CAIE Further Paper 1 2022 November — Question 2 9 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2022
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) uses the substitution y=1/x² to find the new equation; part (b) applies Vieta's formulas to recognize the sum of products of roots taken three at a time; part (c) uses Newton's identities or sum-of-squares techniques. While requiring multiple steps and careful algebra, these are well-practiced techniques in Further Maths syllabi with no novel insight needed.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

2 The equation \(x ^ { 4 } + 3 x ^ { 2 } + 2 x + 6 = 0\) has roots \(\alpha , \beta , \gamma , \delta\).
  1. Find a quartic equation whose roots are \(\frac { 1 } { \alpha ^ { 2 } } , \frac { 1 } { \beta ^ { 2 } } , \frac { 1 } { \gamma ^ { 2 } } , \frac { 1 } { \delta ^ { 2 } }\) and state the value of \(\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } } + \frac { 1 } { \delta ^ { 2 } }\).
  2. Find the value of \(\beta ^ { 2 } \gamma ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \gamma ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 } \gamma ^ { 2 }\).
  3. Find the value of \(\frac { 1 } { \alpha ^ { 4 } } + \frac { 1 } { \beta ^ { 4 } } + \frac { 1 } { \gamma ^ { 4 } } + \frac { 1 } { \delta ^ { 4 } }\).

Question 2(a):
AnswerMarks Guidance
\(y = x^{-2}\) leading to \(x = y^{-\frac{1}{2}}\)B1 Correct substitution.
\(y^{-2} + 3y^{-1} + 2y^{-\frac{1}{2}} + 6 = 0\) leading to \(1 + 3y + 2y^{\frac{3}{2}} + 6y^2 = 0\)M1 Obtains an equation not involving radicals.
\(36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0\)A1
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = -\frac{32}{36} = -\frac{8}{9}\)B1 FT
Total: 4
Question 2(b):
AnswerMarks Guidance
\(\alpha\beta\gamma\delta = 6\)B1 SOI
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2\beta^2\delta^2 + \alpha^2\beta^2\gamma^2 + \beta^2\gamma^2\delta^2 + \alpha^2\gamma^2\delta^2}{\alpha^2\beta^2\gamma^2\delta^2}\)M1 Relates to coefficients.
\(\beta^2\gamma^2\delta^2 + \alpha^2\gamma^2\delta^2\alpha^2\beta^2\delta^2 + \alpha^2\beta^2\gamma^2 = -32\)A1
Total: 3
Question 2(c):
AnswerMarks Guidance
\(\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2\left(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2}\right) = \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)\)M1 Uses formula for sum of squares.
\(-\frac{61}{162}\)A1
Total: 2
## Question 2(a):

$y = x^{-2}$ leading to $x = y^{-\frac{1}{2}}$ | B1 | Correct substitution.

$y^{-2} + 3y^{-1} + 2y^{-\frac{1}{2}} + 6 = 0$ leading to $1 + 3y + 2y^{\frac{3}{2}} + 6y^2 = 0$ | M1 | Obtains an equation not involving radicals.

$36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0$ | A1 |

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = -\frac{32}{36} = -\frac{8}{9}$ | B1 FT |

**Total: 4**

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## Question 2(b):

$\alpha\beta\gamma\delta = 6$ | B1 | SOI

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2\beta^2\delta^2 + \alpha^2\beta^2\gamma^2 + \beta^2\gamma^2\delta^2 + \alpha^2\gamma^2\delta^2}{\alpha^2\beta^2\gamma^2\delta^2}$ | M1 | Relates to coefficients.

$\beta^2\gamma^2\delta^2 + \alpha^2\gamma^2\delta^2\alpha^2\beta^2\delta^2 + \alpha^2\beta^2\gamma^2 = -32$ | A1 |

**Total: 3**

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## Question 2(c):

$\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2\left(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2}\right) = \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)$ | M1 | Uses formula for sum of squares.

$-\frac{61}{162}$ | A1 |

**Total: 2**

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2 The equation $x ^ { 4 } + 3 x ^ { 2 } + 2 x + 6 = 0$ has roots $\alpha , \beta , \gamma , \delta$.
\begin{enumerate}[label=(\alph*)]
\item Find a quartic equation whose roots are $\frac { 1 } { \alpha ^ { 2 } } , \frac { 1 } { \beta ^ { 2 } } , \frac { 1 } { \gamma ^ { 2 } } , \frac { 1 } { \delta ^ { 2 } }$ and state the value of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } } + \frac { 1 } { \delta ^ { 2 } }$.
\item Find the value of $\beta ^ { 2 } \gamma ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \gamma ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 } \delta ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 } \gamma ^ { 2 }$.
\item Find the value of $\frac { 1 } { \alpha ^ { 4 } } + \frac { 1 } { \beta ^ { 4 } } + \frac { 1 } { \gamma ^ { 4 } } + \frac { 1 } { \delta ^ { 4 } }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q2 [9]}}