| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 12 |
| Topic | Groups |
| Type | Matrix groups |
| Difficulty | Hard +2.3 This question requires deep understanding of abstract algebra (group theory, isomorphisms), matrix operations, and function composition at a level well beyond standard A-level. Students must understand why f^6=x doesn't guarantee order 6, work with matrix groups and determinants, recognize the connection between matrices and rational functions via Möbius transformations, and troubleshoot when an isomorphism fails. The multi-layered conceptual demands and need for mathematical maturity place this significantly above typical Further Maths questions. |
| Spec | 4.03c Matrix multiplication: properties (associative, not commutative)4.03n Inverse 2x2 matrix8.03c Group definition: recall and use, show structure is/isn't a group8.03h Generators: of cyclic and non-cyclic groups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) If \(f^3(x) = x\) or \(f^3(a) = x\) then it will generate a group of order 2 or 3. | B1, B1 | |
| (ii) Since \(\det(\text{product}) = \text{product of dets}\) we need \(\det(M^3) = (\det M)^3 = \det(I) = 1 \Rightarrow \det(M) = \pm 1\) | E1, E1 | |
| (iii) Want \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{2\pi}{6} & -\sin\frac{2\pi}{6} \\ \sin\frac{2\pi}{6} & \cos\frac{2\pi}{6} \end{pmatrix}\) | B1 | i.e. a generator of the group of rotations through multiples of \(\frac{\pi}{3}\) |
| (iv) (a) \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\) gives \(f(x) = \frac{\frac{1}{x} - \sqrt{3}}{\frac{\sqrt{3}}{2}x + \frac{1}{2}}\) or \(\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\) | B1 | |
| (iv) (b) \(M^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\) gives \(f^3(x) = \frac{-x+0}{0x-1} = x\), so \(f\) not of order 6 | B1, B1 | * BC Must state the conclusion and not stop with the working |
| (iv) (c) However, taking \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{\pi}{6} & -\sin\frac{\pi}{6} \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} \end{pmatrix}\) | M1, A1 | Another rotation matrix; A correct suggestion |
| gives \(f(x) = \frac{x\sqrt{3} - 1}{x + \sqrt{3}}\) for which \(f^2(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}\) and \(f^3(x) = \frac{-1}{x}\) \(\Rightarrow f^6(x) = x\) | M1, M1 | Attempt at powers of \(f\) up to \(f^6\) |
| Must state, or have shown, that both \(f^2(x)\) and \(f^3(x) \neq x\) | A1 |
**(i)** If $f^3(x) = x$ or $f^3(a) = x$ then it will generate a group of order 2 or 3. | B1, B1 |
**(ii)** Since $\det(\text{product}) = \text{product of dets}$ we need $\det(M^3) = (\det M)^3 = \det(I) = 1 \Rightarrow \det(M) = \pm 1$ | E1, E1 |
**(iii)** Want $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{2\pi}{6} & -\sin\frac{2\pi}{6} \\ \sin\frac{2\pi}{6} & \cos\frac{2\pi}{6} \end{pmatrix}$ | B1 | i.e. a generator of the group of rotations through multiples of $\frac{\pi}{3}$
**(iv) (a)** $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$ gives $f(x) = \frac{\frac{1}{x} - \sqrt{3}}{\frac{\sqrt{3}}{2}x + \frac{1}{2}}$ or $\frac{x - \sqrt{3}}{x\sqrt{3} + 1}$ | B1 |
**(iv) (b)** $M^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ gives $f^3(x) = \frac{-x+0}{0x-1} = x$, so $f$ not of order 6 | B1, B1 | * BC Must state the conclusion and not stop with the working
**(iv) (c)** However, taking $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos\frac{\pi}{6} & -\sin\frac{\pi}{6} \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} \end{pmatrix}$ | M1, A1 | Another rotation matrix; A correct suggestion
gives $f(x) = \frac{x\sqrt{3} - 1}{x + \sqrt{3}}$ for which $f^2(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$ and $f^3(x) = \frac{-1}{x}$ $\Rightarrow f^6(x) = x$ | M1, M1 | Attempt at powers of $f$ up to $f^6$
Must state, or have shown, that both $f^2(x)$ and $f^3(x) \neq x$ | A1 |A class of students is set the task of finding a group of functions, under composition of functions, of order 6.
Student P suggests that this can be achieved by finding a function $f$ for which $f^6(x) = x$ and using this as a generator for the group.
\begin{enumerate}[label=(\roman*)]
\item Explain why the suggestion by Student P might not work. [2]
\end{enumerate}
Student Q observes that their class has already found a group of order 6 in a previous task; a group consisting of the powers of a particular, non-singular $2 \times 2$ real matrix $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, under the operation of matrix multiplication.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Explain why such a group is only possible if $\det(\mathbf{M}) = 1$ or $-1$. [2]
\item Write down values of $a$, $b$, $c$ and $d$ that would give a suitable matrix $\mathbf{M}$ for which $\mathbf{M}^6 = \mathbf{I}$ and $\det(\mathbf{M}) = 1$. [1]
\end{enumerate}
Student Q believes that it is possible to construct a rational function $f$ in the form $f(x) = \frac{ax + b}{cx + d}$ so that the group of functions is isomorphic to the matrix group which is generated by the matrix $\mathbf{M}$ of part (iii).
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item \begin{enumerate}[label=(\alph*)]
\item Write down and simplify the function $f$ that, according to Student Q, corresponds to $\mathbf{M}$. [1]
\item By calculating $\mathbf{M}^2$, show that Student Q's suggestion does not work. [2]
\item Find a different function $f$ that will satisfy the requirements of the task. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q6 [12]}}