| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Complex roots with real coefficients |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring systematic application of complex conjugate root theorem, Vieta's formulas, and polynomial relationships. While the techniques are standard for FP1 (complex roots, sum/product of roots), the question requires careful coordination across four parts with multiple algebraic steps, making it moderately challenging but still within typical Further Maths scope. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Sum of all roots \(= \frac{9}{2}\); complex roots contribute \(6\), so \(\alpha+\beta = \frac{9}{2}-6 = -\frac{3}{2}\) | M1 A1 | |
| \( | \gamma | ^2 = 13\), product of complex pair \(= 13\); \(\alpha\beta = \frac{-26}{2\cdot13} = -1\) |
| Answer | Marks |
|---|---|
| \(\alpha,\beta\) satisfy \(t^2+\frac{3}{2}t-1=0 \Rightarrow 2t^2+3t-2=0\) | M1 |
| \(\alpha = \frac{1}{2}, \beta = -2\) | A1 A1 |
| Answer | Marks |
|---|---|
| \(f(z) = 2(z^2+\frac{3}{2}z-1)(z^2-6z+13)\) | M1 |
| \(A = -6, B = -10\)... expanding gives \(A\) and \(B\) | A1 A1 |
| Answer | Marks |
|---|---|
| Replace \(z\) with \(\frac{w}{\text{j}}\): roots are \(\frac{1}{2}\text{j}, -2\text{j}, (3+2\text{j})\text{j}, (3-2\text{j})\text{j}\) i.e. \(\frac{\text{j}}{2}, -2\text{j}, -2+3\text{j}, -2-3\text{j}\) | B1 B1 |
## Question 7:
**(i)**
Sum of all roots $= \frac{9}{2}$; complex roots contribute $6$, so $\alpha+\beta = \frac{9}{2}-6 = -\frac{3}{2}$ | M1 A1 |
$|\gamma|^2 = 13$, product of complex pair $= 13$; $\alpha\beta = \frac{-26}{2\cdot13} = -1$ | M1 A1 A1 |
**(ii)**
$\alpha,\beta$ satisfy $t^2+\frac{3}{2}t-1=0 \Rightarrow 2t^2+3t-2=0$ | M1 |
$\alpha = \frac{1}{2}, \beta = -2$ | A1 A1 |
**(iii)**
$f(z) = 2(z^2+\frac{3}{2}z-1)(z^2-6z+13)$ | M1 |
$A = -6, B = -10$... expanding gives $A$ and $B$ | A1 A1 |
**(iv)**
Replace $z$ with $\frac{w}{\text{j}}$: roots are $\frac{1}{2}\text{j}, -2\text{j}, (3+2\text{j})\text{j}, (3-2\text{j})\text{j}$ i.e. $\frac{\text{j}}{2}, -2\text{j}, -2+3\text{j}, -2-3\text{j}$ | B1 B1 |
---7 The function $\mathrm { f } ( z ) = 2 z ^ { 4 } - 9 z ^ { 3 } + A z ^ { 2 } + B z - 26$ has real coefficients. The equation $\mathrm { f } ( z ) = 0$ has two real roots, $\alpha$ and $\beta$, where $\alpha > \beta$, and two complex roots, $\gamma$ and $\delta$, where $\gamma = 3 + 2 \mathrm { j }$.\\
(i) Show that $\alpha + \beta = - \frac { 3 } { 2 }$ and find the value of $\alpha \beta$.\\
(ii) Hence find the two real roots $\alpha$ and $\beta$.\\
(iii) Find the values of $A$ and $B$.\\
(iv) Write down the roots of the equation $\mathrm { f } \left( \frac { w } { \mathrm { j } } \right) = 0$.
\hfill \mbox{\textit{OCR MEI FP1 2016 Q7 [13]}}