| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| 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 Newton's identities and Vieta's formulas. Part (a) uses the substitution y=x² method, parts (b-c) involve routine manipulation of symmetric functions. While it requires multiple steps and careful algebra, the techniques are well-practiced in Further Pure syllabi with no novel insight needed. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = x^2\) so \(x = y^{\frac{1}{2}}\) | B1 | Correct substitution |
| \(y^2 - y + 2y^{\frac{1}{2}} + 5 = 0\) so \((2y^{\frac{1}{2}})^2 = (-y^2+y-5)^2\) | M1 | Obtains an equation which eliminates radicals |
| \(y^4 - 2y^3 + 11y^2 - 14y + 25 = 0\) | A1 | |
| \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 2\) | B1FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha^2\beta^2\gamma^2\delta^2 = 25\) | B1FT | |
| \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}+\frac{1}{\delta^2} = \frac{\alpha^2\beta^2\delta^2+\alpha^2\beta^2\gamma^2+\beta^2\gamma^2\delta^2+\alpha^2\gamma^2\delta^2}{\alpha^2\beta^2\gamma^2\delta^2}\) | M1 | Relates to coefficients |
| \(\frac{14}{25}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha^4+\beta^4+\gamma^4+\delta^4 = (\alpha^2+\beta^2+\gamma^2+\delta^2)^2 - 2(\alpha^2\beta^2+\alpha^2\gamma^2+\alpha^2\delta^2+\beta^2\gamma^2+\beta^2\delta^2+\gamma^2\delta^2) = 2^2 - 2(11)\) | M1 | Uses formula for sum of squares or uses original equation |
| \(-18\) | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = x^2$ so $x = y^{\frac{1}{2}}$ | B1 | Correct substitution |
| $y^2 - y + 2y^{\frac{1}{2}} + 5 = 0$ so $(2y^{\frac{1}{2}})^2 = (-y^2+y-5)^2$ | M1 | Obtains an equation which eliminates radicals |
| $y^4 - 2y^3 + 11y^2 - 14y + 25 = 0$ | A1 | |
| $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 2$ | B1FT | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha^2\beta^2\gamma^2\delta^2 = 25$ | B1FT | |
| $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}+\frac{1}{\delta^2} = \frac{\alpha^2\beta^2\delta^2+\alpha^2\beta^2\gamma^2+\beta^2\gamma^2\delta^2+\alpha^2\gamma^2\delta^2}{\alpha^2\beta^2\gamma^2\delta^2}$ | M1 | Relates to coefficients |
| $\frac{14}{25}$ | A1 | CAO |
## Question 3(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha^4+\beta^4+\gamma^4+\delta^4 = (\alpha^2+\beta^2+\gamma^2+\delta^2)^2 - 2(\alpha^2\beta^2+\alpha^2\gamma^2+\alpha^2\delta^2+\beta^2\gamma^2+\beta^2\delta^2+\gamma^2\delta^2) = 2^2 - 2(11)$ | M1 | Uses formula for sum of squares or uses original equation |
| $-18$ | A1 | |3 The equation $x ^ { 4 } - x ^ { 2 } + 2 x + 5 = 0$ has roots $\alpha , \beta , \gamma , \delta$.
\begin{enumerate}[label=(\alph*)]
\item Find a quartic equation whose roots are $\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 } , \delta ^ { 2 }$ and state the value of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 }$.
\item Find the value of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } } + \frac { 1 } { \delta ^ { 2 } }$.
\item Find the value of $\alpha ^ { 4 } + \beta ^ { 4 } + \gamma ^ { 4 } + \delta ^ { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q3 [9]}}