| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| 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 known techniques: substituting y=x³ to find the new equation, using Vieta's formulas, and applying the recurrence relation from the original equation. While it involves multiple steps and careful algebraic manipulation, the methods are well-established textbook procedures without requiring novel insight or particularly complex problem-solving. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^3\) | B1 | Correct substitution |
| \(y^{\frac{4}{3}} - 2y - 1 = 0 \Rightarrow y^4 = (2y+1)^3 = 8y^3+12y^2+6y+1\) | M1 | Obtains an equation not involving radicals |
| \(y^4 - 8y^3 - 12y^2 - 6y - 1 = 0\) | A1 | |
| \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = 8\) | B1 FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{\alpha^3}+\frac{1}{\beta^3}+\frac{1}{\gamma^3}+\frac{1}{\delta^3} = \frac{\alpha^3\beta^3\delta^3+\alpha^3\beta^3\gamma^3+\beta^3\gamma^3\delta^3+\alpha^3\gamma^3\delta^3}{\alpha^3\beta^3\gamma^3\delta^3} = \frac{6}{-1}\) | M1 A1 FT | Relates \(\frac{1}{\alpha^3}+\frac{1}{\beta^3}+\frac{1}{\gamma^3}+\frac{1}{\delta^3}\) to coefficients |
| \(-6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha^4+\beta^4+\gamma^4+\delta^4 = 2(\alpha^3+\beta^3+\gamma^3+\delta^3)+4\) | M1 | Uses original equation |
| \(= 20\) | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^3$ | B1 | Correct substitution |
| $y^{\frac{4}{3}} - 2y - 1 = 0 \Rightarrow y^4 = (2y+1)^3 = 8y^3+12y^2+6y+1$ | M1 | Obtains an equation not involving radicals |
| $y^4 - 8y^3 - 12y^2 - 6y - 1 = 0$ | A1 | |
| $\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = 8$ | B1 FT | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\alpha^3}+\frac{1}{\beta^3}+\frac{1}{\gamma^3}+\frac{1}{\delta^3} = \frac{\alpha^3\beta^3\delta^3+\alpha^3\beta^3\gamma^3+\beta^3\gamma^3\delta^3+\alpha^3\gamma^3\delta^3}{\alpha^3\beta^3\gamma^3\delta^3} = \frac{6}{-1}$ | M1 A1 FT | Relates $\frac{1}{\alpha^3}+\frac{1}{\beta^3}+\frac{1}{\gamma^3}+\frac{1}{\delta^3}$ to coefficients |
| $-6$ | A1 | |
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^4+\beta^4+\gamma^4+\delta^4 = 2(\alpha^3+\beta^3+\gamma^3+\delta^3)+4$ | M1 | Uses original equation |
| $= 20$ | A1 | |3 The equation $x ^ { 4 } - 2 x ^ { 3 } - 1 = 0$ has roots $\alpha , \beta , \gamma , \delta$.
\begin{enumerate}[label=(\alph*)]
\item Find a quartic equation whose roots are $\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 } , \delta ^ { 3 }$ and state the value of $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } + \delta ^ { 3 }$. [4]
\item Find the value of $\frac { 1 } { \alpha ^ { 3 } } + \frac { 1 } { \beta ^ { 3 } } + \frac { 1 } { \gamma ^ { 3 } } + \frac { 1 } { \delta ^ { 3 } }$.
\item Find the value of $\alpha ^ { 4 } + \beta ^ { 4 } + \gamma ^ { 4 } + \delta ^ { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q3 [9]}}