| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.3 This is a Further Maths question on symmetric functions of transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (i) involves a non-trivial substitution to derive a new polynomial, parts (ii-iii) require computing symmetric functions of the transformed roots. While methodical, it demands careful bookkeeping and multiple steps beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | −7y ( −7y )+2 ( −7y )+ −7y +7=0 | |
| ⇒ −7y ( −7y+1 )=14y−7⇒ −7y ( −7y+1 )2 =( 14y−7 )2 | M1 | Uses given substitution and eliminates radical. |
| ⇒49y3+14y2 −27y+7=0 | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| −7 αβγ | M1 | αβγ=−7. |
| Answer | Marks | Guidance |
|---|---|---|
| αβγ βγ αβγ αγ αβγ αβ | A1 | AG |
| Answer | Marks |
|---|---|
| 7(ii) | α β γ 2 1 1 1 27 |
| Answer | Marks | Guidance |
|---|---|---|
| βγ αγ αβ 7 γ2 β2 α2 49 | B1 | α'β'+α'γ'+β'γ'. |
| Answer | Marks | Guidance |
|---|---|---|
| β2γ2 α2γ2 α2β2 7 49 49 | M1 A1 | Uses |
| Answer | Marks |
|---|---|
| 7(iii) | α3 β3 γ3 58 2 |
| Answer | Marks | Guidance |
|---|---|---|
| β3γ3 α3γ3 α3β3 49 7 | M1 | Uses 49α' 3 =−14α′ 2 +27α′−7. |
| Answer | Marks |
|---|---|
| β3γ3 α3γ3 α3β3 343 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(i) ---
7(i) | −7y ( −7y )+2 ( −7y )+ −7y +7=0
⇒ −7y ( −7y+1 )=14y−7⇒ −7y ( −7y+1 )2 =( 14y−7 )2 | M1 | Uses given substitution and eliminates radical.
⇒49y3+14y2 −27y+7=0 | A1 | AG
x2 x2
y = =
−7 αβγ | M1 | αβγ=−7.
Uses
α2 α β2 β γ2 γ
So roots are = , = , =
αβγ βγ αβγ αγ αβγ αβ | A1 | AG
4
--- 7(ii) ---
7(ii) | α β γ 2 1 1 1 27
+ + =− , + + =−
βγ αγ αβ 7 γ2 β2 α2 49 | B1 | α'β'+α'γ'+β'γ'.
States sum of roots and
α2 β2 γ2 2 2 27 58
+ + = − −2 − =
β2γ2 α2γ2 α2β2 7 49 49 | M1 A1 | Uses
α' 2 +β' 2 +γ' 2 =( α'+β'+γ' )2 −2 ( α'β'+α'γ'+β'γ' )
AG
3
--- 7(iii) ---
7(iii) | α3 β3 γ3 58 2
49 + + =−14 +27 − −21
β3γ3 α3γ3 α3β3 49 7 | M1 | Uses 49α' 3 =−14α′ 2 +27α′−7.
α3 β3 γ3 317
⇒ + + =−
β3γ3 α3γ3 α3β3 343 | A1
2
Question | Answer | Marks | GuidanceThe equation $x^3 + 2x^2 + x + 7 = 0$ has roots $\alpha$, $\beta$, $\gamma$.
\begin{enumerate}[label=(\roman*)]
\item Use the relation $x^2 = -7y$ to show that the equation
$$49y^3 + 14y^2 - 27y + 7 = 0$$
has roots $\frac{\alpha}{\beta \gamma}$, $\frac{\beta}{\gamma \alpha}$, $\frac{\gamma}{\alpha \beta}$. [4]
\item Show that $\frac{\alpha^2}{\beta^2 \gamma^2} + \frac{\beta^2}{\gamma^2 \alpha^2} + \frac{\gamma^2}{\alpha^2 \beta^2} = \frac{58}{49}$. [3]
\item Find the exact value of $\frac{\alpha^2}{\beta^3 \gamma^3} + \frac{\beta^2}{\gamma^3 \alpha^3} + \frac{\gamma^2}{\alpha^3 \beta^3}$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2019 Q7 [9]}}