Question 9 - A-Level Maths

CAIE FP1 2019 June — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2019
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Roots of polynomials
Type	Finding polynomial from root properties
Difficulty	Challenging +1.2 This is a structured multi-part question on symmetric functions of roots requiring systematic application of Vieta's formulas and algebraic manipulation. While it involves several steps and the cubic identity for sum of cubes, each part guides the student through the process with clear signposting. The techniques (converting 1/α + 1/β + 1/γ, using α³+β³+γ³ identity, solving the auxiliary cubic) are standard Further Maths content. Part (iv) requires recognizing complex conjugate roots but is essentially given. Moderately above average difficulty due to length and FM content, but not requiring novel insight.
Spec	4.01a Mathematical induction: construct proofs 4.02j Cubic/quartic equations: conjugate pairs and factor theorem 4.05a Roots and coefficients: symmetric functions

9 A cubic equation $x ^ { 3 } + b x ^ { 2 } + c x + d = 0$ has real roots $\alpha , \beta$ and $\gamma$ such that $$\begin{aligned} \frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma } & = - \frac { 5 } { 12 } \\ \alpha \beta \gamma & = - 12 \\ \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } & = 90 \end{aligned}$$

Find the values of $c$ and $d$.
Express $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ in terms of $b$.
Show that $b ^ { 3 } - 15 b + 126 = 0$.
Given that $3 + \mathrm { i } \sqrt { } ( 12 )$ is a root of $y ^ { 3 } - 15 y + 126 = 0$, deduce the value of $b$.

Show mark scheme Show mark scheme source

Question 9(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\alpha\gamma}{\alpha\beta\gamma} \Rightarrow c = \alpha\beta+\beta\gamma+\alpha\gamma = 5$	M1 A1	Uses $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\alpha\gamma}{\alpha\beta\gamma}$
$d = 12$	B1
Total: 3

Question 9(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$	M1
$(\alpha+\beta+\gamma)^2 - 10 = b^2 - 10$	A1
Alternative: $90 + bS_2 - 5b + 36 = 0 \Rightarrow S_2 = \frac{5b-126}{b}$	M1 A1	Sums over roots
Total: 2

Question 9(iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$x^3 + bx^2 + 5x + 12 = 0 \quad (-b = \alpha+\beta+\gamma)$	M1	Formulates equation
$90 + b(b^2-10) - 5b + 36 = 0$	M1 A1	Sums over roots and uses 9(ii)
$b^3 - 15b + 126 = 0$	A1
Alternative 1: $(\alpha+\beta+\gamma)^3 = \alpha^3+\beta^3+\gamma^3 + 3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$	M1 A1	Uses expansion of $(\alpha+\beta+\gamma)^3$
$-b^3 = 90 - 15b + 36$	(M1)	Substitutes known values
$b^3 - 15b + 126 = 0$	A1
Alternative 2: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = b^2-10$	M1 A1	Finds sum of squares in terms of $b$
$\frac{5b-126}{b} = b^2 - 10 \Rightarrow b^3 - 15b + 126 = 0$	M1 A1	Equates expressions
Total: 4

Question 9(iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
Real root is $b = -6$	M1 A1	Gives the real root for $b$

## Question 9(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\alpha\gamma}{\alpha\beta\gamma} \Rightarrow c = \alpha\beta+\beta\gamma+\alpha\gamma = 5$ | M1 A1 | Uses $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\alpha\gamma}{\alpha\beta\gamma}$ |
| $d = 12$ | B1 | |
| **Total: 3** | | |

## Question 9(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$ | M1 | |
| $(\alpha+\beta+\gamma)^2 - 10 = b^2 - 10$ | A1 | |
| **Alternative:** $90 + bS_2 - 5b + 36 = 0 \Rightarrow S_2 = \frac{5b-126}{b}$ | M1 A1 | Sums over roots |
| **Total: 2** | | |

## Question 9(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^3 + bx^2 + 5x + 12 = 0 \quad (-b = \alpha+\beta+\gamma)$ | M1 | Formulates equation |
| $90 + b(b^2-10) - 5b + 36 = 0$ | M1 A1 | Sums over roots and uses 9(ii) |
| $b^3 - 15b + 126 = 0$ | A1 | |
| **Alternative 1:** $(\alpha+\beta+\gamma)^3 = \alpha^3+\beta^3+\gamma^3 + 3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$ | M1 A1 | Uses expansion of $(\alpha+\beta+\gamma)^3$ |
| $-b^3 = 90 - 15b + 36$ | (M1) | Substitutes known values |
| $b^3 - 15b + 126 = 0$ | A1 | |
| **Alternative 2:** $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = b^2-10$ | M1 A1 | Finds sum of squares in terms of $b$ |
| $\frac{5b-126}{b} = b^2 - 10 \Rightarrow b^3 - 15b + 126 = 0$ | M1 A1 | Equates expressions |
| **Total: 4** | | |

## Question 9(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Real root is $b = -6$ | M1 A1 | Gives the real root for $b$ |

Show LaTeX source

9 A cubic equation $x ^ { 3 } + b x ^ { 2 } + c x + d = 0$ has real roots $\alpha , \beta$ and $\gamma$ such that

$$\begin{aligned}
\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma } & = - \frac { 5 } { 12 } \\
\alpha \beta \gamma & = - 12 \\
\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } & = 90
\end{aligned}$$

(i) Find the values of $c$ and $d$.\\

(ii) Express $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ in terms of $b$.\\

(iii) Show that $b ^ { 3 } - 15 b + 126 = 0$.\\

(iv) Given that $3 + \mathrm { i } \sqrt { } ( 12 )$ is a root of $y ^ { 3 } - 15 y + 126 = 0$, deduce the value of $b$.\\

\hfill \mbox{\textit{CAIE FP1 2019 Q9 [11]}}

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