2014 Paper 1 Q5
Year: 2014
Paper: 1
Question Number: 5
Course: LFM Pure
Section: Proof
Problem
- Let \(f(x) = (x+2a)^3 -27 a^2 x\), where \(a\ge 0\). By sketching \(f(x)\), show that \(f(x)\ge 0\) for \(x \ge0\).
- Use part (i) to find the greatest value of \(xy^2\) in the region of the \(x\)-\(y\) plane given by \(x\ge0\), \(y\ge0\) and \(x+2y\le 3\,\). For what values of \(x\) and \(y\) is this greatest value achieved?
- Use part (i) to show that \((p+q+r)^3 \ge 27pqr\) for any non-negative numbers \(p\), \(q\) and \(r\). If \((p+q+r)^3 = 27pqr\), what relationship must \(p\), \(q\) and \(r\) satisfy?
Solution
- Note that \(f(x) = (x+2a)^3 - 27a^2x\) so \(f'(x) = 3(x+2a)^2-27a^2 = 3x^2+12ax-15a^2 = (x-a)(3x+15a)\) so the turning points are at \(x = a, x = -5a\). But \(f(a) = 0\), so the curve just touches the \(x\)-axis. Note also that \(f(0) = 8a^3 \geq 0\) so:
- \(\,\) \begin{align*} && 0 &\leq (x+2y)^3 - 27y^2 x \\ &&& \leq 3^3 - 27 y^2 x \\ &&&= 27(1-y^2x) \\ \Rightarrow && xy^2 &\leq1 \end{align*} with equality when \(x = y = 1\)
- Notice that \begin{align*} && 0 &\leq (p+q+r)^3 - 27\left (\frac{q+r}{2}\right)^2 p \\ \Rightarrow && (p+q+r)^3 &\geq 27\left (\frac{q+r}{2}\right)^2 p \\ &&&\underbrace{\geq}_{AM-GM} 27\left (\sqrt{qr}\right)^2 p \\ &&&= 27pqr \end{align*} Equality can only hold if \(p = \frac{q+r}{2}\), but by symmetry we must also have \(q = \frac{r+p}{2}, r = \frac{p+q}{2}\) ie \(p = q = r\). And indeed equality does hold in this case.
Examiner's report
— 2014 STEP 1, Question 5More than 1800 candidates sat this paper, which represents another increase in uptake for this STEP paper. The impression given, however, is that many of these extra candidates are just not sufficiently well prepared for questions which are not structured in the same way as are the A-level questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give all able candidates "a bit of an intro." into each question, they are not intended to be easy, and (at some point) imagination and real flair (as well as determination) are required if one is to score well on them. In general, it is simply not possible to get very far into a question without making some attempt to think about what is actually going on in the situation presented therein; and those students who expect to be told exactly what to do at each stage of a process are in for a shock. Too many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks on each, rather than trying to get to grips with substantial portions of work – the readiness to give up and try to find something else that is "easy pickings" seldom allows such candidates to acquire more than 40 marks (as was the case with almost half of this year's candidature, in fact). Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those students who took the paper. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item
Let $f(x) = (x+2a)^3 -27 a^2 x$, where $a\ge 0$.
By sketching $f(x)$, show that $f(x)\ge 0$ for $x \ge0$.
\item Use part (i) to find the greatest value of $xy^2$ in the region of the $x$-$y$ plane given by $x\ge0$, $y\ge0$ and $x+2y\le 3\,$. For what values of $x$ and $y$ is this greatest value achieved?
\item Use part (i) to show that $(p+q+r)^3 \ge 27pqr$ for any non-negative numbers $p$, $q$ and $r$.
If $(p+q+r)^3 = 27pqr$, what relationship must $p$, $q$ and $r$ satisfy?
\end{questionparts}Solution source
\begin{questionparts}
\item Note that $f(x) = (x+2a)^3 - 27a^2x$ so $f'(x) = 3(x+2a)^2-27a^2 = 3x^2+12ax-15a^2 = (x-a)(3x+15a)$ so the turning points are at $x = a, x = -5a$. But $f(a) = 0$, so the curve just touches the $x$-axis. Note also that $f(0) = 8a^3 \geq 0$ so:
\begin{center}
\begin{tikzpicture}
\def\a{.6};
\def\functionf(#1){(#1 + 2*\a)^3 - 27*\a*\a*(#1)};
% \def\functiong(#1){sin(deg(6*#1))*exp(-#1)};
\def\xl{-6};
\def\xu{6};
\def\yl{-25};
\def\yu{25};
% 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 reusable styles to keep code clean
\tikzset{
x=\xscale cm, y=\yscale cm,
axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
grid/.style={thin, dashed, gray!30},
curveA/.style={very thick, color=cyan!70!black, smooth},
curveB/.style={very thick, color=orange!90!black, smooth},
curveC/.style={very thick, color=green!70!black, smooth},
dot/.style={circle, fill=black, inner sep=1.2pt},
labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
}
% Draw background grid
\draw[grid] (\xl,\yl) grid (\xu,\yu);
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
% Define the bounding region with clip
\begin{scope}
\clip (\xl,\yl) rectangle (\xu,\yu);
\draw[curveA, domain=\xl:\xu, samples=450]
plot ({\x},{\functionf(\x)});
% \draw[curveB, domain=\xl:\xu, samples=450]
% plot ({\x},{\functiong(\x)});
\end{scope}
\end{tikzpicture}
\end{center}
\item $\,$ \begin{align*}
&& 0 &\leq (x+2y)^3 - 27y^2 x \\
&&& \leq 3^3 - 27 y^2 x \\
&&&= 27(1-y^2x) \\
\Rightarrow && xy^2 &\leq1
\end{align*}
with equality when $x = y = 1$
\item Notice that \begin{align*}
&& 0 &\leq (p+q+r)^3 - 27\left (\frac{q+r}{2}\right)^2 p \\
\Rightarrow && (p+q+r)^3 &\geq 27\left (\frac{q+r}{2}\right)^2 p \\
&&&\underbrace{\geq}_{AM-GM} 27\left (\sqrt{qr}\right)^2 p \\
&&&= 27pqr
\end{align*}
Equality can only hold if $p = \frac{q+r}{2}$, but by symmetry we must also have $q = \frac{r+p}{2}, r = \frac{p+q}{2}$ ie $p = q = r$. And indeed equality does hold in this case.
\end{questionparts}
As already indicated in the comments for Q3, candidates generally do not like working with inequalities and this question is riddled with them. Q5 thus turned out to be the second least popular of the pure questions, and scored poorly with a mean score of under 5/20, again largely due to a lack of progress beyond the first part of the question. Even in (i), there was a tendency to dive straight in to the sketch without having worked out any useful points on the curve, including missing the obvious point (a, 0). Using this would have led easily towards a factorisation of f(x) into three linear factors, though most preferred to find the turning points instead (which approach works equally well). Though not crucial to the following working, the special case a = 0 remained almost universally unaddressed. Each of parts (ii) and (iii) can be approached via the AM-GM Inequality though very few did so as we had instructed candidates to use the result of part (i). It was disappointing to see so few serious attempts at part (ii) since all that is required is to set a = y and then the "x + 2y" is practically waving at you. Part (iii) required a bit more thought (setting p = x and then q + r = 2a) but both imagination and determination seemed in short supply by this stage of the question.