A2 June 2023 Q7
7. The set \(G = \mathbb{R} - \left\{-\dfrac{3}{2}\right\}\) with the operation of \(x \bullet y = 3(x + y + 1) + 2xy\) forms a group.
| Scheme | Marks | AO |
|---|---|---|
| \(x \bullet e = x \Rightarrow 3(x + e + 1) + 2xe = x\) \(\Rightarrow 3e + 2xe = x - 3x - 3 \Rightarrow e = \ldots\) Or Let \(e = -1 \Rightarrow e \bullet y = 3(-1 + y + 1) + 2(-1)y = \ldots\) | M1 | 2.1 |
| \[\Rightarrow e = -\frac{2x + 3}{2x + 3} \Rightarrow e = -1\]Or \(e \bullet y = 3(-1 + y + 1) + 2(-1)y = y\) therefore \(e = -1\) | A1 | 2.2a |
| (2) |
Notes
M1: Sets up and solves a correct equation for the identity element.
Alternatively spots the identity is \(-1\) and attempts to show this satisfies the identity property.
E.g. \(-1 \bullet x = 3(-1 + x + 1) + 2(-1)x = \ldots\)
A1: Correct identity / correct demonstration that \(-1\) is the identity with appropriate conclusion. (Note only one side needs to be checked as we are told \(G\) is a group so one side is sufficient by uniqueness).
Special case: states \(e = -1\) with no working scores M1A0
| Scheme | Marks | AO |
|---|---|---|
| Let the inverse of \(x\) be \(y\) then \(x \bullet y = e \Rightarrow 3(x + y + 1) + 2xy = -1\) | M1 | 3.1a |
| \(\Rightarrow 3y + 2xy = -1 - 3x - 3 \Rightarrow y = \ldots\) | M1 | 2.1 |
| \[\Rightarrow x^{-1} = -\frac{3x + 4}{2x + 3} \text{ oe such as } -\frac{3}{2} + \frac{1}{4x + 6}\] | A1 | 2.2a |
| (3) |
Notes
M1: Sets up a correct equation for the inverse using their identity element. May use \(x^{-1}\) throughout, or may use \(y\) or other variable.
M1: Expands, rearranges and by factorising out the inverse element from two terms to find an expression for the inverse element.
A1: Correct inverse element. Accept equivalent forms.
| Scheme | Marks | AO |
|---|---|---|
| E.g. for \(x = -\tfrac{3}{2}\) the identity property would fail since we cannot divide by \(2x + 3 = 0\) or shows \(x^{-1} = -\tfrac{-\tfrac{1}{2}}{0}\) so no inverse or can’t divide by 0 Or \(-\tfrac{3}{2} \bullet x = 3\left(-\tfrac{3}{2} + x + 1\right) - 3x = -\tfrac{3}{2}\) so the identity property fails as not unique as any \(x\) would be an identity for \(-\tfrac{3}{2}\). | B1 | 2.4 |
| (1) | ||
| (6 marks) |
Notes
B1: Explains that the identity property would fail for \(x = -\dfrac{3}{2}\). Alternatively accept a similar argument that \(x = -\dfrac{3}{2}\) could not have an inverse due to dividing by zero. May refer back to work in (b), if done there, without need to repeat it.
Just saying undefined is B0 without a reason why.
Note: Reasons saying closure or associativity would fail are incorrect so B0, as both of these properties are satisfied on all of \(\mathbb{R}\)