Year: 2024
Paper: 2
Question Number: 5
Course: LFM Stats And Pure
Section: Polynomials
Many candidates produced good solutions to the questions, with the majority of candidates opting to focus on the pure questions of the paper. Candidates demonstrated very good ability, particularly in the area of manipulating algebra. Many candidates produced clear diagrams which in many cases meant that they were more successful in their attempts at their questions than those who did not do so. The paper also contained a number of places where the answer to be reached was given in the question. In such cases, candidates must be careful to ensure that they provide sufficient evidence of the method used to reach the result in order to gain full credit.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The functions $\mathrm{f}_1$ and $\mathrm{F}_1$, each with domain $\mathbb{Z}$, are defined by
\[ \mathrm{f}_1(n) = n^2 + 6n + 11, \]
\[ \mathrm{F}_1(n) = n^2 + 2. \]
Show that $\mathrm{F}_1$ has the same range as $\mathrm{f}_1$.
\item The function $\mathrm{g}_1$, with domain $\mathbb{Z}$, is defined by
\[ \mathrm{g}_1(n) = n^2 - 2n + 5. \]
Show that the ranges of $\mathrm{f}_1$ and $\mathrm{g}_1$ have empty intersection.
\item The functions $\mathrm{f}_2$ and $\mathrm{g}_2$, each with domain $\mathbb{Z}$, are defined by
\[ \mathrm{f}_2(n) = n^2 - 2n - 6, \]
\[ \mathrm{g}_2(n) = n^2 - 4n + 2. \]
Find any integers that lie in the intersection of the ranges of the two functions.
\item Show that $p^2 + pq + q^2 \geqslant 0$ for all real $p$ and $q$.
The functions $\mathrm{f}_3$ and $\mathrm{g}_3$, each with domain $\mathbb{Z}$, are defined by
\[ \mathrm{f}_3(n) = n^3 - 3n^2 + 7n, \]
\[ \mathrm{g}_3(n) = n^3 + 4n - 6. \]
Find any integers that lie in the intersection of the ranges of the two functions.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& f_1(n) &= n^2 + 6n + 11 \\
&&&= (n+3)^2 + 2 \\
&&&=F_1(n+3)
\end{align*}
Since $n \mapsto n+3$ is a bijection on $\mathbb{Z}$ both functions must have exactly the same range.
\item $g_1(n) = n^2-2n+5 = (n-1)^2 + 4$. Since squares are always $0, 1 \pmod{4}$ it's impossible for $f_1$ and $g_1$ to take the same value therefore the ranges have empty intersection.
\item $\,$ \begin{align*}
&& f_2(n) &= n^2-2n - 6 \\
&&&= (n-1)^2-7 \\
&& g_2(n) &= n^2-4n+2 \\
&&&= (n-2)^2 - 2
\end{align*} so suppose $x^2 - 7 = y^2 - 2$ then
\begin{align*}
&& x^2 - 7 &= y^2 -2 \\
\Rightarrow && 5 &= y^2 - x^2 \\
&&&= (y-x)(y+x)
\end{align*}
So we have cases:
$y-x = -5, y + x = -1 \Rightarrow y = -3$ and the output is $7$
$y-x=-1, y+x = -5 \Rightarrow y = -3$ same output
$y-x=1, y+x = 5 \Rightarrow y = 3$ same output
$y-x=5, y-x = 1 \Rightarrow y = 3$ same ouput.
\item \begin{align*}
&& 0 &\leq \frac12(p^2+q^2)+\frac12(p+q)^2 \\
&&&= p^2 + q^2 + pq
\end{align*}
Looking at $f_3$ we see
\begin{align*}
&& f_3(n) &= n^3 - 3n^2 + 7n \\
&&&= (n-1)^3 -3n + 7n +1 \\
&&&= (n-1)^3 +4(n-1) -3 \\
&&&= g_3(n-1) + 3
\end{align*} So suppose we have two values which are equal, ie
\begin{align*}
&& x^3 + 4x -3 &= y^3 +4y -6 \\
\Rightarrow && 3 &= y^3-x^3+4y-4x \\
&&&= (y-x)(y^2+xy+x^2+4)
\end{align*}
Since $x^2+xy+y^2 \geq 0$ then the right hand factor is always a positive integer bigger than $3$ and in particular there will be no solutions and hence no integers in the intersection of the ranges.
\end{questionparts}
This was the second most popular question, but many candidates did not recognise the significance of the domain being used for the functions in this question. This meant that many applied techniques relevant to functions where the domain is the set of real numbers and therefore reached incorrect answers. The first requirement of part (iv) was not affected by this misunderstanding and most candidates were able to prove the required result successfully, usually by completing the square. A number of good solutions were seen, however. Those who completed the square and used the difference of two squares were often successful in parts (i), (ii) and (iii). Some of these were also able to make progress on the end of part (iv).