| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas. While it involves multiple parts and algebraic manipulation, the techniques are routine for FP1 students: finding symmetric functions of roots, then using these to construct a new equation. The transformations (reciprocals of products of pairs) follow predictable patterns without requiring novel insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Working | Marks |
| \(\alpha+\beta+\gamma=7\), \(\alpha\beta+\beta\gamma+\gamma\alpha=2\), \(\alpha\beta\gamma=3\) | B1 | Stated or implied by working |
| (i) | \(\frac{1}{(\alpha\beta)(\beta\gamma)(\gamma\alpha)} = \frac{1}{(\alpha\beta\gamma)^2} = \frac{1}{9}\) | B1 |
| (ii) | \(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} = \frac{7}{3}\) | M1A1 |
| (iii) | \(\frac{1}{\alpha^2\beta\gamma} + \frac{1}{\alpha\beta^2\gamma} + \frac{1}{\alpha\beta\gamma^2} = \frac{1}{\alpha\beta\gamma}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right) = \frac{1}{\alpha\beta\gamma}\left(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\right) = \frac{2}{9}\) | M1A1 |
| \(\Rightarrow x^3 - \frac{7}{3}x^2 + \frac{2}{9}x - \frac{1}{9} = 0 \Rightarrow 9x^3 - 21x^2 + 2x - 1 = 0\) | M1A1\(\checkmark\) | (2) |
| Total: 8 |
## Question 4:
| Part | Working | Marks | Notes |
|------|---------|-------|-------|
| | $\alpha+\beta+\gamma=7$, $\alpha\beta+\beta\gamma+\gamma\alpha=2$, $\alpha\beta\gamma=3$ | B1 | Stated or implied by working |
| **(i)** | $\frac{1}{(\alpha\beta)(\beta\gamma)(\gamma\alpha)} = \frac{1}{(\alpha\beta\gamma)^2} = \frac{1}{9}$ | B1 | |
| **(ii)** | $\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} = \frac{7}{3}$ | M1A1 | (6) |
| **(iii)** | $\frac{1}{\alpha^2\beta\gamma} + \frac{1}{\alpha\beta^2\gamma} + \frac{1}{\alpha\beta\gamma^2} = \frac{1}{\alpha\beta\gamma}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right) = \frac{1}{\alpha\beta\gamma}\left(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\right) = \frac{2}{9}$ | M1A1 | |
| | $\Rightarrow x^3 - \frac{7}{3}x^2 + \frac{2}{9}x - \frac{1}{9} = 0 \Rightarrow 9x^3 - 21x^2 + 2x - 1 = 0$ | M1A1$\checkmark$ | (2) |
| | | **Total: 8** | |
---4 The roots of the cubic equation $x ^ { 3 } - 7 x ^ { 2 } + 2 x - 3 = 0$ are $\alpha , \beta$ and $\gamma$. Find the values of\\
(i) $\frac { 1 } { ( \alpha \beta ) ( \beta \gamma ) ( \gamma \alpha ) }$,\\
(ii) $\frac { 1 } { \alpha \beta } + \frac { 1 } { \beta \gamma } + \frac { 1 } { \gamma \alpha }$,\\
(iii) $\frac { 1 } { \alpha ^ { 2 } \beta \gamma } + \frac { 1 } { \alpha \beta ^ { 2 } \gamma } + \frac { 1 } { \alpha \beta \gamma ^ { 2 } }$.
Deduce a cubic equation, with integer coefficients, having roots $\frac { 1 } { \alpha \beta } , \frac { 1 } { \beta \gamma }$ and $\frac { 1 } { \gamma \alpha }$.
\hfill \mbox{\textit{CAIE FP1 2015 Q4 [8]}}