| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.8 This is a Further Maths group theory question requiring verification of all four group axioms for a non-standard binary operation on complex numbers. While the individual verifications are methodical, part (d) requires insight to identify which elements lack inverses (those with a=0), making it significantly harder than routine A-level questions but not exceptionally difficult for Further Maths students who have studied abstract algebra. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Closed … YES since the ‘product’ of 2 complex numbers is a third |
| [1] | 2.4 | ‘Yes’ with valid reason |
| (b) | [(a + ib) (c + id)] (e + if) = [ac + i(b + ad)] (e + if) |
| Answer | Marks |
|---|---|
| = ace + i(b + a[d + cf]) = … | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at first of (pq)r or p(qr), correct to here |
| Answer | Marks |
|---|---|
| (c) | 1 + i0 is the identity |
| since (1 + i0) (a + ib) = 1∙a + i(0 + 1b) = (a + ib) (1 + i0) | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.4 | Demonstrated (condone one-side only) |
| (d) | (a + ib) (c + id) = 1 ac + i(b + ad) = 1 |
| Answer | Marks |
|---|---|
| i.e. the strictly imaginary numbers)) | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 3.2a | Test for an inverse of a general element |
Question 6:
6 | (a) | Closed … YES since the ‘product’ of 2 complex numbers is a third | B1
[1] | 2.4 | ‘Yes’ with valid reason
(b) | [(a + ib) (c + id)] (e + if) = [ac + i(b + ad)] (e + if)
= ace + i(b + ad + acf)
(a + ib) [(c + id) (e + if)] = (a + ib) [ce + i(d + cf)]
= ace + i(b + a[d + cf]) = … | M1
A1
M1
A1
[4] | 1.2
1.1
2.1
1.1 | Attempt at first of (pq)r or p(qr), correct to here
Attempt at the other, correct to here
Must be convincingly shown equal to first answer
(c) | 1 + i0 is the identity
since (1 + i0) (a + ib) = 1∙a + i(0 + 1b) = (a + ib) (1 + i0) | B1
B1
[2] | 2.2a
2.4 | Demonstrated (condone one-side only)
(d) | (a + ib) (c + id) = 1 ac + i(b + ad) = 1
1 −b
c = and d =
a a
S is the set of complex numbers with non-zero real part
(The ‘exclusions’ are those complex numbers with zero real part;
i.e. the strictly imaginary numbers)) | M1
A1
B1
[3] | 3.1a
2.2a
3.2a | Test for an inverse of a general element
Both components of inverse element found
Explained6 The binary operation ◇ is defined on the set $\mathbb { C }$ of complex numbers by\\
$( a + i b ) \diamond ( c + i d ) = a c + i ( b + a d )$\\
where $a , b , c$ and $d$ are real numbers.
\begin{enumerate}[label=(\alph*)]
\item Is $\mathbb { C }$ closed under △ ? Justify your answer.
\item Prove that ◇ is associative on $\mathbb { C }$.
\item Determine the identity element of $\mathbb { C }$ under $\diamond$.
\item Determine the largest subset S of $\mathbb { C }$ such that $( \mathrm { S } , \diamond )$ is a group.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q6 [10]}}