| Date | May 2011 | Marks available | 4 | Reference code | 11M.2.hl.TZ2.7 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate | Question number | 7 | Adapted from | N/A |
Question
Consider the functions \(f(x) = {x^3} + 1\) and \(g(x) = \frac{1}{{{x^3} + 1}}\). The graphs of \(y = f(x)\) and \(y = g(x)\) meet at the point (0, 1) and one other point, P.
Find the coordinates of P.
Calculate the size of the acute angle between the tangents to the two graphs at the point P.
Markscheme
\({x^3} + 1 = \frac{1}{{{x^3} + 1}}\)
\(( - 1.26, - 1)\,\,\,\,\,\left( { = \left( { - \sqrt[3]{2}, - 1} \right)} \right)\) A1
[1 mark]
\(f'( - 1.259...) = 4.762…\) \((3 \times {2^{\frac{2}{3}}})\) A1
\(g'( - 1.259...) = - 4.762…\) \(( - 3 \times {2^{\frac{2}{3}}})\) A1
required angle \( = 2\arctan \left( {\frac{1}{{4.762...}}} \right)\) M1
\( = 0.414\) (accept 23.7 ) A1
Note: Accept alternative methods including finding the obtuse angle first.
[4 marks]
Examiners report
In part (a) almost all candidates obtained the correct answer, either in numerical form or in exact form. Although many candidates scored one mark in (b), for one gradient, few scored any more. Successful candidates almost always adopted a vector approach to finding the angle between the two tangents, rather than using trigonometry.
In part (a) almost all candidates obtained the correct answer, either in numerical form or in exact form. Although many candidates scored one mark in (b), for one gradient, few scored any more. Successful candidates almost always adopted a vector approach to finding the angle between the two tangents, rather than using trigonometry.
