| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring knowledge of Vieta's formulas, polynomial transformation techniques, and algebraic manipulation with symmetric functions. Part (a) is routine, part (b) requires systematic substitution to find the transformed equation, and part (c) involves non-trivial deduction using the given constraint—requiring several steps of algebraic reasoning beyond standard exercises. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d = 1\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x + 1 \Rightarrow x = y - 1\) | B1 | Uses correct substitution |
| \(y^3 + (b-3)y^2 + (c-2b+3)y + b - c = 0\) | M1 A1 | Substitutes and expands |
| Alternative method: | ||
| \((\alpha+1)(\beta+1) + (\alpha+1)(\gamma+1) + (\beta+1)(\gamma+1) = c - 2b + 3\) | B1 | |
| \((\alpha+1+\beta+1+\gamma+1) = 3-b\), \(\ (\alpha+1)(\beta+1)(\gamma+1) = c-b\) | M1 | Using these relationships |
| \(y^3 + (b-3)y^2 + (c-2b+3)y + b - c = 0\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\beta + 1 = -(b-3)\) | B1 | Uses sum of roots |
| \(-(\alpha+1)(\alpha+1) = c - 2b + 3\) | B1 | Uses sum of products in pairs |
| \(-(\alpha+1)(\beta+1)(\alpha+1) = -(b-c)\) | M1 | Applies product of roots |
| \(\Rightarrow (c-2b+3)(b-3) = b-c\) | A1 | AG |
| Total: 4 |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d = 1$ | **B1** | |
| | **Total: 1** | |
---
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x + 1 \Rightarrow x = y - 1$ | **B1** | Uses correct substitution |
| $y^3 + (b-3)y^2 + (c-2b+3)y + b - c = 0$ | **M1 A1** | Substitutes and expands |
| **Alternative method:** | | |
| $(\alpha+1)(\beta+1) + (\alpha+1)(\gamma+1) + (\beta+1)(\gamma+1) = c - 2b + 3$ | **B1** | |
| $(\alpha+1+\beta+1+\gamma+1) = 3-b$, $\ (\alpha+1)(\beta+1)(\gamma+1) = c-b$ | **M1** | Using these relationships |
| $y^3 + (b-3)y^2 + (c-2b+3)y + b - c = 0$ | **A1** | |
| | **Total: 3** | |
---
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\beta + 1 = -(b-3)$ | **B1** | Uses sum of roots |
| $-(\alpha+1)(\alpha+1) = c - 2b + 3$ | **B1** | Uses sum of products in pairs |
| $-(\alpha+1)(\beta+1)(\alpha+1) = -(b-c)$ | **M1** | Applies product of roots |
| $\Rightarrow (c-2b+3)(b-3) = b-c$ | **A1** | AG |
| | **Total: 4** | |1 The cubic equation $\mathrm { x } ^ { 3 } + \mathrm { bx } ^ { 2 } + \mathrm { cx } + \mathrm { d } = 0$, where $b , c$ and $d$ are constants, has roots $\alpha , \beta , \gamma$. It is given that $\alpha \beta \gamma = - 1$.
\begin{enumerate}[label=(\alph*)]
\item State the value of $d$.
\item Find a cubic equation, with coefficients in terms of $b$ and $c$, whose roots are $\alpha + 1 , \beta + 1 , \gamma + 1$.
\item Given also that $\gamma + 1 = - \alpha - 1$, deduce that $( \mathrm { c } - 2 \mathrm {~b} + 3 ) ( \mathrm { b } - 3 ) = \mathrm { b } - \mathrm { c }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q1 [8]}}