# 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 |