2008 Paper 2 Q3
Year: 2008
Paper: 2
Question Number: 3
Course: LFM Pure and Mechanics
Section: Differentiation from first principles
Problem
- Find the coordinates of the turning points of the curve \(y=27x^3-27x^2+4\). Sketch the curve and deduce that \(x^2(1-x)\le 4/27\) for all \(x\ge0\,\). Given that each of the numbers \(a\), \(b\) and \(c\) lies between \(0\) and \(1\), prove by contradiction that at least one of the numbers \(bc(1-a)\), \(ca(1-b)\) and \(ab(1-c)\) is less than or equal to \(4/27\).
- Given that each of the numbers \(p\) and \(q\) lies between \(0\) and \(1\), prove that at least one of the numbers \(p(1-q)\) and \(q(1-p)\) is less than or equal to \(1/4\).
Solution
- \(\,\)
\begin{align*}
&& y & = 27x^3 - 27x^2 + 4 \\
\Rightarrow && \frac{\d y}{\d x} &= 81x^2 - 54x \\
\Rightarrow && x &= 0, \frac23 \\
\Rightarrow && (x,y) &= (0, 4), \left (\frac23, 0 \right)
\end{align*}
Since \(f(x) \geq 0\) for \(x \geq 0\) we must have \(27x^2(1-x) \leq 4 \Rightarrow x^2(1-x) \leq \frac{4}{27}\) Suppose for contradiction that \(bc(1-a) > \frac{4}{27}, ca(1-b) > \frac{4}{27}, ab(1-c) > \frac{4}{27}\) then taking the product we see \begin{align*} && \left ( \frac{4}{27} \right)^3 &< bc(1-a) \cdot ca(1-b) \cdot ab(1-c) \\ &&&= a^2(1-c) \cdot b^2(1-b) \cdot c^2(1-c) \leq \left ( \frac{4}{27}\right)^3 \end{align*} which is a contradiction.
- Notice that \(f(x) = x(1-x)\) has a turning point at \((\frac12, \frac14)\), and so \(f(x) \leq \frac14\). Suppose for contradiction that both \(p(1-q)\) and \(q(1-p)\) are larger than \(1/4\) \begin{align*} && \left ( \frac14 \right)^2 &< p(1-q) \cdot q(1-p) \\ &&&= p(1-p) \cdot q(1-q) \\ &&&\leq \left ( \frac14 \right)^2 \end{align*} which is a contradiction.
Examiner's report
— 2008 STEP 2, Question 3There were around 850 candidates for this paper – a slight increase on the 800 of the past two years – and the scripts received covered the full range of marks (and beyond!). The questions on this paper in recent years have been designed to be a little more accessible to all top A-level students, and this has been reflected in the numbers of candidates making good attempts at more than just a couple of questions, in the numbers making decent stabs at the six questions required by the rubric, and in the total scores achieved by candidates. Most candidates made attempts at five or more questions, and most genuinely able mathematicians would have found the experience a positive one in some measure at least. With this greater emphasis on accessibility, it is more important than ever that candidates produce really strong, essentially-complete efforts to at least four questions. Around half marks are required in order to be competing for a grade 2, and around 70 for a grade 1. The range of abilities on show was still quite wide. Just over 100 candidates failed to score a total mark of at least 30, with a further 100 failing to reach a total of 40. At the other end of the scale, more than 70 candidates scored a mark in excess of 100, and there were several who produced completely (or nearly so) successful attempts at more than six questions; if more than six questions had been permitted to contribute towards their paper totals, they would have comfortably exceeded the maximum mark of 120. While on the issue of the "best-six question-scores count" rubric, almost a third of candidates produced efforts at more than six questions, and this is generally a policy not to be encouraged. In most such cases, the seventh, eighth, or even ninth, question-efforts were very low scoring and little more than a waste of time for the candidates concerned. Having said that, it was clear that, in many of these cases, these partial attempts represented an abandonment of a question after a brief start, with the candidates presumably having decided that they were unlikely to make much successful further progress on it, and this is a much better employment of resources. As in recent years, most candidates' contributing question-scores came exclusively from attempts at the pure maths questions in Section A. Attempts at the mechanics and statistics questions were very much more of a rarity, although more (and better) attempts were seen at these than in other recent papers.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item
Find the coordinates of the turning points of the curve $y=27x^3-27x^2+4$. Sketch the curve and deduce that $x^2(1-x)\le 4/27$ for all $x\ge0\,$.
Given that each of the numbers $a$, $b$ and $c$ lies between $0$ and $1$, prove by contradiction that at least one of the numbers $bc(1-a)$, $ca(1-b)$ and $ab(1-c)$ is less than or equal to $4/27$.
\item Given that each of the numbers $p$ and $q$ lies between $0$ and $1$, prove that at least one of the numbers $p(1-q)$ and $q(1-p)$ is less than or equal to $1/4$.
\end{questionparts}Solution source
\begin{questionparts}
\item $\,$
\begin{align*}
&& y & = 27x^3 - 27x^2 + 4 \\
\Rightarrow && \frac{\d y}{\d x} &= 81x^2 - 54x \\
\Rightarrow && x &= 0, \frac23 \\
\Rightarrow && (x,y) &= (0, 4), \left (\frac23, 0 \right)
\end{align*}
\begin{center}
\begin{tikzpicture}
\def\a{-0.8};
\def\functionf(#1){27*(#1)^3-27*(#1)^2+4};
\def\xl{-1.5};
\def\xu{1.5};
\def\yl{-10};
\def\yu{10};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
\draw[blue, smooth, thick, domain=\xl:\xu, samples=101]
plot({\x}, {\functionf(\x)});
\node[blue, below, rotate=75] at (1, {\functionf(1)}) {\tiny $y=27x^3-27x^2+4$};
% \filldraw (0.5, 0) circle (1.5pt) node[below right] {$a$};
% \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});
\filldraw (0, 4) circle (1.5pt) node[right] {$4$};
\filldraw ({2/3}, 0) circle (1.5pt) node[below] {$\frac23$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
Since $f(x) \geq 0$ for $x \geq 0$ we must have $27x^2(1-x) \leq 4 \Rightarrow x^2(1-x) \leq \frac{4}{27}$
Suppose for contradiction that $bc(1-a) > \frac{4}{27}, ca(1-b) > \frac{4}{27}, ab(1-c) > \frac{4}{27}$ then taking the product we see
\begin{align*}
&& \left ( \frac{4}{27} \right)^3 &< bc(1-a) \cdot ca(1-b) \cdot ab(1-c) \\
&&&= a^2(1-c) \cdot b^2(1-b) \cdot c^2(1-c) \leq \left ( \frac{4}{27}\right)^3
\end{align*}
which is a contradiction.
\item Notice that $f(x) = x(1-x)$ has a turning point at $(\frac12, \frac14)$, and so $f(x) \leq \frac14$. Suppose for contradiction that both $p(1-q)$ and $q(1-p)$ are larger than $1/4$
\begin{align*}
&& \left ( \frac14 \right)^2 &< p(1-q) \cdot q(1-p) \\
&&&= p(1-p) \cdot q(1-q) \\
&&&\leq \left ( \frac14 \right)^2
\end{align*}
which is a contradiction.
\end{questionparts}
This, the third most popular question on the paper, producing a mixed bag of responses. It strikes me that, although the A-level specifications require candidates to understand the process of proof by contradiction, this is never actually tested anywhere by any of the exam. boards. Nonetheless, it was very pleasing to see that so many candidates were able to grasp the basic idea of what to do, and many did so very successfully. The impartial observer might well note that the situation in (i) is very much tougher (in terms of degree) than that in (ii). However, candidates were very much more closely guided in (i) and then left to make their own way in (ii). Apart from the standard, expected response to (i), many other candidates produced a very pleasing alternative which they often dressed up as proof by contradiction but which was, in fact, a direct proof. It was, however, so mathematically sound and appealing an argument (and a legitimate imitation of a p by c) that we gave it all but one of the marks available in this part of the question. It ran like this: Suppose w.l.o.g. that 0 < a ≤ b ≤ c < 1. Then ab(1 – c) ≤ b²(1 – b) ≤ 4/27 by the previous result (namely x²(1 – x) ≤ 4/27 for all x ≥ 0). QED. [Note that we could have used ab(1 – c) ≤ c²(1 – c) ≤ 4/27 also.] It has to be said that most other inequality arguments were rather poorly constructed and unconvincing, leaving the markers with little option but to put a line through (often) several pages of circular arguments, faulty assumptions, dubious conclusions, and occasionally correct statements with either no supporting reasoning or going nowhere useful. There was one remarkable alternative which was produced by just a couple of candidates (that I know of) and is not included in the SOLUTIONS because it is such a rarity. However, for those who know of the AM – GM Inequality, it is sufficiently appealing to include it here for novelty value. It ran like this: Assume that bc(1 – a), ca(1 – b), ab(1 – c) > 4/27. Using the previous result, we have a²(1 – a), b²(1 – b), c²(1 – c) ≤ 4/27. Then, since all terms are positive, it follows that a² ≤ bc, b² ≤ ca, c² ≤ ab so that a² + b² + c² ≤ bc + ca + ab. (*) However, by the AM – GM Inequality (or directly by the Cauchy-Schwarz Inequality), a² + b² ≥ 2ab, b² + c² ≥ 2bc and c² + a² ≥ 2ca. Adding and dividing by two then gives a² + b² + c² ≥ ab + bc + ca, which contradicts the conclusion (*), etc., etc.