Year: 2007
Paper: 2
Question Number: 1
Course: LFM Stats And Pure
Section: Generalised Binomial Theorem
Although the paper was by no means an easy one, it was generally found a more accessible paper than last year's, with most questions clearly offering candidates an attackable starting-point. The candidature represented the usual range of mathematical talents, with a pleasingly high number of truly outstanding students; many more who were able to demonstrate a thorough grasp of the material in at least three questions; and the few whose three-hour long experience was unlikely to have been a particularly pleasant one. However, even for these candidates, many were able to make some progress on at least two of the questions chosen. Really able candidates generally produced solid attempts at five or six questions, and quite a few produced outstanding efforts at up to eight questions. In general, it would be best if centres persuaded candidates not to spend valuable time needlessly in this way – it is a practice that is not to be encouraged, as it uses valuable examination time to little or no avail. Weaker brethren were often to be found scratching around at bits and pieces of several questions, with little of substance being produced on more than a couple. It is an important examination skill – now more so than ever, with most candidates now not having to employ such a skill on the modular papers which constitute the bulk of their examination experience – for candidates to spend a few minutes at some stage of the examination deciding upon their optimal selection of questions to attempt. As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly constructed. In the event, at least one – and usually both – of these two questions were among candidates' chosen questions. These, along with questions 3 and 6, were by far the most popularly chosen questions to attempt. The majority of candidates only attempted questions in Section A (Pure Maths), and there were relatively few attempts at the Applied Maths questions in Sections B & C, with Mechanics proving the more popular of the two options. It struck me that, generally, the working produced on the scripts this year was rather better set-out, with a greater logical coherence to it, and this certainly helps the markers identify what each candidate thinks they are doing. Sadly, this general remark doesn't apply to the working produced on the Mechanics questions, such as they were. As last year, the presentation was usually appalling, with poorly labelled diagrams, often with forces missing from them altogether, and little or no attempt to state the principles that the candidates were attempting to apply.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\textit{In this question, you are not required to justify the accuracy of the approximations.}
\begin{questionparts}
\item Write down the binomial expansion of $\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}$in ascending powers of $k$, up to and including the $k^3$ term.
\begin{enumerate}
\item Use the value $k=8$ to find an approximation to five decimal places for $\sqrt{3}\,$.
\item By choosing a suitable integer value of $k$, find an approximation to five decimal places for $\sqrt6\,$.
\end{enumerate}
\item
By considering the first two terms of the binomial expansion of $\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}$, show that $\dfrac{3029}{2100}$ is an approximation to $\sqrt[3]{3}$.
\end{questionparts}\begin{questionparts}
\item Using the generalise binomial theorem
\begin{align*}
\left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\
&= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots
\end{align*}
\begin{enumerate}
\item If $k = 8$, \begin{align*}
&& \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\
\Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\
&&&= 1.039232\\
\Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.})
\end{align*}
\item If $k = -4$, \begin{align*}
&& \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\
\Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\
&&&= 0.979796\\
\Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.})
\end{align*}
\end{enumerate}
\item $\,$ \begin{align*}
&& \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\
&&&= 1 + \frac{k}{3\,000} + \cdots \\
&& 3 \times 7^3 &= 1029 \\
\Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\
\Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\
\Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100}
\end{align*}
\end{questionparts}
Most candidates attempted this question and the majority coped fairly well with the algebraic demands. Surprisingly, it was when the work went numerical that candidates tended to let themselves down; poor arithmetic providing the main difficulty. The final three marks available in (i) parts (a) and (b) were the marks most frequently scorned, generally being lost by candidates' unwillingness or inability to simplify fractions and/or turn them into decimals. In many cases, candidates had difficulty deciding on a suitable value for k in (i) (b) and (ii). In (b), the value k = 50 was often selected, rather than the intended value of – 4. Although this does lead to a similar set of working, the ultimate approximation is relatively poor, and they lost the final mark here. It is, rather more tellingly, indicative of the way in which many modern A-level mathematicians have great difficulty in thinking only in terms of positive integers! A small, but significant, number of candidates offered a value of k that exceeded 100 (the denominator of the "x" term), and these were penalised all four of the available marks for this part of the question, on the not unreasonable grounds that they really should have appreciated the general convergence condition |k/100| < 1 for binomial series of this kind.