2025 Paper 3 Q2
Year: 2025
Paper: 3
Question Number: 2
Course: LFM Pure
Section: Modulus function
Problem
- Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(x))\).
- Find all solutions of the equation \(f(f(x)) = x\).
- Find the values of \(a\) for which the sequence \(u_0, u_1, u_2, \ldots\) has period 2.
- Show that, if \(a = \frac{28}{5}\), then the sequence \(u_2, u_3, u_4, \ldots\) has period 2, but neither \(u_0\) or \(u_1\) is equal to either of \(u_2\) or \(u_3\).
- Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(f(x)))\).
- Consider the sequence \(u_0, u_1, u_2, \ldots\) in the cases \(a = 1\) and \(a = -\tfrac79\). Hence find all the solutions of the equation \(f(f(f(x))) = x\).
- Find a value of \(a\) such that the sequence \(u_3, u_4, u_5, \ldots\) has period 3, but where none of \(u_0, u_1\) or \(u_2\) is equal to any of \(u_3, u_4\) or \(u_5\).
Solution
- If \(a = 1\) then \(u_1 = f(a) = 7-2 = 5\), \(u_2 = f(5) = -3\), \(u_3 = f(-3) = 7-6 = 1\). Therefore it must be the case that \(f(f(f(x))) = x\) for \(x = 1, 5, -3\). Similarly, if \(a = -\tfrac79\) then \(u_1 = f(-\tfrac79) = \tfrac{49}{9}\), \(u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}\) and \(u_3 = f(-\tfrac{35}{9}) = -\tfrac79\). Therefore we must also have roots \(x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9\). We also have the roots \(x = -7, \tfrac73\) from the first part so we have found all \(8\) roots.
- We need \(f(f(f(x))) = 1\) but \(f(f(x)) \neq -3, f(x) \neq 5, x \neq 1\). Suppose \(f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3\). So \(y = 3\), ie \(f(f(x)) = 3\). Suppose \(f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2\). Finally we need \(f(x) = \pm 2\), so say \(7-2|x| = 2 \Rightarrow x = \tfrac52\), so we have the sequence \(\tfrac52, 2, 3, 1, 5, -3, 1, \cdots\)as required.
Examiner's report
— 2025 STEP 3, Question 2The majority of candidates focused solely on the pure questions, with questions 1, 2 and 8 the most popular. The statistics questions were more popular than the mechanics questions in this exam series. Candidates who did well on this paper generally: were careful to explain and justify the steps in their arguments, explaining what they had done rather than expecting the examiner to infer what had been done from disjointed groups of calculations; paid close attention to what was required by the questions; made fewer unnecessary mistakes with calculations; thought carefully about how to present rigorous arguments involving trig functions and their inverse functions, especially in relation to domain considerations; understood that questions set on the STEP papers require sufficient justification to earn full credit; knew the difference between 'positive' and 'non-negative'; attempted all parts of a question, picking up marks for later parts even when they had not necessarily attempted or completed previous parts. Candidates who did less well on this paper generally: did not pay attention to 'Hence' instructions: this means that you must use the previous part; presented explanations that were not precise enough (e.g. in Question 3 describing the transformations but not in the context of the graphs or in Question 8 not explaining use of trigonometric relationships sufficiently well); made additional assumptions, e.g. that a function was differentiable when this had not been given; tried to present if and only if arguments in a single argument when dealing with each direction separately would have been more appropriate and safer (note that this is not always the case; in general candidates need to consider what is the most appropriate presentation of an if and only if argument); tried to carry out too many steps in one go, resulting in them not justifying the key steps sufficiently; did not take sufficient care with graphs/curve sketching.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
Let $f(x) = 7 - 2|x|$.
A sequence $u_0, u_1, u_2, \ldots$ is defined by $u_0 = a$ and $u_n = f(u_{n-1})$ for $n > 0$.
\begin{questionparts}
\item \begin{enumerate}
\item Sketch, on the same axes, the graphs with equations $y = f(x)$ and $y = f(f(x))$.
\item Find all solutions of the equation $f(f(x)) = x$.
\item Find the values of $a$ for which the sequence $u_0, u_1, u_2, \ldots$ has period 2.
\item Show that, if $a = \frac{28}{5}$, then the sequence $u_2, u_3, u_4, \ldots$ has period 2, but neither $u_0$ or $u_1$ is equal to either of $u_2$ or $u_3$.
\end{enumerate}
\item \begin{enumerate}
\item Sketch, on the same axes, the graphs with equations $y = f(x)$ and $y = f(f(f(x)))$.
\item Consider the sequence $u_0, u_1, u_2, \ldots$ in the cases $a = 1$ and $a = -\tfrac79$. Hence find all the solutions of the equation $f(f(f(x))) = x$.
\item Find a value of $a$ such that the sequence $u_3, u_4, u_5, \ldots$ has period 3, but where none of $u_0, u_1$ or $u_2$ is equal to any of $u_3, u_4$ or $u_5$.
\end{enumerate}
\end{questionparts}Solution source
\begin{questionparts}
\begin{enumerate}
\item \begin{center}
\begin{tikzpicture}
\def\functionf(#1){7-2*abs(#1)};
\def\xl{-15};
\def\xu{15};
\def\yl{-15};
\def\yu{15};
% 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 a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, red, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\functionf(\x))});
\draw[thick, green, dashed] (\xl, \xl) -- (\xu, \xu) node [below left] {$y=x$};
% \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
% \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\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}
\item From our graph we can see there are $4$ solutions, and they will correspond to each possibility in the cases of $|\cdot |$
\begin{align*}
&& 7-2(7-2x) &= x \\
\Rightarrow && 3x &= 7 \\
\Rightarrow && x &= \frac73 \\
&& 7-2(2x-7) &= x \\
\Rightarrow && 5x &= 21\\
\Rightarrow && x &= \frac{21}{5} \\
&& 7 - 2(7+2x) &= x \\
\Rightarrow && 5x &= -7 \\
\Rightarrow && x &= -\frac75 \\
&& 7+2(7+2x) &= x \\
\Rightarrow && 3x &= -21 \\
\Rightarrow && x &= -7 \\
\\
\Rightarrow && x &= -7, -\tfrac75, \tfrac73,\tfrac{21}5
\end{align*}
\item Our sequence will have period $2$ if $f(f(a)) = a$ and $f(a) \neq a$, ie we need to find the points of intersection of $y = x$ with $f(f(x))$ which aren't points of intersection between $x = f(x)$. From our chart we can see these are $x = -\tfrac75, \tfrac{21}{5}$.
\item If $a = \tfrac{23}5$ then
\begin{align*}
&& u_1 &= f(a) \\
&&&= 7 - 2 \cdot \frac{28}{5} \\
&&&= - \frac{21}{5} \\
&& u_2 &= f(u_1) \\
&&&= 7 - 2 \cdot \frac{21}{5} \\
&&&= -\frac{7}{5}
\end{align*}
Since $f(f(u_2)) = u_2$ and $f(f(u_1)) \neq u_1$ the sequence has period $2$ from this point onwards.
\end{enumerate}
\item \begin{enumerate}
\item \begin{center}
\begin{tikzpicture}
\def\functionf(#1){7-2*abs(#1)};
\def\xl{-15};
\def\xu{15};
\def\yl{-15};
\def\yu{15};
% 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 a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, red, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\functionf(\functionf(\x)))});
\draw[thick, green, dashed] (\xl, \xl) -- (\xu, \xu) node [below left] {$y=x$};
% \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
% \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\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}
\item If $a = 1$ then $u_1 = f(a) = 7-2 = 5$, $u_2 = f(5) = -3$, $u_3 = f(-3) = 7-6 = 1$. Therefore it must be the case that $f(f(f(x))) = x$ for $x = 1, 5, -3$. Similarly, if $a = -\tfrac79$ then $u_1 = f(-\tfrac79) = \tfrac{49}{9}$, $u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}$ and $u_3 = f(-\tfrac{35}{9}) = -\tfrac79$. Therefore we must also have roots $x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9$. We also have the roots $x = -7, \tfrac73$ from the first part so we have found all $8$ roots.
\item We need $f(f(f(x))) = 1$ but $f(f(x)) \neq -3, f(x) \neq 5, x \neq 1$. Suppose $f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3$. So $y = 3$, ie $f(f(x)) = 3$. Suppose $f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2$. Finally we need $f(x) = \pm 2$, so say $7-2|x| = 2 \Rightarrow x = \tfrac52$, so we have the sequence $\tfrac52, 2, 3, 1, 5, -3, 1, \cdots$as required.
\end{enumerate}
\end{questionparts}
In terms of attempts, this was the third most popular question. Most candidates who attempted this question were able to make good progress with many of the parts. Candidates were generally able to sketch the graph of y = f(x), but sketches of y = f(f(x)) often had some features missing or incorrect. Many candidates opted to work out the equation for each of the straight-line segments before sketching the graph and, while this generally resulted in the correct overall shape, important points such as the vertical positioning of the points where the two graphs cross were often incorrect. A number of candidates would have benefitted from making their sketches larger. A small number of candidates did not sketch the two graphs on the same set of axes, which meant that some of the marks for this part of the question were not accessible. Part (c) was not answered well, with many candidates simply restating their solutions to part (b) without considering the fact that some of the solutions would lead to sequences with a period of 1. Allowance was made here for those candidates who stated that a solution with period 1 also has period 2 and listed all their solutions to (b). In part (d) candidates successfully calculated the terms of the sequence and many identified the connection with the previous parts to explain that the remainder of the sequence would have a period of 2. A small number of candidates only checked that u0 ≠ u2 and u1 ≠ u3 and therefore did not fully answer this part of the question. Those who had made good sketches for the graphs in part (i) (a) generally made good attempts at the sketches in part (ii) (a), although similar issues were encountered with the positioning of the intersections of the two graphs. Almost all candidates who attempted part (ii) (b) were able to calculate the sequences starting with the two given values, although many did not realise that all three values that appeared in each solution would also be solutions. The question was posed using 'Hence' and so this approach was required. Some candidates simply stated their set of solutions without providing any explanation. While some candidates commented on the fact that there must be eight solutions based on their sketch, a significant number of candidates did not realise that the two period 1 values that lead to a constant sequence would also be solutions.