## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1 & 5 & 1\\1 & -2 & -2\\2 & 3 & \theta\end{pmatrix} \rightarrow \begin{pmatrix}1 & 5 & 1\\0 & -7 & -3\\0 & 0 & \theta+1\end{pmatrix}$ | M1 A1 | Reduces to echelon form. At least one row operation for M1 |
| $r(\mathbf{A}) = 3$ if $\theta \neq -1$ | A1 | |
| $r(\mathbf{A}) = 2$ if $\theta = -1$ | B1 | |
| | **4** | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 5y + z = -1$, $-7y - 3z = 1$, $(\theta+1)z = (\theta+1)$ | M1 | Uses reduced form of augmented matrix or eliminates variables from scratch |
| $z=1$, $y = -\frac{4}{7}$, $x = \frac{6}{7}$ | A1, A1 | One correct. All three correct |
| | **3** | |

## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 5y + z = -1$, $-7y - 3z = 1$, $(\theta+1)z = (\theta+1)$ | | |
| $z = t$ | M1 | Uses parameter |
| $y = -\frac{3t+1}{7}$, $x = \frac{8t-2}{7}$ | A1 A1 | |
| | **3** | |

## Question 10(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 5y + \quad z = -1$, $-7y \quad -3z = 1$, $(\theta + 1)z = \phi + 1$ | M1 | Uses reduced form of augmented matrix or eliminates variables from scratch |
| $\theta = -1 \Rightarrow \phi = -1$ so no solution (inconsistent) | A1 | |
| **Total: 2** | | |

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## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $w = \cos y \Rightarrow \frac{dw}{dx} = -\sin y \frac{dy}{dx}$ | B1 | |
| $\frac{d^2w}{dx^2} = -\sin y \frac{d^2y}{dx^2} - \cos y \left(\frac{dy}{dx}\right)^2$ | B1 | |
| $\frac{d^2w}{dx^2} + 2\frac{dw}{dx} + w = -\sin y \frac{d^2y}{dx^2} - \cos y \left(\frac{dy}{dx}\right)^2 - 2\sin y \frac{dy}{dx} + \cos y$ | M1 | Uses substitution to obtain $w$-$x$ equation, AG |
| $= -\cos y(e^{-2x} \sec y) = -e^{-2x}$ | A1 | |
| **Total: 4** | | |

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## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + 2m + 1 = 0 \Rightarrow m = -1$ | M1 | Finds CF |
| CF: $w = (Ax + B)e^{-x}$ | A1 | |
| PI: $w = ke^{-2x} \Rightarrow w' = -2ke^{-2x} \Rightarrow w'' = 4ke^{-2x}$ | M1 | Forms PI and differentiates |
| $4k - 4k + k = -1 \Rightarrow k = -1$ | A1 | |
| $w = (Ax + B)e^{-x} - e^{-2x}$ | A1 | States general solution |
| $x = 0,\ y = \frac{1}{3}\pi,\ w = \frac{1}{2} \Rightarrow B = \frac{3}{2}$ | B1 | Uses initial conditions to find constants |
| $w' = -(Ax + B)e^{-x} + Ae^{-x} + 2e^{-2x}$ | M1 | Differentiates general solution |
| $x = 0,\ y = \frac{1}{3}\pi,\ y' = \frac{\sqrt{3}}{3},\ w' = -\frac{1}{2} \Rightarrow -\frac{1}{2} = -\frac{3}{2} + A + 2 \Rightarrow A = -1$ | M1 A1 | Substitutes initial conditions |
| $y = \cos^{-1}\!\left(\left(\frac{3}{2} - x\right)e^{-x} - e^{-2x}\right)$ | A1 | States particular solution for $y$ in terms of $x$ |
| **Total: 10** | | |

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## Question 11O(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{2\alpha} - e^{-2\alpha} = 2(e^\alpha + e^{-\alpha}) \Rightarrow e^\alpha - e^{-\alpha} = 2$ | M1 | Sets equations equal and divides by $e^\alpha + e^{-\alpha}$ |
| $e^{2\alpha} - 2e^\alpha - 1 = 0 \Rightarrow e^\alpha = 1 + \sqrt{2}$ | M1 A1 | Forms quadratic in $e^\alpha$, AG |
| $\alpha = \ln(1 + \sqrt{2})$ | A1 | Must be exact |
| $r = 2(1 + \sqrt{2} + \sqrt{2} - 1) = 4\sqrt{2}$ | M1 A1 | Substitutes to find $r$ |
| **Total: 6** | | |

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## Question 11O(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| [sketch] | B1 | $C_1$ has correct shape |
| [sketch] | B1 | $C_2$ has correct shape |
| [sketch] | B1 | Intersection points positioned correctly |
| **Total: 3** | | |

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## Question 11O(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\int_0^{\ln(1+\sqrt{2})} (e^\theta + e^{-\theta})^2\, d\theta - \frac{1}{2}\int_0^{\ln(1+\sqrt{2})} (e^{2\theta} - e^{-2\theta})^2\, d\theta$ $= \int_0^{\ln(1+\sqrt{2})} 5 + 2e^{2\theta} + 2e^{-2\theta} - \frac{1}{2}e^{4\theta} - \frac{1}{2}e^{-4\theta}\, d\theta$ | M1 A1 | Uses $\frac{1}{2}\int r^2\, d\theta$ to formulate correct area |
| $= \left[5\theta + e^{2\theta} - e^{-2\theta} - \frac{1}{8}e^{4\theta} + \frac{1}{8}e^{-4\theta}\right]_0^{\ln(1+\sqrt{2})}$ | M1 A1 | Expands and integrates |
| $= 5\ln(1+\sqrt{2}) + (1+\sqrt{2})^2 - (1+\sqrt{2})^{-2} - \frac{1}{8}\left((1+\sqrt{2})^4 - (1+\sqrt{2})^{-4}\right) = 5.82$ | A1 | |
| **Total: 5** | | |