Year: 2008
Paper: 1
Question Number: 5
Course: LFM Pure
Section: Proof
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
The polynomial $\p(x)$ is given by
\[
\ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,,
\]
where $a_0$, $a_1$, $\ldots$ , $a_{n-1}$ are fixed real
numbers and $n\ge1$. Let $M$ be the greatest value of $\big\vert \p(x) \big\vert$ for $\vert x \vert\le
1$. Then \textit{Chebyshev's theorem} states that $M\ge 2^{1-n}$.
\begin{questionparts}
\item Prove Chebyshev's theorem in the case $n=1$ and verify that Chebyshev's theorem holds in the following cases:
\begin{enumerate}
\item $ \p(x) = x^2 - \frac12\,$;
\item $\p(x) = x^3 - x \,$.
\end{enumerate}
\item Use Chebyshev's theorem to show that the curve
$ \ y= 64x^5+25x^4-66x^3-24x^2+3x+1
\ $
has at least one turning point in the interval $-1\le x \le 1$.
\end{questionparts}\begin{questionparts}
\item If $n = 1$ the theorem is $\max_{x \in [-1,1]} \left ( |x + a_0 |\right) \geq 1$, but clearly $\max(1+a_0, |a_0 - 1|) \geq 1$ (taking according to the sign of $a_0$)
\begin{enumerate}
\item $ \p(x) = x^2 - \frac12\,$ - take $x = 0$ then $|p(0)| = \frac12 \geq 2^{1-2} = \frac12$
\item $\p(x) = x^3 - x \,$. take $x = \frac1{\sqrt{2}}$, then $|p\left ( \frac1{\sqrt{2}}\right)| = |\frac12 \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}| = \frac{1}{2\sqrt{2}} > \frac14 = 2^{1-3} $
\end{enumerate}
\item Consider $p(x) = \frac{1}{64} \left ( 64x^5+25x^4-66x^3-24x^2+3x+1\right)$, then $p$ satisfies the conditions of the theorem, therefore $\max |p(x)| \geq 2^{1-5} = \frac1{16} = \frac{4}{64}$.
However, $p(-1) = \frac{1}{64}$ and $p(1) = \frac{3}{64}$, so it cannot be strictly increasing or decreasing and there must be at turning point to achieve $\frac{4}{64}$
\end{questionparts}
This was the least popular of the Pure Mathematics questions. There was a fair amount of confusion as to the meaning of the summation, with the majority of attempts at the n = 1 case in part (i) failing to understand that the polynomial would be p(x) = x + a₀ rather than just p(x) = x. A small number thought that the summation indicated a geometric series, and proceeded to claim that p(x) = (1 − xⁿ)/(1 − x) or other such things. Nonetheless, there were many good answers to the rest of part (i), with candidates showing that they understood the statement of Chebyshev's theorem. A small number of strong candidates had a mature enough understanding of mathematics to use the alternative method given in the sample solutions; most were content with finding the maxima and minima. Some forgot to check the value of p(x) at the ends of the interval, which was not penalised as long as they did not incorrectly assert that they had found the value of M. One recurrent incorrect assertion was that from the inequality p(x) ≥ ½, it necessarily follows that M = ½, without showing that equality is obtained for some value of x. There were few serious attempts at part (ii), but most of those achieved full marks or very close to it. Several candidates had difficulty in explaining their reasoning: a sketch would certainly have helped clarify why a maximum value of |p(x)| occurring in the interval −1 < x < 1 necessarily forces this point to be a turning point. Of the other attempts, many could not see the relevance of Chebyshev's theorem to this situation, or even if they did, then failed to divide the given polynomial by 64. Arguments which did not invoke Chebyshev's theorem were not given any credit (the main alternative being to differentiate, then to find points where the derivative was positive and negative, and use the intermediate value theorem to assert that there is a point where the derivative must be zero).