| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Product expressions with roots |
| Difficulty | Standard +0.8 This is a Further Maths question requiring manipulation of symmetric functions of roots beyond basic Vieta's formulas. Part (i) requires expanding the product and substituting Vieta's relations, while part (ii) needs recognition that the expression equals (sum of roots taken two at a time)² minus (product of roots)(sum of roots), or similar algebraic manipulation. These are non-routine applications requiring algebraic insight beyond standard A-level, though the techniques are systematic once learned. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha+\beta+\gamma = \frac{3}{2},\quad \alpha\beta+\beta\gamma+\gamma\alpha = 2,\quad \alpha\beta\gamma = 5+\beta+\gamma = \frac{3}{2}\alpha\beta+\beta\gamma+\gamma\alpha = 2\alpha\beta\gamma = 5\) | B1 | Can be awarded in (ii) if not seen here. SOI |
| \((\alpha+1)(\beta+1)(\gamma+1) = \alpha\beta\gamma + (\alpha\beta+\beta\gamma+\gamma\alpha) + (\alpha+\beta+\gamma) + 1\) | M1A1 | Multiply out and group for M1 |
| \(= 5 + 2 + 1\frac{1}{2} + 1 = 9\frac{1}{2}\) | A1FT | Alt: Let \(x = y-1\), sub and expand \(2y^3 - 9y^2 - 16y - 19 = 0\); product of roots \(= 19/2\) |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\beta+\gamma)(\gamma+\alpha)(\alpha+\beta) = \left(\frac{3}{2}-\alpha\right)\left(\frac{3}{2}-\beta\right)\left(\frac{3}{2}-\gamma\right)\) | M1 | Alt methods: \(= (\Sigma\alpha)(\Sigma\alpha\beta) - \alpha\beta\gamma\) or \(\Sigma\alpha^2\Sigma\alpha + 2\alpha\beta\gamma - \Sigma\alpha^3\) |
| \(= \frac{27}{8} - \frac{9}{4}(\alpha+\beta+\gamma) + \frac{3}{2}(\alpha\beta+\beta\gamma+\gamma\alpha) - \alpha\beta\gamma\) | A1 | |
| \(= \frac{27}{8} - \frac{9}{4}\times\frac{3}{2} + \frac{3}{2}\times 2 - 5 = -2\) | M1A1 | |
| Total | 4 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha+\beta+\gamma = \frac{3}{2},\quad \alpha\beta+\beta\gamma+\gamma\alpha = 2,\quad \alpha\beta\gamma = 5+\beta+\gamma = \frac{3}{2}\alpha\beta+\beta\gamma+\gamma\alpha = 2\alpha\beta\gamma = 5$ | B1 | Can be awarded in (ii) if not seen here. SOI |
| $(\alpha+1)(\beta+1)(\gamma+1) = \alpha\beta\gamma + (\alpha\beta+\beta\gamma+\gamma\alpha) + (\alpha+\beta+\gamma) + 1$ | M1A1 | Multiply out **and** group for M1 |
| $= 5 + 2 + 1\frac{1}{2} + 1 = 9\frac{1}{2}$ | A1FT | Alt: Let $x = y-1$, sub and expand $2y^3 - 9y^2 - 16y - 19 = 0$; product of roots $= 19/2$ |
| **Total** | **4** | |
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## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\beta+\gamma)(\gamma+\alpha)(\alpha+\beta) = \left(\frac{3}{2}-\alpha\right)\left(\frac{3}{2}-\beta\right)\left(\frac{3}{2}-\gamma\right)$ | M1 | Alt methods: $= (\Sigma\alpha)(\Sigma\alpha\beta) - \alpha\beta\gamma$ or $\Sigma\alpha^2\Sigma\alpha + 2\alpha\beta\gamma - \Sigma\alpha^3$ |
| $= \frac{27}{8} - \frac{9}{4}(\alpha+\beta+\gamma) + \frac{3}{2}(\alpha\beta+\beta\gamma+\gamma\alpha) - \alpha\beta\gamma$ | A1 | |
| $= \frac{27}{8} - \frac{9}{4}\times\frac{3}{2} + \frac{3}{2}\times 2 - 5 = -2$ | M1A1 | |
| **Total** | **4** | |4 The cubic equation $2 x ^ { 3 } - 3 x ^ { 2 } + 4 x - 10 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
(i) Find the value of $( \alpha + 1 ) ( \beta + 1 ) ( \gamma + 1 )$.\\
(ii) Find the value of $( \beta + \gamma ) ( \gamma + \alpha ) ( \alpha + \beta )$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q4 [8]}}