Date | None Specimen | Marks available | 4 | Reference code | SPNone.2.hl.TZ0.13 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Sketch | Question number | 13 | Adapted from | N/A |
Question
The function f is defined on the domain [0, 2] by \(f(x) = \ln (x + 1)\sin (\pi x)\) .
Obtain an expression for \(f'(x)\) .
Sketch the graphs of f and \(f'\) on the same axes, showing clearly all x-intercepts.
Find the x-coordinates of the two points of inflexion on the graph of f .
Find the equation of the normal to the graph of f where x = 0.75 , giving your answer in the form y = mx + c .
Consider the points \({\text{A}}\left( {a{\text{ }},{\text{ }}f(a)} \right)\) , \({\text{B}}\left( {b{\text{ }},{\text{ }}f(b)} \right)\) and \({\text{C}}\left( {c{\text{ }},{\text{ }}f(c)} \right)\) where a , b and c \((a < b < c)\) are the solutions of the equation \(f(x) = f'(x)\) . Find the area of the triangle ABC.
Markscheme
\(f'(x) = \frac{1}{{x + 1}}\sin (\pi x) + \pi \ln (x + 1)\cos (\pi x)\) M1A1A1
[3 marks]
A4
Note: Award A1A1 for graphs, A1A1 for intercepts.
[4 marks]
0.310, 1.12 A1A1
[2 marks]
\(f'(0.75) = - 0.839092\) A1
so equation of normal is \(y - 0.39570812 = \frac{1}{{0.839092}}(x - 0.75)\) M1
\(y = 1.19x - 0.498\) A1
[3 marks]
\({\text{A}}(0,{\text{ }}0)\)
\({\text{B(}}\overbrace {0.548 \ldots }^c,\overbrace {0.432 \ldots }^d)\) A1
\({\text{C(}}\overbrace {1.44 \ldots }^e,\overbrace { - 0.881 \ldots }^f)\) A1
Note: Accept coordinates for B and C rounded to 3 significant figures.
area \(\Delta {\text{ABC}} = \frac{1}{2}|\)(ci + dj) \( \times \) (ei + fj)\(|\) M1A1
\( = \frac{1}{2}(de - cf)\) A1
\( = 0.554\) A1
[6 marks]