| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring sophisticated use of Newton's identities and recurrence relations for power sums. Part (a) demands finding a transformed equation via substitution (y = x^4), part (b) requires using the recurrence relation S_n = -2S_{n-1} + S_{n-4}, and part (c) builds on previous results. While the techniques are standard for Further Maths, the multi-step nature, algebraic manipulation required, and need to track multiple power sums across three interconnected parts makes this substantially harder than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | y= x4 | B1 |
| y+2y 3 4 −1=016y3 =(1−y)4 | M1 | Substitutes and obtains an equation not |
| Answer | Marks | Guidance |
|---|---|---|
| 16y3 =1−4y+6y2 −4y3+ y4 | M1 | Uses binomial expansion. |
| y4 −20y3 +6y2 −4y+1=0 | A1 | Must be an equation. |
| 4 +4 +4 +4 =20 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(b) | +++=−2 | B1 |
| x5+2x4−x=05+5+5+5 =−2(20)+(−2) | M1 | Multiplies original equation by x and substitutes. |
| −42 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(c) | 8+8+8+8 =202−2(6) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 388 | A1 | CAO. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | y= x4 | B1 | Uses correct substitution.
y+2y 3 4 −1=016y3 =(1−y)4 | M1 | Substitutes and obtains an equation not
involving radicals.
16y3 =1−4y+6y2 −4y3+ y4 | M1 | Uses binomial expansion.
y4 −20y3 +6y2 −4y+1=0 | A1 | Must be an equation.
4 +4 +4 +4 =20 | B1
5
--- 3(b) ---
3(b) | +++=−2 | B1
x5+2x4−x=05+5+5+5 =−2(20)+(−2) | M1 | Multiplies original equation by x and substitutes.
−42 | A1
3
--- 3(c) ---
3(c) | 8+8+8+8 =202−2(6) | M1 | Uses formula for sum of squares. Or alternative
complete method eg another substitution.
388 | A1 | CAO.
2
Question | Answer | Marks | GuidanceThe quartic equation $x^4 + 2x^3 - 1 = 0$ has roots $\alpha, \beta, \gamma, \delta$.
\begin{enumerate}[label=(\alph*)]
\item Find a quartic equation whose roots are $\alpha^4, \beta^4, \gamma^4, \delta^4$ and state the value of $\alpha^4 + \beta^4 + \gamma^4 + \delta^4$. [5]
\item Find the value of $\alpha^5 + \beta^5 + \gamma^5 + \delta^5$. [3]
\item Find the value of $\alpha^8 + \beta^8 + \gamma^8 + \delta^8$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q3 [10]}}