2009 Paper 2 Q1
Year: 2009
Paper: 2
Question Number: 1
Course: LFM Pure
Section: Simultaneous equations
Problem
Solution
- \(y = 0\)
- \(x = 0\)
- \(y = x\)
- \(y = -x\)
- \(y = x\)
- \(y = -x\)
Examiner's report
— 2009 STEP 2, Question 1Of the 1000+ entries for this paper, around 920 scripts actually arrived for marking, giving another slight increase in the take-up for this paper. Of this number, five candidates scored a maximum and seventy-five achieved a scoring total of 100 or more. At the other end of the scale, almost two hundred candidates failed to reach the 40-mark mark. Otherwise, marks were spread reasonably normally across the mark range, though there were two peaks at about 45 and 65 in the distribution. It is comforting to find that the 'post-match analysis' bears out the view that I gained, quite firmly, during the marking process that there were several quantum states of mark-scoring ability amongst the candidature. Many (about one-fifth of the entry) struggled to find anything very much with which they were comfortable, and marks for these candidates were scored in 3s and 4s, with such folk often making eight or nine poor efforts at different questions without ever getting to grips with the content of any one of them. The next "ability band" saw those who either scored moderately well on a handful of questions or managed one really successful question plus a few bits-'n'-pieces in order to get up to a total in the mid-forties. To go much beyond that score required a little bit of extra talent that could lead them towards the next mark-hurdle in the mid-sixties. Thereafter, totals seemed to decline almost linearly on the distribution. Once again, it is clear that candidates need to give the questions at least a couple of minutes' worth of thought before commencing answering. Making attempts at more than the six scoring efforts permitted is a waste of valuable time, and the majority of those who do so are almost invariably the weaker brethren in the game. Many such candidates begin their efforts to individual questions promisingly, but get no more than half-a-dozen marks in before abandoning that question in favour of another – often with the replacement faring no better than its predecessor. In many such cases, the candidate's best-scoring question mark would come from their fifth, or sixth, or seventh, or …?, question attempted, and this suggests either that they do not know where their strengths lie, or that they are just not going to be of the view that they are not going to be challenged to think. And, to be fair to the setting panel, we did put some fairly obvious signposts up for those who might take the trouble to look for such things. With the pleasing number of very high totals to be found, it is clear that there are many places in which good marks were available to those with the ability to first identify them and then to persevere long enough to be able to determine what was really going on therein. It is extremely difficult to set papers in which each question is pitched at an equivalent level of difficulty. Apart from any other factors, candidates have widely differing strengths and weaknesses; one student's algebraic nuance can be the final nail in the coffin of many others, for instance. Moreover, it has seemed enormously clear to me – more particularly so since the arrival of modular A-levels – that there is absolutely no substitute for prolonged and determined practice at questions of substance. One moment's recognition of a technique at work can turn several hours of struggle into just a few seconds of polishing off, and a lack of experience is always painfully clear when marking work from candidates who are under-practised at either the art of prolonged mathematics or the science of creative problem-solving. At the other, more successful, end of the scale there were many candidates who managed to produce extraordinary amounts of outstanding work, racking up full-, or nearly full-, marks on question after question. With the marks distributed as they were, it seems that the paper was pitched appropriately at the intended level, and that it successfully managed to distinguish between the different ability-levels to be found among the candidates. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Moreover, many of these were clearly acts of desperation.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show LaTeX source
Problem source
Two curves have equations $\; x^4+y^4=u\;$ and $\; xy = v\;$, where $u$ and $v$ are positive constants. State the equations of the lines of symmetry of each curve.
The curves intersect at the distinct points $A$, $B$, $C$ and $D$ (taken anticlockwise from $A$).
The coordinates of $A$ are $(\alpha,\beta)$, where $\alpha > \beta > 0$. Write down, in terms of $\alpha$ and $\beta$, the coordinates of
$B$, $C$ and $D$.
Show that the quadrilateral $ABCD$ is a rectangle and find its area in terms of $u$ and $v$ only. Verify that, for the case $u=81$ and $v=4$, the area is $14$.Solution source
The curve $x^4 + y^4 = u$ has lines of symmetry:
\begin{itemize}
\item $y = 0$
\item $x = 0$
\item $y = x$
\item $y = -x$
\end{itemize}
The curve $xy = v$ has lines of symmetry:
\begin{itemize}
\item $y = x$
\item $y = -x$
\end{itemize}
\begin{center}
\begin{tikzpicture}
\def\a{2};
\def\b{0.5};
\def\t{(\b)/(\b*\b+1)};
\def\functionf(#1){2*(#1)*((#1)^2-5)/((#1)^2-4)};
\def\xl{-1.5};
\def\xu{1.5};
\def\yl{-1.5};
\def\yu{1.5};
% 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=0:360, samples=101]
plot({sign(cos(\x))*sqrt(abs(cos(\x))}, {sign(sin(\x))*sqrt(abs(sin(\x))});
\draw[blue, smooth, thick, domain=0.1:\xu, samples=101]
plot({\x}, {.5/\x});
\draw[blue, smooth, thick, domain=\xl:-0.1, samples=101]
plot({\x}, {.5/\x});
% \draw[red, thin, dashed] ({-\b*\yl}, {\yl}) -- ({-\b*\yu}, {\yu});
\filldraw (0.508742611840723, 0.982815255421419) circle (1.5pt) node[above right ] {\tiny $B = (\beta, \alpha)$};
\filldraw (0.982815255421419, 0.508742611840723) circle (1.5pt) node[right] {\tiny $A = (\alpha, \beta)$};
\filldraw (-0.508742611840723, -0.982815255421419) circle (1.5pt) node[above right ] {\tiny $D = (-\beta, -\alpha)$};
\filldraw (-0.982815255421419, -0.508742611840723) circle (1.5pt) node[right] {\tiny $C = (-\alpha, -\beta)$};
% \filldraw (0.5, 0) circle (1.5pt) node[below right] {$a$};
% \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});
\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}
The points are $A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)$
$AD$ has gradient $\frac{\beta+\alpha}{\alpha+\beta} = 1$, $BC$ has the same gradient. $AB$ has gradient $\frac{\alpha-\beta}{\beta-\alpha} = -1$, as does $CD$. Therefore it has two sets of perpendicular and parallel sides, hence a rectangle.
The area is $|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)$
The squared area is $4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)$ ie the area is $2\sqrt{u-2v^2}$
When $u = 81, v = 4$ we have the area is $2 \sqrt{81 - 2 \cdot 16} = 14$ as required.
The first question is usually intended to be a gentle introduction to the paper, and to allow all candidates to gain some marks without making great demands on either memory or technical skills. This year, however, and for the first time that I can recall in recent years, the obviously algebraic nature of the question was enough to deter half the candidates from attempting it. Indeed, apart from Q8, it was the least popular pure maths question. This was a great pity: the helpful structure really did guide folks in the right direction, and any half-decent candidates who did try it usually scored very highly on it. There were, nonetheless, a couple of stumbling-blocks along the way for the less wary, and many candidates tripped over them. The point at which most of the less successful students started to go astray was when asked to show that ABCD is a rectangle. Lots of these folk elected to do so by working out distances … when the use of gradients would have been much simpler. The really disappointing thing was that many simply showed both pairs of opposite sides to be equal in length without realising that this only proved the quadrilateral a parallelogram. The next major difficulty was to be found in the algebra, in turning the area, 2|αβ − γδ|², into something to do with u and v. It was quite apparent that many were unable to do so because they failed to appreciate that α and β were particular values of x and y that satisfy the two original curve equations, so that α⁴ + β⁴ = u and αβ = v. Then, squaring the area expression does the trick. Some got part of the way to grasping this idea, but approached from the direction of solving x⁴ + y⁴ = u and xy = v as simultaneous equations; the resulting surds-within-surds expressions for α and β were too indigestible for almost anyone to cope with.