| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Standard quadratic with real coefficients |
| Difficulty | Moderate -0.8 Part (a) is a straightforward application of the quadratic formula to find complex roots, requiring only substitution and simplification of √(-100). Part (b) is basic algebraic manipulation to isolate ω. Both are routine exercises testing fundamental complex number operations with no problem-solving insight required, making this easier than average even for Further Maths. |
| Spec | 4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4\times2\times25}}{2\times2}\) | M1 | 2.1 — Correct substitution into formula. If formula quoted allow one slip. Or completing the square — one slip allowed. NB: This question required detailed reasoning |
| \(z = \dfrac{5}{2} \pm \dfrac{5}{2}i\) | A1 | 1.1 — Allow \(z = \dfrac{5\pm5i}{2}\) or equivalent fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3\omega - 2 = 5i + 2i\omega \Rightarrow 3\omega - 2i\omega = 2 + 5i\) | M1 | 1.1 — Expanding and rearranging. Must rearrange to isolate \(\omega\) terms on one side. NB: This question required detailed reasoning |
| \((3-2i)\omega = 2+5i \Rightarrow \omega = \dfrac{2+5i}{3-2i}\) | M1 | 1.1 — Factorising and dividing by two term complex number |
| \(\omega = \dfrac{2+5i}{3-2i} \times \dfrac{3+2i}{3+2i} = \dfrac{6+4i+15i-10}{9+4}\) | M1 | 2.1 — Multiplying top and bottom by conjugate of bottom |
| \(\omega = -\dfrac{4}{13} + \dfrac{19}{13}i\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\omega = a+bi \Rightarrow 3a+3bi-2 = 5i+2ai-2b\) | M1 | Assigning real and imaginary parts to \(\omega\), expanding and rearranging |
| \(3a-2=-2b\) and \(3b=5+2a\) | M1 | Comparing real and imaginary parts |
| \(9a-6+10+4a=0 \Rightarrow a = -\dfrac{4}{13}\) | M1 | Using valid algebra to eliminate one unknown and finding the other |
| \(\Rightarrow b = \dfrac{19}{13} \Rightarrow \omega = -\dfrac{4}{13}+\dfrac{19}{13}i\) | A1 |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4\times2\times25}}{2\times2}$ | M1 | 2.1 — Correct substitution into formula. If formula quoted allow one slip. Or completing the square — one slip allowed. **NB: This question required detailed reasoning** |
| $z = \dfrac{5}{2} \pm \dfrac{5}{2}i$ | A1 | 1.1 — Allow $z = \dfrac{5\pm5i}{2}$ or equivalent fractions |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\omega - 2 = 5i + 2i\omega \Rightarrow 3\omega - 2i\omega = 2 + 5i$ | M1 | 1.1 — Expanding and rearranging. Must rearrange to isolate $\omega$ terms on one side. **NB: This question required detailed reasoning** |
| $(3-2i)\omega = 2+5i \Rightarrow \omega = \dfrac{2+5i}{3-2i}$ | M1 | 1.1 — Factorising and dividing by two term complex number |
| $\omega = \dfrac{2+5i}{3-2i} \times \dfrac{3+2i}{3+2i} = \dfrac{6+4i+15i-10}{9+4}$ | M1 | 2.1 — Multiplying top and bottom by conjugate of bottom |
| $\omega = -\dfrac{4}{13} + \dfrac{19}{13}i$ | A1 | 1.1 |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega = a+bi \Rightarrow 3a+3bi-2 = 5i+2ai-2b$ | M1 | Assigning real and imaginary parts to $\omega$, expanding and rearranging |
| $3a-2=-2b$ and $3b=5+2a$ | M1 | Comparing real and imaginary parts |
| $9a-6+10+4a=0 \Rightarrow a = -\dfrac{4}{13}$ | M1 | Using valid algebra to eliminate one unknown and finding the other |
| $\Rightarrow b = \dfrac{19}{13} \Rightarrow \omega = -\dfrac{4}{13}+\dfrac{19}{13}i$ | A1 | |
---6 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $2 z ^ { 2 } - 10 z + 25 = 0$ giving your answers in the form $\mathrm { a } + \mathrm { bi }$.
\item Solve the equation $3 \omega - 2 = \mathrm { i } ( 5 + 2 \omega )$ giving your answer in the form $\mathrm { a } + \mathrm { bi }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2021 Q6 [6]}}