Question 11 EITHER - A-Level Maths

CAIE FP1 2011 November — Question 11 EITHER

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Roots of unity with trigonometric identities
Difficulty	Challenging +1.8 This is a substantial Further Maths question requiring multiple techniques: roots of unity, De Moivre's theorem, complex conjugates, and forming quadratic equations from roots. While each individual step follows standard methods (showing ω^5=-1, manipulating complex expressions, using conjugate properties), the extended multi-part nature and need to synthesize several concepts to reach the exact value of cos(π/5) places it well above average difficulty. However, it's a relatively standard FP1 roots-of-unity question without requiring particularly novel insights.
Spec	4.02q De Moivre's theorem: multiple angle formulae 4.02r nth roots: of complex numbers 4.05a Roots and coefficients: symmetric functions

Let $\omega = \cos \frac { 1 } { 5 } \pi + \mathrm { i } \sin \frac { 1 } { 5 } \pi$. Show that $\omega ^ { 5 } + 1 = 0$ and deduce that $$\omega ^ { 4 } - \omega ^ { 3 } + \omega ^ { 2 } - \omega = - 1$$ Show further that $$\omega - \omega ^ { 4 } = 2 \cos \frac { 1 } { 5 } \pi \quad \text { and } \quad \omega ^ { 3 } - \omega ^ { 2 } = 2 \cos \frac { 3 } { 5 } \pi$$ Hence find the values of $$\cos \frac { 1 } { 5 } \pi + \cos \frac { 3 } { 5 } \pi \quad \text { and } \quad \cos \frac { 1 } { 5 } \pi \cos \frac { 3 } { 5 } \pi$$ Find a quadratic equation having roots $\cos \frac { 1 } { 5 } \pi$ and $\cos \frac { 3 } { 5 } \pi$ and deduce the exact value of $\cos \frac { 1 } { 5 } \pi$.

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Question 11 (EITHER – Complex Numbers):

Part 1: Verification and Factorisation

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)^5+1=\cos\pi+i\sin\pi+1=0$	B1	Verifies $\omega$ is a root
$(\omega^5+1)=(\omega+1)(\omega^4-\omega^3+\omega^2-\omega+1)=0$		Factorisation
$\omega\neq-1\Rightarrow\omega^4-\omega^3+\omega^2-\omega+1=0\Rightarrow\omega^4-\omega^3+\omega^2-\omega=-1$	B1	2 marks total

Part 2: Finding $\omega-\omega^4$ and $\omega^3-\omega^2$

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\omega^4=\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}=-\cos\frac{\pi}{5}+\sin\frac{\pi}{5}$	M1	Finds $\omega^4$
$\Rightarrow\omega-\omega^4=2\cos\frac{\pi}{5}$	A1
$\omega^3=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5}$		Finds $\omega^3$
$\omega^2=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}=\cos\frac{3\pi}{5}-i\sin\frac{3\pi}{5}$	M1	Finds $\omega^2$
$\omega^3-\omega^2=2\cos\frac{3\pi}{5}$	A1	4 marks total

Part 3: Sum and Product of $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$-2\cos\frac{\pi}{5}-2\cos\frac{3\pi}{5}=-1\Rightarrow\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}=\frac{1}{2}$	M1A1
$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac{1}{4}(\omega-\omega^4)(\omega^3-\omega^2)$	M1
$=\frac{1}{4}(\omega^4-\omega^3-\omega^7+\omega^6)$
$=\frac{1}{4}(\omega^4-\omega^3+\omega^2-\omega)=-\frac{1}{4}$	A1	4 marks total

Part 4: Quadratic and Final Value

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x^2-\frac{1}{2}x-\frac{1}{4}=0$ or $4x^2-2x-1=0$	M1	Finds required quadratic
$\Rightarrow x=\frac{2\pm2\sqrt{5}}{8}$	M1A1	Solves for $x$
$\Rightarrow\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$ (since $0<\cos\frac{\pi}{5}<1$)	A1	4 marks total

Question 11 (OR – Differential Equations):

Part 1: Substitution and Transformation

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$z=x^2y\Rightarrow\frac{dz}{dx}=x^2\frac{dy}{dx}+2xy$	M1	Differentiates
$\Rightarrow\frac{d^2z}{dx^2}=x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+2y$	A1	Differentiates twice
$\therefore x^2\frac{d^2y}{dx^2}+4x(1+x)\frac{dy}{dx}+(2+8x+4x^2)y=\left(x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+2y\right)+4\left(x^2\frac{dy}{dx}+2xy\right)+4x^2y$	M1	Rearranges LHS of DE
$=\frac{d^2z}{dx^2}+4\frac{dz}{dx}+4z=8x^2$ (AG)	A1	4 marks total

Part 2: Solving the ODE

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$m^2+4m+4=0\Rightarrow(m+2)^2=0\Rightarrow m=-2$	M1	Finds and solves AQE
CF: $z=Ae^{-2x}+Bxe^{-2x}$	A1	States CF
PI: $z=ax^2+bx+c$		States form of PI
$\Rightarrow z'=2ax+b\Rightarrow z''=2a$	M1	Differentiates twice
$2a+8ax+4b+4ax^2+4bx+4c=8x^2$	A1	Substitutes
$2a+4b+4c=0$; $8a+4b=0$; $4a=8$	M1	Equates coefficients
$a=2\quad b=-4\quad c=3$	A1	Solves
$z=Ae^{-2x}+Bxe^{-2x}+2x^2-4x+3$	M1	States GS for $z$–$x$
$y=\frac{A}{x^2}e^{-2x}+\frac{B}{x}e^{-2x}+2-\frac{4}{x}+\frac{3}{x^2}$	A1	8 marks total

Part 3: Behaviour as $x\to\infty$

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
As $x\to\infty$, $e^{-2x}$, $\frac{1}{x}$ and $\frac{1}{x^2}\to 0$	M1	Considers effect of $x\to\infty$
$\therefore y\to 2$	A1	2 marks total

# Question 11 (EITHER – Complex Numbers):

## Part 1: Verification and Factorisation

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)^5+1=\cos\pi+i\sin\pi+1=0$ | B1 | Verifies $\omega$ is a root |
| $(\omega^5+1)=(\omega+1)(\omega^4-\omega^3+\omega^2-\omega+1)=0$ | | Factorisation |
| $\omega\neq-1\Rightarrow\omega^4-\omega^3+\omega^2-\omega+1=0\Rightarrow\omega^4-\omega^3+\omega^2-\omega=-1$ | B1 | 2 marks total |

## Part 2: Finding $\omega-\omega^4$ and $\omega^3-\omega^2$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\omega^4=\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}=-\cos\frac{\pi}{5}+\sin\frac{\pi}{5}$ | M1 | Finds $\omega^4$ |
| $\Rightarrow\omega-\omega^4=2\cos\frac{\pi}{5}$ | A1 | |
| $\omega^3=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5}$ | | Finds $\omega^3$ |
| $\omega^2=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}=\cos\frac{3\pi}{5}-i\sin\frac{3\pi}{5}$ | M1 | Finds $\omega^2$ |
| $\omega^3-\omega^2=2\cos\frac{3\pi}{5}$ | A1 | 4 marks total |

## Part 3: Sum and Product of $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2\cos\frac{\pi}{5}-2\cos\frac{3\pi}{5}=-1\Rightarrow\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}=\frac{1}{2}$ | M1A1 | |
| $\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac{1}{4}(\omega-\omega^4)(\omega^3-\omega^2)$ | M1 | |
| $=\frac{1}{4}(\omega^4-\omega^3-\omega^7+\omega^6)$ | | |
| $=\frac{1}{4}(\omega^4-\omega^3+\omega^2-\omega)=-\frac{1}{4}$ | A1 | 4 marks total |

## Part 4: Quadratic and Final Value

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2-\frac{1}{2}x-\frac{1}{4}=0$ or $4x^2-2x-1=0$ | M1 | Finds required quadratic |
| $\Rightarrow x=\frac{2\pm2\sqrt{5}}{8}$ | M1A1 | Solves for $x$ |
| $\Rightarrow\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$ (since $0<\cos\frac{\pi}{5}<1$) | A1 | 4 marks total |

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# Question 11 (OR – Differential Equations):

## Part 1: Substitution and Transformation

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z=x^2y\Rightarrow\frac{dz}{dx}=x^2\frac{dy}{dx}+2xy$ | M1 | Differentiates |
| $\Rightarrow\frac{d^2z}{dx^2}=x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+2y$ | A1 | Differentiates twice |
| $\therefore x^2\frac{d^2y}{dx^2}+4x(1+x)\frac{dy}{dx}+(2+8x+4x^2)y=\left(x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+2y\right)+4\left(x^2\frac{dy}{dx}+2xy\right)+4x^2y$ | M1 | Rearranges LHS of DE |
| $=\frac{d^2z}{dx^2}+4\frac{dz}{dx}+4z=8x^2$ (AG) | A1 | 4 marks total |

## Part 2: Solving the ODE

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2+4m+4=0\Rightarrow(m+2)^2=0\Rightarrow m=-2$ | M1 | Finds and solves AQE |
| CF: $z=Ae^{-2x}+Bxe^{-2x}$ | A1 | States CF |
| PI: $z=ax^2+bx+c$ | | States form of PI |
| $\Rightarrow z'=2ax+b\Rightarrow z''=2a$ | M1 | Differentiates twice |
| $2a+8ax+4b+4ax^2+4bx+4c=8x^2$ | A1 | Substitutes |
| $2a+4b+4c=0$; $8a+4b=0$; $4a=8$ | M1 | Equates coefficients |
| $a=2\quad b=-4\quad c=3$ | A1 | Solves |
| $z=Ae^{-2x}+Bxe^{-2x}+2x^2-4x+3$ | M1 | States GS for $z$–$x$ |
| $y=\frac{A}{x^2}e^{-2x}+\frac{B}{x}e^{-2x}+2-\frac{4}{x}+\frac{3}{x^2}$ | A1 | 8 marks total |

## Part 3: Behaviour as $x\to\infty$

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $x\to\infty$, $e^{-2x}$, $\frac{1}{x}$ and $\frac{1}{x^2}\to 0$ | M1 | Considers effect of $x\to\infty$ |
| $\therefore y\to 2$ | A1 | 2 marks total |

Show LaTeX source

Let $\omega = \cos \frac { 1 } { 5 } \pi + \mathrm { i } \sin \frac { 1 } { 5 } \pi$. Show that $\omega ^ { 5 } + 1 = 0$ and deduce that

$$\omega ^ { 4 } - \omega ^ { 3 } + \omega ^ { 2 } - \omega = - 1$$

Show further that

$$\omega - \omega ^ { 4 } = 2 \cos \frac { 1 } { 5 } \pi \quad \text { and } \quad \omega ^ { 3 } - \omega ^ { 2 } = 2 \cos \frac { 3 } { 5 } \pi$$

Hence find the values of

$$\cos \frac { 1 } { 5 } \pi + \cos \frac { 3 } { 5 } \pi \quad \text { and } \quad \cos \frac { 1 } { 5 } \pi \cos \frac { 3 } { 5 } \pi$$

Find a quadratic equation having roots $\cos \frac { 1 } { 5 } \pi$ and $\cos \frac { 3 } { 5 } \pi$ and deduce the exact value of $\cos \frac { 1 } { 5 } \pi$.

\hfill \mbox{\textit{CAIE FP1 2011 Q11 EITHER}}

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Q1 5 Q2 5 Q3 7 Q4 7 Q5 7 Q6 8 Q7 11 Q8 11 Q9 12 Q10 13 Q11 EITHER Q11 OR