Year: 2019
Paper: 2
Question Number: 5
Course: LFM Pure
Section: Proof
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The sequence $u_0, u_1, \ldots$ is said to be a constant sequence if $u_n = u_{n+1}$ for $n = 0, 1, 2, \ldots$.
The sequence is said to be a sequence of period 2 if $u_n = u_{n+2}$ for $n = 0, 1, 2, \ldots$ and the sequence is not constant.
\begin{questionparts}
\item A sequence of real numbers is defined by $u_0 = a$ and $u_{n+1} = f(u_n)$ for $n = 0, 1, 2, \ldots$, where
$$f(x) = p + (x - p)x,$$
and $p$ is a given real number.
Find the values of $a$ for which the sequence is constant.
Show that the sequence has period 2 for some value of $a$ if and only if $p > 3$ or $p < -1$.
\item A sequence of real numbers is defined by $u_0 = a$ and $u_{n+1} = f(u_n)$ for $n = 0, 1, 2, \ldots$, where
$$f(x) = q + (x - p)x,$$
and $p$ and $q$ are given real numbers.
Show that there is no value of $a$ for which the sequence is constant if and only if $f(x) > x$ for all $x$.
Deduce that, if there is no value of $a$ for which the sequence is constant, then there is no value of $a$ for which the sequence has period 2.
Is it true that, if there is no value of $a$ for which the sequence has period 2, then there is no value of $a$ for which the sequence is constant?
\end{questionparts}\begin{questionparts}
\item If $f(a) = a$ then the sequence is constant, ie $a = p+a^2-pa \Rightarrow 0 = (a-p)(a-1)$. Therefore $a = 1, p$
If there sequence has period $2$ then there must be a solution to $f(f(x)) = x$, ie \begin{align*}
&& x &= p+(f(x)-p)f(x) \\
&&&= p+(p+(x-p)x-p)(p+(x-p)x) \\
&&&= p + (x-p)x(p+(x-p)x) \\
&&&= p+(x^2-px)(x^2-px+p) \\
\Rightarrow && 0 &= x^4-2px^3+(p+p^2)x^2-(p^2+1)x+p \\
&&&= (x-1)(x-p)(x^2-(p-1)x+1)
\end{align*}
The first two roots ($x = 1, p$) are constant sequences, so we need the second quadratic to have a root, ie $(p-1)^2-4 \geq 0 \Rightarrow p \geq 3 , p \leq -1$. We also need this root not to be $1$ or $p$, ie $1-(p-1)+1 = 3-p \neq 0$ and $p^2-(p-1)p + 1 = 1+p \neq 0$ so $p \neq -1, 3$. Therefore $p > 3$ or $p < -1$.
\item There exists a constant sequence iff there is a solution to $f(x) = x$, ie
\begin{align*}
&& x &= f(x) \\
&&&= q + (x-p)x \\
\Leftrightarrow && 0 &= x^2-(p+1)x + q \tag{has a solution} \\
\end{align*}
But if it doesn't have a solution, clearly the RHS is always larger, and if it does have a solution then there is some point where the inequality doesn't hold.
Suppose $f(x) > x$ for all $x$ then $f(f(x)) > f(x) > x$ therefore there is no value where $f(f(x)) = x$ which is required for any sequence of period 2.
No, consider $p = q = 0$ so $f(x) = x^2$ then there cannot be a period $2$ sequence by the first part, but also clearly $u_n = 1$ is a valid constant sequence.
\end{questionparts}
It was difficult to get full marks on this question, with most candidates struggling to correctly prove 'if and only if' statements in both directions. Mostly, the two constant sequences were successfully found and then correctly rejected for sequences of period 2, but few thought to check that the other two solutions to the quartic did not also coincide with the constant sequences. Most candidates were able to use the discriminant to produce bounds on p, but many could not justify the strictness of the inequality, which was best done by considering the boundary cases separately. The first request of the second part was answered well, with most using only the fact that it was a positive quadratic and a minority delving into the details of f(x). Most candidates who reached this part of the questions correctly used the result f(x) ≠ x to show that f(f(x)) ≠ x has no solutions, but many overlooked the connection between the final part and part (i).