| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Standard +0.8 This is a Further Maths question requiring systematic application of Newton's identities and power sum formulas to find sums of powers of roots, plus a transformation to find a new equation with shifted roots. While the techniques are standard for Further Maths students, the multi-part structure, algebraic manipulation required, and the need to carefully track coefficients across three parts makes this moderately challenging. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_2 = S_1^2 - 2\Sigma\alpha\beta = 1^2 - 2\times(-1) = 3\) | (M1A1) | 2 marks |
| \(S_3 = S_2 + S_1 + 15 = 3 + 1 + 15 = 19\) | M1A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(\Sigma\alpha^3 = (\Sigma\alpha)(\Sigma\alpha\beta) - 3(\Sigma\alpha\beta) + 3\alpha\beta\gamma\) | (M1) | 1 mark |
| \(\Sigma\alpha = 1\), \(\Sigma\alpha\beta = -1\), \(\alpha\beta\gamma = 5\) | B1 | 1 mark |
| \(\Sigma\alpha^2 = 1 - 3\times1\times-1 + 3\times5\) | M1 | 1 mark |
| \(= 19\) | A1 | 1 mark |
| Available marks | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_3 = S_2 + 5S_1 = 19 + 3 + 5 = 27\) | M1A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = z + 1 \Rightarrow z = x - 1\) | B1 | 1 mark |
| Required eqn: \((x-1)^3 - (x-1)^2 - 5 = 0\) | M1 | 1 mark |
| So, \(x^3 - 4x^2 + 4x - 6 = 0\) | A1 | 1 mark |
| 3 |
## Question 4(a):
**EITHER Solution 1:**
| $S_2 = S_1^2 - 2\Sigma\alpha\beta = 1^2 - 2\times(-1) = 3$ | (M1A1) | 2 marks |
| $S_3 = S_2 + S_1 + 15 = 3 + 1 + 15 = 19$ | M1A1 | 2 marks |
**OR Solution 2:**
| Use $\Sigma\alpha^3 = (\Sigma\alpha)(\Sigma\alpha\beta) - 3(\Sigma\alpha\beta) + 3\alpha\beta\gamma$ | (M1) | 1 mark |
| $\Sigma\alpha = 1$, $\Sigma\alpha\beta = -1$, $\alpha\beta\gamma = 5$ | B1 | 1 mark |
| $\Sigma\alpha^2 = 1 - 3\times1\times-1 + 3\times5$ | M1 | 1 mark |
| $= 19$ | A1 | 1 mark |
| Available marks | | **4** |
## Question 4(b):
| $S_3 = S_2 + 5S_1 = 19 + 3 + 5 = 27$ | M1A1 | **2** |
## Question 4(c):
| $x = z + 1 \Rightarrow z = x - 1$ | B1 | 1 mark |
| Required eqn: $(x-1)^3 - (x-1)^2 - 5 = 0$ | M1 | 1 mark |
| So, $x^3 - 4x^2 + 4x - 6 = 0$ | A1 | 1 mark |
| | | **3** |
---4 The cubic equation
$$z ^ { 3 } - z ^ { 2 } - z - 5 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
(a) Show that the value of $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }$ is 19 .\\
(b) Find the value of $\alpha ^ { 4 } + \beta ^ { 4 } + \gamma ^ { 4 }$.\\
(c) Find a cubic equation with roots $\alpha + 1 , \beta + 1$ and $\gamma + 1$, giving your answer in the form
$$p x ^ { 3 } + q x ^ { 2 } + r x + s = 0 ,$$
where $p , q , r$ and $s$ are constants to be determined.\\
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q4 [9]}}