| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Challenging +1.2 This is a Further Maths question on polynomial roots with constraints. Part (a) requires deriving the discriminant condition for a repeated root using Vieta's formulas—a standard technique but requiring careful algebraic manipulation. Part (b) adds an extra constraint requiring simultaneous equations with the relationships from part (a). While systematic, it demands multiple steps and comfort with symmetric functions beyond typical A-level, placing it moderately above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\alpha + \gamma = -b\) and \(\alpha^2 + 2\alpha\gamma = 0\) | B1 | |
| \(\alpha = -2\gamma\) leading to \(-4\gamma + \gamma = -b\) | M1 | Solves simultaneous equations, or, express \(b\) and \(d\) in terms of \(\alpha\) and \(\gamma\) |
| \(\gamma = \frac{1}{3}b\), \(\alpha = -\frac{2}{3}b\) | A1 | \(b = 3\gamma = -\frac{3}{2}\alpha\) and \(d = -\alpha^2\gamma\) |
| \(\alpha^2\gamma = -d\) leading to \(\frac{4}{27}b^3 = -d\) leading to \(4b^3 + 27d = 0\) | M1 A1 | Substitutes into third equation, AG |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3b = b^2\) leading to \(b = 3\) | M1 A1 | Uses \(2\alpha^2 + \gamma^2 = (2\alpha + \gamma)^2 - 2(\alpha^2 + 2\alpha\gamma)\) or substitutes for \(\alpha, \gamma\) in terms of \(b\) |
| \(d = -4\) | A1 | |
| 3 |
## Question 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\alpha + \gamma = -b$ and $\alpha^2 + 2\alpha\gamma = 0$ | **B1** | |
| $\alpha = -2\gamma$ leading to $-4\gamma + \gamma = -b$ | **M1** | Solves simultaneous equations, or, express $b$ and $d$ in terms of $\alpha$ and $\gamma$ |
| $\gamma = \frac{1}{3}b$, $\alpha = -\frac{2}{3}b$ | **A1** | $b = 3\gamma = -\frac{3}{2}\alpha$ and $d = -\alpha^2\gamma$ |
| $\alpha^2\gamma = -d$ leading to $\frac{4}{27}b^3 = -d$ leading to $4b^3 + 27d = 0$ | **M1 A1** | Substitutes into third equation, AG |
| | **5** | |
## Question 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3b = b^2$ leading to $b = 3$ | **M1 A1** | Uses $2\alpha^2 + \gamma^2 = (2\alpha + \gamma)^2 - 2(\alpha^2 + 2\alpha\gamma)$ or substitutes for $\alpha, \gamma$ in terms of $b$ |
| $d = -4$ | **A1** | |
| | **3** | |1 The cubic equation $x ^ { 3 } + b x ^ { 2 } + d = 0$ has roots $\alpha , \beta , \gamma$, where $\alpha = \beta$ and $d \neq 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $4 b ^ { 3 } + 27 d = 0$.
\item Given that $2 \alpha ^ { 2 } + \gamma ^ { 2 } = 3 b$, find the values of $b$ and $d$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q1 [8]}}