| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Quadratic from one complex root |
| Difficulty | Easy -1.2 This is a straightforward application of the complex conjugate root theorem for polynomials with real coefficients, followed by routine use of sum and product of roots. It requires only direct recall of standard theory with minimal calculation—significantly easier than average A-level questions which typically involve multi-step problem-solving. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 2 + 3i |
| [1] | 1.2 | |
| 1 | (b) | (x − 2 − 3i)(x − 2 + 3i) = 0 |
| Answer | Marks |
|---|---|
| x2 − 4x + 13 = 0 [so a = −4 and b = 13] | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | factors x − (2 + 3i) and x − (2 − 3i) |
| Answer | Marks | Guidance |
|---|---|---|
| (2 – 3i)2 + a(2 – 3i) + b = 0 | M1 | use of factor theorem |
| −5 + 2a + b = 0, –12 – 3a = 0 | A1 | |
| a = −4, b = 13 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | M1 |
| a2 − 4b = −36, −a/2 = 2 | A1 | |
| a = −4, b = 13 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| a = −( + ) | M1 | sum or product of roots formulae used, condone sign errors |
| = −4 | A1 | |
| b = = (2 + 3i)(2 − 3i) = 13 | A1 |
Question 1:
1 | (a) | 2 + 3i | B1
[1] | 1.2
1 | (b) | (x − 2 − 3i)(x − 2 + 3i) = 0
(x − 2)2 + 9 = 0
x2 − 4x + 13 = 0 [so a = −4 and b = 13] | M1
A1
A1 | 1.1
1.1
1.1 | factors x − (2 + 3i) and x − (2 − 3i)
or x2 − 2x +3ix − 2x + 4 − 6i − 3ix + 6i − 9i2 or better
Alternative solution 1
(2 – 3i)2 + a(2 – 3i) + b = 0 | M1 | use of factor theorem
−5 + 2a + b = 0, –12 – 3a = 0 | A1
a = −4, b = 13 | A1
Alternative solution 2
− a a 2 − 4 b
= 2 3 i
2 | M1 | M1 | use of quadratic formula | use of quadratic formula
a2 − 4b = −36, −a/2 = 2 | A1
a = −4, b = 13 | A1
Alternative solution 3
a = −( + ) | M1 | sum or product of roots formulae used, condone sign errors | sum or product of roots formulae used, condone sign errors
= −4 | A1
b = = (2 + 3i)(2 − 3i) = 13 | A1
[3]1 The quadratic equation $\mathrm { x } ^ { 2 } + \mathrm { ax } + \mathrm { b } = 0$, where $a$ and $b$ are real constants, has a root 2-3.
\begin{enumerate}[label=(\alph*)]
\item Write down the other root.
\item Hence or otherwise determine the values of $a$ and $b$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2024 Q1 [4]}}