2025 Paper 3 Q4

Year: 2025
Paper: 3
Question Number: 4

Course: LFM Pure
Section: Linear transformations

Difficulty: 1500.0 Banger: 1500.0
linear transformations rotation matrix ellipse reflection invariant line curve sketching implicit differentiation conics

Problem

  1. \(x_2\) and \(y_2\) are defined in terms of \(x_1\) and \(y_1\) by the equation $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ \(G_1\) is the graph with equation $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ and \(G_2\) is the graph with equation $$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$ Show that, if \((x_1, y_1)\) is a point on \(G_1\), then \((x_2, y_2)\) is a point on \(G_2\). Show that \(G_2\) is an anti-clockwise rotation of \(G_1\) through \(45^\circ\) about the origin.
    1. The matrix $$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$ represents a reflection. Find the line of invariant points of this matrix.
    2. Sketch, on the same axes, the graphs with equations $$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
  3. Sketch, on the same axes, for \(0 \leq x \leq 2\pi\), the graphs with equations $$y = \sin x \text{ and } y = \sin(x - 2y)$$ You should determine the exact co-ordinates of the points on the graph with equation \(y = \sin(x - 2y)\) where the tangent is horizontal and those where it is vertical.

Solution

  1. Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if \(\frac{x_1^2}9+\frac{y_1^2}{4} = 1\) we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that \((x_1, y_1)\) is on \(G_1\) is equivalent to \((x_2,y_2)\) being on the \(G_2\). Since the point \((x_2,y_2)\) is a \(45^{\circ}\) rotation of \((x_1,y_1)\) anticlockwise about the origin, this means \(G_2\) is a \(45^{\circ}\) anticlockwise rotation of \(G_1\).
    1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
    2. TikZ diagram
  3. Consider the transformation \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) which is a shear, leaving the \(x\)-axis invariant. Then we must have:
    TikZ diagram
    Since the shear leaves lines of the form \(y = k\) invariant, the points where \(\frac{\d y}{\d x} = 0\) must also map to points where this is true, ie \((\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)\) map to points \((\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)\) where the tangent is horizontal. The line \(x = c\) map back to lines \(\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}\), ie \(y = -\frac12 x- \frac{c}{2}\). Therefore we are interested in points on the original curve where the gradient is \(-\frac12\), ie \((\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})\), these map to \((\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})\)
Examiner's report
— 2025 STEP 3, Question 4
Mean: ~7.5 / 20 (inferred) Below Average Inferred ~7.5/20: less popular pure Q but good number of attempts; part (i) approach known but detail lacking, variable success in later parts, only most successful completed part (ii)(b) fully

This question was one of the less popular pure questions but still had a good number of attempts. In the question, candidates are led through an example of how to apply the mathematics they know to a new context and then are expected to apply what they have learnt to other problems. For part (i) most candidates knew how to approach the problem but showed insufficient detail in their working for full credit, usually by not making the link (x1, y1) on G1 ⟹ x1²/9 + y1²/4 = 1 when trying to show that (x2, y2) is on G2. A handful did not realise the significance of the indexing on the points and instead tried to show that the equations of the two curves were equivalent. Almost all candidates recognised that the given matrix was a rotation matrix, but some did not make the link between this and the relationship between the points on the curves clear. In part (ii) (a) it became evident that a number of candidates do not know the difference between a line of invariant points and an invariant line (in the second case points can move under the transformation but must stay on the same line). This meant some candidates did a lot more working than was necessary and often ended up with an extra answer, meaning that they could not gain full credit for this part. A few candidates used the general form of a reflection matrix in the line y = tan θ and a t substitution to find the required line. This method also required candidates to reject one solution, which was usually done by those taking this route. In part (ii) (b) only the most successful candidates showed convincingly that if (x1, y1) was on the graph of y = 2x then (x2, y2) = (−0.6 0.8; 0.8 0.6)(x1; y1) was on the other graph, but most realised that the two graphs were reflections of each other and so could make an attempt at the sketch. The most common mistakes here were assuming that the second graph was asymptotic to the y axis and not showing the two graphs intersecting twice. Attempts were variable for part (iii). A lot of candidates found a matrix connecting points on the two curves, but often had the relationship the 'wrong way around' with (x2; y2) = (1 −2; 0 1)(x1; y1) rather than the correct version (x2; y2) = (1 2; 0 1)(x1; y1). Many candidates could differentiate the implicit equation y = sin(x − 2y) successfully and some successfully went on to find the points where the tangent was horizontal and vertical. Some who had found the correct transformation matrix could use this to find the points with horizontal tangents but struggled to use a similar argument convincingly for the vertical tangents. Some candidates successfully set dy/dx = 0 and dx/dy = 0 to get x − 2y = π/2 or x − 2y = 2π/3 but were then uncertain how they could use this to find the coordinates of the relevant points. Many of the candidates who found the points with horizontal or vertical tangents could 'join the dots' to complete the sketch, but some joined them in the wrong order. Many candidates laboured under the misunderstanding that it is not possible for an implicit function to be one-to-many valued which caused a variety of different mistakes.

The 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.

Source: Cambridge STEP 2025 Examiner's Report · 2025-p3.pdf
Rating Information

Difficulty Rating: 1500.0

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Banger Rating: 1500.0

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Show LaTeX source
Problem source
\begin{questionparts}
\item $x_2$ and $y_2$ are defined in terms of $x_1$ and $y_1$ by the equation
$$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$
$G_1$ is the graph with equation 
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
and $G_2$ is the graph with equation
$$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$
Show that, if $(x_1, y_1)$ is a point on $G_1$, then $(x_2, y_2)$ is a point on $G_2$.
Show that $G_2$ is an anti-clockwise rotation of $G_1$ through $45^\circ$ about the origin.
\item \begin{enumerate}
\item The matrix 
$$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$
represents a reflection. Find the line of invariant points of this matrix.
\item Sketch, on the same axes, the graphs with equations 
$$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
\end{enumerate}
\item Sketch, on the same axes, for $0 \leq x \leq 2\pi$, the graphs with equations 
$$y = \sin x \text{ and } y = \sin(x - 2y)$$
You should determine the exact co-ordinates of the points on the graph with equation $y = \sin(x - 2y)$ where the tangent is horizontal and those where it is vertical.
\end{questionparts}
Solution source
\begin{questionparts}
\item Suppose \begin{align*}
&& \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\
\Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2}
\end{align*}
Therefore if $\frac{x_1^2}9+\frac{y_1^2}{4} = 1$ we must have

\begin{align*}
\frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1
\end{align*}

but this is precisely the statement that $(x_1, y_1)$ is on $G_1$ is equivalent to $(x_2,y_2)$ being on the $G_2$. Since the point $(x_2,y_2)$ is a $45^{\circ}$ rotation of $(x_1,y_1)$ anticlockwise about the origin, this means $G_2$ is a $45^{\circ}$ anticlockwise rotation of $G_1$.

\item \begin{enumerate}
\item
\begin{align*}
&& \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\
\Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\
\Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\
\Rightarrow && y &=2 x
\end{align*}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2^(#1)};
    \def\xl{-7};
    \def\xu{7};
    \def\yl{-7};
    \def\yu{7};
    
    % 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, black, dashed] (\xl, 2*\xl) -- (\xu, 2*\xu) node [below left] {$y=2x$};
        \draw[thick, red, smooth, domain=\xl:\xu, samples=100] 
            plot ({-0.6*\x+0.8*\functionf(\x)}, {0.8*\x+0.6*\functionf(\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}
\end{enumerate}
\item Consider the transformation $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ which is a shear, leaving the $x$-axis invariant. Then we must have:

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(#1)};
    \def\xl{-1};
    \def\xu{7};
    \def\yl{-2};
    \def\yu{2};
    
    % 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=0:2*pi, samples=100] 
            plot (\x, {\functionf(deg(\x))});
        \draw[thick, red, smooth, domain=0:2*pi, samples=100] 
            plot ({\x+2*\functionf(deg(\x))}, {\functionf(deg(\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}

Since the shear leaves lines of the form $y = k$ invariant, the points where $\frac{\d y}{\d x} = 0$ must also map to points where this is true, ie $(\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)$ map to points $(\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)$ where the tangent is horizontal.

The line $x = c$ map back to lines $\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}$, ie $y = -\frac12 x- \frac{c}{2}$. Therefore we are interested in points on the original curve where the gradient is $-\frac12$, ie $(\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})$, these map to $(\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}),  (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})$

\end{questionparts}
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