}} \right) = \frac{{5\pi }}{6} - \sqrt 3 \) (or \(0.886\)) (M1)
\(f(2\pi ) = 2\pi + 2\) (or \(8.28\)) (M1)
the range is \(\left[ {\frac{{5\pi }}{6} - \sqrt 3 ,{\text{ }}2\pi + 2} \right]\) (or [\(0.886\), \(8.28\)]) A1
[3 marks]
(c) \(f'(x) = 1 - 2\sin x\) (M1)
\(f'\left( {\frac{{3\pi }}{6}} \right) = 3\) A1
gradient of normal \( = - \frac{1}{3}\) (M1)
equation of the normal is \(y - \frac{{3\pi }}{2} = - \frac{1}{3}\left( {x - \frac{{3\pi }}{2}} \right)\) (M1)
\(y = - \frac{1}{3}x + 2\pi \) (or equivalent decimal values) A1 N4
[5 marks]
(d) (i) \(V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} - {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x\) (or equivalent) A1A1
Note: Award A1 for limits and A1 for \(\pi \) and integrand.
(ii) \(V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} - {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x\)
\( = - \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {4x\cos x + 4{{\cos }^2}x} \right)} {\text{d}}x\)
using integration by parts M1
and the identity \(4{\cos ^2}x = 2\cos 2x + 2\) , M1
\(V = - \pi \left[ {\left( {4x\sin x + 4\cos x} \right) + \left( {\sin 2x + 2x} \right)} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}\) A1A1
Note: Award A1 for \({4x\sin x + 4\cos x}\) and A1 for sin \({2x + 2x}\) .
\( = - \pi \left[ {\left( {6\pi \sin \frac{{3\pi }}{2} + 4\cos \frac{{3\pi }}{2} + \sin 3\pi + 3\pi } \right) - \left( {2\pi sin\frac{\pi }{2} + 4\cos \frac{\pi }{2} + \sin \pi + \pi } \right)} \right]\) A1
\( = - \pi \left( { - 6\pi + 3\pi - \pi } \right)\)
\( = 6{\pi ^2}\) AG N0
Note: Do not accept numerical answers.
[7 marks]
Total [19 marks]
Examiners report
Generally there were many good attempts to this, more difficult, question. A number of students found \(b\) to be equal to 1, rather than 5. In the final part few students could successfully work through the entire integral successfully.
