| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find line of invariant points |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard techniques: matrix multiplication (part a), finding eigenvalues/eigenvectors for invariant lines (part b), and determinant calculation (part c). Each part requires routine application of learned methods with no novel insight, making it easier than average even for Further Maths content. |
| Spec | 4.03g Invariant points and lines4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| When \(p = 0\): \(\mathbf{M}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}1 & 12\\2 & -3\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}64\\-7\end{bmatrix}\); \(\therefore\) image of \((4,5)\) is \((64,-7)\) | M1, R1 | May use letter M or M in terms of \(p\), or with \(p=0\); condone no conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}-5 & 12\\0 & 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x\\y\end{bmatrix}\); \(\therefore -5x + 12y = x\) and \(0x + y = y\); \(12y = 6x\); \(y = \frac{1}{2}x\); gradient is \(\frac{1}{2}\) | M1, M1, A1 | First M1: multiplies M (or \(\mathbf{M}^{-1}\)) by \(\begin{bmatrix}x\\y\end{bmatrix}\) and equates to \(\begin{bmatrix}x\\y\end{bmatrix}\); accept \(y\) replaced with \(mx\) or \(mx+c\); second M1: multiplying top row of M by \(\begin{bmatrix}x\\y\end{bmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det\mathbf{M} = 0\); \(\det\mathbf{M} = (3p+1)(p^2-3) - 12(p+2) = 3p^3 + p^2 - 9p - 3 - 12p - 24 = 3p^3 + p^2 - 21p - 27\); \(\mathbf{M}\) is singular when \(3p^3 + p^2 - 21p - 27 = 0\); \(\Rightarrow p = 3\) or \(p = \frac{-5 \pm i\sqrt{2}}{3}\); \(\therefore p = 3\) is the only real value | B1, M1, A1, A1, M1, R1 | Condone \(ad + bc\) for det; last M1: uses correct method to deduce det \(\mathbf{M} = 0\) has exactly one real root |
## Question 9:
### Part 9(a):
When $p = 0$: $\mathbf{M}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}1 & 12\\2 & -3\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}64\\-7\end{bmatrix}$; $\therefore$ image of $(4,5)$ is $(64,-7)$ | M1, R1 | May use letter **M** or **M** in terms of $p$, or with $p=0$; condone no conclusion
### Part 9(b):
$\begin{bmatrix}-5 & 12\\0 & 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x\\y\end{bmatrix}$; $\therefore -5x + 12y = x$ and $0x + y = y$; $12y = 6x$; $y = \frac{1}{2}x$; gradient is $\frac{1}{2}$ | M1, M1, A1 | First M1: multiplies **M** (or $\mathbf{M}^{-1}$) by $\begin{bmatrix}x\\y\end{bmatrix}$ and equates to $\begin{bmatrix}x\\y\end{bmatrix}$; accept $y$ replaced with $mx$ or $mx+c$; second M1: multiplying top row of **M** by $\begin{bmatrix}x\\y\end{bmatrix}$
### Part 9(c):
$\det\mathbf{M} = 0$; $\det\mathbf{M} = (3p+1)(p^2-3) - 12(p+2) = 3p^3 + p^2 - 9p - 3 - 12p - 24 = 3p^3 + p^2 - 21p - 27$; $\mathbf{M}$ is singular when $3p^3 + p^2 - 21p - 27 = 0$; $\Rightarrow p = 3$ or $p = \frac{-5 \pm i\sqrt{2}}{3}$; $\therefore p = 3$ is the only real value | B1, M1, A1, A1, M1, R1 | Condone $ad + bc$ for det; last M1: uses correct method to deduce det $\mathbf{M} = 0$ has exactly one real root
---9 The matrix $\mathbf { M }$ represents the transformation T and is given by
$$\mathbf { M } = \left[ \begin{array} { c c }
3 p + 1 & 12 \\
p + 2 & p ^ { 2 } - 3
\end{array} \right]$$
9
\begin{enumerate}[label=(\alph*)]
\item In the case when $p = 0$ show that the image of the point $( 4,5 )$ under T is the point $( 64 , - 7 )$\\
9
\item In the case when $p = - 2$ find the gradient of the line of invariant points under $T$\\
9
\item Show that $p = 3$ is the only real value of $p$ for which $\mathbf { M }$ is singular.\\
The curve $C$ has equation
$$y = \frac { 3 x ^ { 2 } + m x + p } { x ^ { 2 } + p x + m }$$
where $m$ and $p$ are integers.\\
The vertical asymptotes of $C$ are $x = - 4$ and $x = - 1$\\
The curve $C$ is shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b37e2ee7-1cde-4d75-895a-381b32f4e95a-12_867_1102_733_463}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q9 [11]}}