CAIE FP1 2019 November — Question 7 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSum of powers of roots
DifficultyChallenging +1.8 This is a sophisticated Further Maths question requiring deep understanding of symmetric functions and polynomial transformations. Part (i) demands recognizing how the substitution relates roots through Vieta's formulas, part (ii) requires manipulating symmetric functions of transformed roots, and part (iii) builds on previous results. While systematic, it requires significant algebraic maturity and non-routine problem-solving beyond standard A-level techniques.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
The equation \(x^3 + 2x^2 + x + 7 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\).
  1. 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]
  2. 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]
  3. Find the exact value of \(\frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\gamma^3\alpha^3} + \frac{\gamma^3}{\alpha^3\beta^3}\). [2]
Question 7:
AnswerMarks Guidance
7(i)−7y ( −7y )+2 ( −7y )+ −7y +7=0
⇒ −7y ( −7y+1 )=14y−7⇒ −7y ( −7y+1 )2 =( 14y−7 )2M1 Uses given substitution and eliminates radical.
⇒49y3+14y2 −27y+7=0A1 AG
x2 x2
y = =
AnswerMarks Guidance
−7 αβγM1 αβγ=−7.
Uses
α2 α β2 β γ2 γ
So roots are = , = , =
AnswerMarks Guidance
αβγ βγ αβγ αγ αβγ αβA1 AG
4
AnswerMarks
7(ii)α β γ 2 1 1 1 27
+ + =− , + + =−
AnswerMarks Guidance
βγ αγ αβ 7 γ2 β2 α2 49B1 α'β'+α'γ'+β'γ'.
States sum of roots and
α2 β2 γ2  2 2  27 58
+ + = − −2 − =
   
AnswerMarks Guidance
β2γ2 α2γ2 α2β2  7  49 49M1 A1 Uses
α' 2 +β' 2 +γ' 2 =( α'+β'+γ' )2 −2 ( α'β'+α'γ'+β'γ' )
AG
3
AnswerMarks
7(iii) α3 β3 γ3  58  2
49 + +  =−14   +27  −  −21
AnswerMarks Guidance
β3γ3 α3γ3 α3β3  49  7M1 Uses 49α' 3 =−14α′ 2 +27α′−7.
α3 β3 γ3 317
⇒ + + =−
AnswerMarks
β3γ3 α3γ3 α3β3 343A1
2
AnswerMarks Guidance
QuestionAnswer 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 | Guidance
The 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 $\frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\gamma^3\alpha^3} + \frac{\gamma^3}{\alpha^3\beta^3}$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q7 [9]}}
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