| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and Newton's identities. While it involves multiple parts and careful algebraic manipulation, the techniques are well-established and follow predictable patterns taught in Further Pure syllabi. The substitution y = x² leads to a straightforward method for part (a), and parts (b) and (c) use standard power sum formulas, making this 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 | Marks | Guidance |
| \(y = x^2 + 1\) | B1 | Uses correct substitution. May use a different letter. |
| \((y-1)^{\frac{3}{2}} + 2(y-1) + 3(y-1)^{\frac{1}{2}} + 1 = 0 \Rightarrow (y-1)^{\frac{1}{2}}((y-1)+3) = -2(y-1)-1\) \((y-1)(y+2)^2 = (-2y+1)^2 \Rightarrow (y-1)(y^2+4y+4) = 4y^2-4y+1\) OR \((x^3+3x)^2 = (-2x^2-1)^2 \Rightarrow x^6+2x^4+5x^2-1=0\) \((y-1)^3+2(y-1)^2+5(y-1)-1=0\) | M1 | Substitutes and obtains an equation not involving radicals. |
| \(y^3 - y^2 + 4y - 5 = 0\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\alpha^2+1)^2 + (\beta^2+1)^2 + (\gamma^2+1)^2 = 1 - 2(4)\) | M1 | Uses \(\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\), on *their* answer for part 2(a). |
| \(-7\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\alpha^2+1)^3 + (\beta^2+1)^3 + (\gamma^2+1)^3 = -7 - 4 + 5(3)\) | M1 | Uses *their* equation from part 2(a). |
| \(4\) | A1 | CAO |
| Alternative: \(\sum(\alpha^2+1)^3 = \left(\sum(\alpha^2+1)\right)^3 - 3\left(\sum(\alpha^2+1)\right)\left(\sum(\alpha^2+1)(\beta^2+1)\right) + 3\left((\alpha^2+1)(\beta^2+1)(\gamma^2+1)\right)\) \(= 1^3 - 3\times1\times4 + 3\times5\) | M1 | Uses *their* equation from part 2(a) in a valid formula. |
| \(4\) | A1 | CAO |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^2 + 1$ | **B1** | Uses correct substitution. May use a different letter. |
| $(y-1)^{\frac{3}{2}} + 2(y-1) + 3(y-1)^{\frac{1}{2}} + 1 = 0 \Rightarrow (y-1)^{\frac{1}{2}}((y-1)+3) = -2(y-1)-1$ $(y-1)(y+2)^2 = (-2y+1)^2 \Rightarrow (y-1)(y^2+4y+4) = 4y^2-4y+1$ OR $(x^3+3x)^2 = (-2x^2-1)^2 \Rightarrow x^6+2x^4+5x^2-1=0$ $(y-1)^3+2(y-1)^2+5(y-1)-1=0$ | **M1** | Substitutes and obtains an equation not involving radicals. |
| $y^3 - y^2 + 4y - 5 = 0$ | **A1** | OE |
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\alpha^2+1)^2 + (\beta^2+1)^2 + (\gamma^2+1)^2 = 1 - 2(4)$ | **M1** | Uses $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$, on *their* answer for part 2(a). |
| $-7$ | **A1** | CAO |
---
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\alpha^2+1)^3 + (\beta^2+1)^3 + (\gamma^2+1)^3 = -7 - 4 + 5(3)$ | **M1** | Uses *their* equation from part 2(a). |
| $4$ | **A1** | CAO |
| **Alternative:** $\sum(\alpha^2+1)^3 = \left(\sum(\alpha^2+1)\right)^3 - 3\left(\sum(\alpha^2+1)\right)\left(\sum(\alpha^2+1)(\beta^2+1)\right) + 3\left((\alpha^2+1)(\beta^2+1)(\gamma^2+1)\right)$ $= 1^3 - 3\times1\times4 + 3\times5$ | **M1** | Uses *their* equation from part 2(a) in a valid formula. |
| $4$ | **A1** | CAO |
---2 The cubic equation $x ^ { 3 } + 2 x ^ { 2 } + 3 x + 1 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation whose roots are $\alpha ^ { 2 } + 1 , \beta ^ { 2 } + 1 , \gamma ^ { 2 } + 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-04_2714_34_143_2012}
\item Find the value of $\left( \alpha ^ { 2 } + 1 \right) ^ { 2 } + \left( \beta ^ { 2 } + 1 \right) ^ { 2 } + \left( \gamma ^ { 2 } + 1 \right) ^ { 2 }$.
\item Find the value of $\left( \alpha ^ { 2 } + 1 \right) ^ { 3 } + \left( \beta ^ { 2 } + 1 \right) ^ { 3 } + \left( \gamma ^ { 2 } + 1 \right) ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q2 [7]}}