| Date | May 2009 | Marks available | 13 | Reference code | 09M.2.hl.TZ0.4 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The function \(f(x)\) is defined by the series \(f(x) = 1 + \frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \) .
Write down the general term.
Find the interval of convergence.
Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) , giving your answer in the form \(u = f(v)\) .
Markscheme
the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\) A1
[1 mark]
\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\) M1A1A1
\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\) A1
\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\) A1R1
the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\) R1
then \( - 3 < x + 2 < 3 \Rightarrow - 5 < x < 1\) A1
if \(x = - 5\) , series is \(1 - 1 + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{{{{( - 1)}^n}}}{n} + \ldots \) which converges M1A1
if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots \) which diverges M1A1
the interval of convergence is \( - 5 \le x < 1\) A1
[13 marks]
\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\) M1A1
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} - \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\) A1
IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\) M1
\( = {v^{\frac{1}{2}}}\) A1
\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\) M1
\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\) A1
\(u = \frac{3}{5}{v^3} + c\sqrt v \) A1
[8 marks]
Examiners report
In (a) the general term was usually found.
Part (b) was completed mostly except for testing the ends of the interval of convergence.
A surprising number of candidates started off their solution by saying "let \(x = u\) and \(y = v\) " as if the world suddenly changed when \(x\) and \(y\) were not being used in a differential equation. Some also after seeing \(u\) and \(v\) thought they had a homogeneous equation and got lost in a maze of algebra that lead nowhere. Find \(\frac{{{\rm{d}}u}}{{{\rm{d}}v}}\) by inverting the given expression was also something that only the best candidates were able to do.
