Year: 2017
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Proof
This year's paper was, perhaps, slightly more straightforward than usual, with more helpful guidance offered in some of the questions. Thus the mark required for a "1", a Distinction, was 80 (out of 120), around ten marks higher than that which would customarily be required to be awarded this grade. Nonetheless, a three‐figure mark is still a considerable achievement and, of the 1330 candidates sitting the paper, there were 89 who achieved this. At the other end of the scale, there were over 350 who scored 40 or below, including almost 150 who failed to exceed a total score of 25. As a general strategy for success in a STEP examination, candidates should be looking to find four "good" questions to work at (which may be chosen freely by the candidates from a total of 13 questions overall). It is unfortunately the case that so many low‐scoring candidates flit from one question to another, barely starting each one before moving on. There needs to be a willingness to persevere with a question until a measure of understanding as to the nature of the question's purpose and direction begins to emerge. Many low‐scoring candidates fail to deal with those parts of questions which cover routine mathematical processes ‐ processes that should be standard for an A‐level candidate. The significance of the "rule of four" is that four high‐scoring questions (15‐20 marks apiece) obtains you up to around the total of 70 that is usually required for a "1"; and with a couple of supporting starts to questions, such a total should not be beyond a good candidate who has prepared adequately. This year, significantly more than 10% of candidates failed to score at least half marks on any one question; and, given that Q1 (and often Q2 also) is (are) specifically set to give all candidates the opportunity to secure some marks, this indicates that these candidates are giving up too easily. Mathematics is about more than just getting to correct answers. It is about communicating clearly and precisely. Particularly with "show that" questions, candidates need to distinguish themselves from those who are just tracking back from given results. They should also be aware that convincing themselves is not sufficient, and if they are using a result from 3 pages earlier, they should make this clear in their working. A few specifics: In answers to mechanics questions, clarity of diagrams would have helped many students. If new variables or functions are introduced, it is important that students clearly define them. One area which is very important in STEP but which was very poorly done is dealing with inequalities. Although a wide range of approaches such as perturbation theory were attempted, at STEP level having a good understanding of the basics – such as changing the inequality if multiplying by a negative number – is more than enough. In fact, candidates who used more advanced methods rarely succeeded.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The sequence of numbers $x_0$, $x_1$, $x_2$, $\ldots$ satisfies
\[
x_{n+1} = \frac{ax_n-1}{x_n+b}
\,.
\]
(You may assume that $a$, $b$ and $x_0$ are such that $x_n+b\ne0\,$.)
Find an expression for $x_{n+2}$ in terms of $a$, $b$ and $x_n$.
\begin{questionparts}
\item
Show that $a+b=0$ is a necessary condition for the sequence to be periodic with period 2.
\textbf{Note: } The sequence is said to be periodic with period $k$ if $x_{n+k} = x_n$ for all $n$, and there is no integer $m$ with $0 < m < k$ such that $x_{n+m} = x_n$ for all $n$.
\item Find necessary and sufficient conditions for the sequence to have period 4.
\end{questionparts}\begin{align*}
x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\
&= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\
&= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\
&= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1}
\end{align*}
\begin{questionparts}
\item If $x_{n+2} = x_n$ then
\begin{align*}
&& x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\
\Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\
&&&= (a+b)(x_n^2+(a-b)x_n + 1)
\end{align*}
If $x_{n+1} = x_n$ then $x_n^2+(a-b)x_n + 1$ and since our sequence has period $2$ rather than $1$ it must be the case this is non-zero. Therefore $a+b =0$.
\item
\begin{align*}
x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\
&= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\
&= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2}
\end{align*}
If $x_{n+4} = x_n$ then
\begin{align*}
x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\
0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\
&= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\
&= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1)
\end{align*}
Since we do not want $x_n$ to be periodic with period $1$ we must have the quadratic in $x_n$ $\neq 0$. If $a+b = 0$ then $x_n$ is periodic with period $2$ since $x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n$. Therefore it is necessary that $a^2+b^2-2 = 0$.
If $a^2+b^2-2= 0$ then
\begin{align*}
x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\
&=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\
&=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\
&= x_n
\end{align*}
Therefore it is sufficient too. So our conditions are $a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0$ and $a^2+b^2-2 = 0$
\end{questionparts}
This was the second most popular question of all, attempted by over 80% of candidates; but scoring relatively poorly with a mean score of under 10. It was, of course, heavily algebraic and this meant that many candidates found it challenging, getting lost in the algebra. In most cases, this was largely avoidable: the simple device of calling the first term "X" (say) would have prevented a lot of unnecessary subscripts from cluttering up the working. A few moments of thought from those candidates who simply embarked on the (potentially) intricate algebra could have saved a lot of trouble. The point of a sequence's periodicity is that it is the smallest cycle over which terms repeat; it should be noted that the condition for each term to be equal (a constant sequence) must clearly be embedded in any condition that gives . Similarly, in order to satisfy , we must automatically have the cases when all terms are the same and every other term equal present somewhere. This makes any ensuing factorisations much easier to deal with. It could be noted that the requirement for can be thought of as a two‐stage sequence using every other term; and this situation has just been sorted out.