Year: 2010
Paper: 1
Question Number: 1
Course: LFM Pure
Section: Simultaneous equations
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
Given that
\[
5x^{2}+2y^{2}-6xy+4x-4y\equiv
a\left(x-y+2\right)^{2}
+b\left(cx+y\right)^{2}+d\,,
\]
find the values of the constants $a$, $b$, $c$
and $d$.
Solve the simultaneous equations
\begin{align*}
5x^{2}+2y^{2}-6xy+4x-4y&=9\,,
\\
6x^{2}+3y^{2}-8xy+8x-8y&=14\,.
\end{align*}
$a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\, \equiv (a + bc^2)x^2 + (a+b)y^2 + (-2a+2bc)xy + (4a)x+(-4ay) + 4a+d$
so we want to solve
\[
\begin{cases}
a + bc^2 &= 5 \\
a+b &= 2 \\
2bc - 2a &= -6 \\
4a &= 4 \\
-4a &= 4 \\
4a+d &= -9
\end{cases} \Rightarrow a = 1, b = 1, c = -2, d = -13
\]
Therefore we have:
$(x-y+2)^2 + (2x+y)^2-13 = 0$
and our simultaneous equations will be:
\[
\begin{cases}
(x-y+2)^2 + (-2x+y)^2 &= 13 \\
2(x-y+2)^2 + (-2x+y)^2 &= 22
\end{cases}
\]
which are simultaneous equations in $(x-y+2)^2$ and $(-2x+y)^2$ which solve to $(x-y+2)^2 = 9, (-2x+y)^2 = 4 $
so we need to solve $4$ sets of simultaneous equations:
\begin{align*}
&\begin{cases}
x - y + 2 &= 3 \\
-2x + y &= 2
\end{cases} &&\Rightarrow (x,y) = (-3, -4) \\
&\begin{cases}
x - y + 2 &= -3 \\
-2x + y &= 2
\end{cases} &&\Rightarrow (x,y) = (3, 8) \\
&\begin{cases}
x - y + 2 &= 3 \\
-2x + y &= -2
\end{cases} &&\Rightarrow (x,y) = (1, 0) \\
&\begin{cases}
x - y + 2 &= -3 \\
-2x + y &= -2
\end{cases} &&\Rightarrow (x,y) = (7, 12) \\
\end{align*}
So $(x,y) = (-3, -4), (3, 8), (1, 0), (7,12)$
This was by far the most popular question, attempted by almost all candidates. It was very pleasing to see so many perfect solutions, showing that many candidates had a good command of basic algebraic manipulation. The first part of the question was answered very well. Some candidates failed to take enough care with their algebra, the most common errors being to either lose the 4a term entirely or to replace it by 4. There were many different approaches used to tackle the second part, the majority of them being effective. The most common conceptual error was assuming that the solutions had to be integers. As the question did not say this, such attempts gained relatively few marks unless supported by a full algebraic solution. There were a few recurring technical errors and other ineffective approaches. The first was to implicitly assume that the second equation could be rewritten in the same way as the first without checking the consistency of all six equations (most notably, checking that a + bc2 = 6). The second was to fail to square root an equation correctly: from (y − 2x)2 = 4, for example, a significant number of students gave only the single solution y − 2x = 2. A significant number of students made arithmetical errors when solving the equations which left them bogged down in messy calculations, costing them only a few accuracy marks but lots of time. Also, several took long-winded methods which involved multiplying out expressions such as (y − 2x)2 = 4 to reach quadratic equations; as mentioned above, candidates should certainly be aware that they ought to simply take square roots. Most candidates who reached a pair of simultaneous equations for (y−2x)2 and (x−y+2)2 correctly solved for one of the terms, but a surprising number then substituted their answer(s) back into one of the original quadratic equations in x and y to eliminate a variable, giving themselves far more algebraic slog than actually necessary. Overall, the mean mark for this question was in excess of 14/20, making it the most successfully answered question on the paper.