| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Standard +0.8 This is a standard group theory verification question requiring students to find the identity element by solving an equation, find the general inverse by solving another equation, and explain why a value must be excluded (closure/inverse existence). While it involves algebraic manipulation with a non-standard operation, these are routine procedures for Further Maths students covering abstract algebra. The question is moderately above average difficulty due to the algebraic complexity and abstract nature of groups, but follows a predictable template. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x \bullet e = x \Rightarrow 3(x+e+1)+2xe = x\) | M1 | Sets up and solves a correct equation for the identity element |
| \(\Rightarrow 3e+2xe = x-3x-3 \Rightarrow e = -\frac{2x+3}{2x+3}\) | A1 | Correct identity \(e = -1\); only one side needs checking |
| \(\Rightarrow e = -1\) | Special case: states \(e=-1\) with no working scores M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let inverse of \(x\) be \(y\): \(x \bullet y = e \Rightarrow 3(x+y+1)+2xy = -1\) | M1 | Sets up correct equation using identity element |
| \(\Rightarrow 3y+2xy = -1-3x-3 \Rightarrow y = \ldots\) | M1 | Expands, rearranges and factorises to find inverse |
| \(\Rightarrow x^{-1} = -\dfrac{3x+4}{2x+3}\) | A1 | Correct inverse element; accept equivalent forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(x = -\frac{3}{2}\), identity property fails since \(2x+3=0\), giving division by zero; or \(x^{-1} = -\frac{1/2}{0}\) so no inverse exists | B1 | Must explain why identity property fails or why inverse doesn't exist; just saying "undefined" scores B0; closure/associativity arguments score B0 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x \bullet e = x \Rightarrow 3(x+e+1)+2xe = x$ | M1 | Sets up and solves a correct equation for the identity element |
| $\Rightarrow 3e+2xe = x-3x-3 \Rightarrow e = -\frac{2x+3}{2x+3}$ | A1 | Correct identity $e = -1$; only one side needs checking |
| $\Rightarrow e = -1$ | | Special case: states $e=-1$ with no working scores M1A0 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let inverse of $x$ be $y$: $x \bullet y = e \Rightarrow 3(x+y+1)+2xy = -1$ | M1 | Sets up correct equation using identity element |
| $\Rightarrow 3y+2xy = -1-3x-3 \Rightarrow y = \ldots$ | M1 | Expands, rearranges and factorises to find inverse |
| $\Rightarrow x^{-1} = -\dfrac{3x+4}{2x+3}$ | A1 | Correct inverse element; accept equivalent forms |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $x = -\frac{3}{2}$, identity property fails since $2x+3=0$, giving division by zero; or $x^{-1} = -\frac{1/2}{0}$ so no inverse exists | B1 | Must explain why identity property fails or why inverse doesn't exist; just saying "undefined" scores B0; closure/associativity arguments score B0 |
---\begin{enumerate}
\item The set $\mathrm { G } = \mathbb { R } - \left\{ - \frac { 3 } { 2 } \right\}$ with the operation of $x \bullet y = 3 ( x + y + 1 ) + 2 x y$ forms a group.\\
(a) Determine the identity element of this group.\\
(b) Determine the inverse of a general element $x$ in this group.\\
(c) Explain why the value $- \frac { 3 } { 2 }$ must be excluded from $G$ in order for this to be a group.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2023 Q7 [6]}}