| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| 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 Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) involves the standard technique of substituting x² into the original equation and eliminating x, while parts (b)(i-ii) use relationships between symmetric functions. Though multi-step, these are well-practiced techniques in Further Maths with clear pathways, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = x^2\) | B1 | |
| \(6y^{\frac{3}{2}} + py - 3y^{\frac{1}{2}} - 5 = 0 \Rightarrow y^{\frac{1}{2}}(6y-3) = -py+5\); \(y(6y-3)^2 = (-py+5)^2 \Rightarrow y(36y^2 - 36y + 9) = p^2y^2 - 10py + 25\) | M1 | Method of eliminating \(y^{\frac{1}{2}}\) |
| \(36y^3 - (p^2+36)y^2 + (10p+9)y - 25 = 0\) | A1 | Correct cubic |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha^2 + \beta^2 + \gamma^2 = \frac{p^2+36}{36}\) | B1 | |
| \(\frac{p^2+36}{36} = -\frac{2p}{6} \Rightarrow p^2 + 12p + 36 = 0\) | M1 | Setting up equation |
| \(p = -6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6(\alpha^3 + \beta^3 + \gamma^3) = 6(\alpha^2+\beta^2+\gamma^2) + 3(\alpha+\beta+\gamma) + 15\) | M1 | Using Newton's identity |
| \(\alpha^3 + \beta^3 + \gamma^3 = 5\) | A1 |
## Question 2:
### Part 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = x^2$ | B1 | |
| $6y^{\frac{3}{2}} + py - 3y^{\frac{1}{2}} - 5 = 0 \Rightarrow y^{\frac{1}{2}}(6y-3) = -py+5$; $y(6y-3)^2 = (-py+5)^2 \Rightarrow y(36y^2 - 36y + 9) = p^2y^2 - 10py + 25$ | M1 | Method of eliminating $y^{\frac{1}{2}}$ |
| $36y^3 - (p^2+36)y^2 + (10p+9)y - 25 = 0$ | A1 | Correct cubic |
### Part 2(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha^2 + \beta^2 + \gamma^2 = \frac{p^2+36}{36}$ | B1 | |
| $\frac{p^2+36}{36} = -\frac{2p}{6} \Rightarrow p^2 + 12p + 36 = 0$ | M1 | Setting up equation |
| $p = -6$ | A1 | |
### Part 2(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6(\alpha^3 + \beta^3 + \gamma^3) = 6(\alpha^2+\beta^2+\gamma^2) + 3(\alpha+\beta+\gamma) + 15$ | M1 | Using Newton's identity |
| $\alpha^3 + \beta^3 + \gamma^3 = 5$ | A1 | |
---2 The cubic equation $6 \mathrm { x } ^ { 3 } + \mathrm { px } ^ { 2 } - 3 \mathrm { x } - 5 = 0$, where $p$ is a constant, has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation whose roots are $\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }$.
\item It is given that $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = 2 ( \alpha + \beta + \gamma )$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $p$.
\item Find the value of $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q2 [8]}}