## Question 11(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| [Sketch of cardioid] | B1 | Sketch of a cardioid |
| [Correct orientation and labelled] | B1 | Correct orientation and labelled |

## Question 11(ii):

$A = \frac{a^2}{2}\int_{-\pi}^{\pi}(1 + 2\sin\theta + \sin^2\theta)\,d\theta$ | M1 |

$= \left(\frac{a^2}{2}\right)\int_{-\pi}^{\pi}\left(1 + 2\sin\theta + \frac{1}{2} - \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 |

$= \left(\frac{a^2}{2}\right)\left[\frac{3\theta}{2} - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{-\pi}^{\pi} = \frac{3\pi a^2}{2}$ | M1A1 |

**Total: 4**

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## Question 11(iii):

Show that when $r=0$, $\theta = -\frac{\pi}{2}$, when $r=2a$, $\theta = \frac{\pi}{2}$ and that $\frac{dr}{d\theta} = a\cos\theta$ | B1 |

$s = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{a^2(1+2\sin\theta + a^2\sin^2\theta) + a^2\cos^2\theta}\,d\theta = \sqrt{2a}\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\sqrt{1+\sin\theta}\,d\theta$ | M1A1 | Uses correct formula for arc length; AG

**Total: 3**

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## Question 11(iv):

$u = 1 + \sin\theta \Rightarrow \frac{du}{d\theta} = \cos\theta = \sqrt{1-(u-1)^2} = \sqrt{2u - u^2}$ | M1 |

$s = \sqrt{2a}\int_{0}^{2}\frac{1}{\sqrt{2-u}}\,du$ AG | A1 | Including limits

$= \sqrt{2a}\left[-2(2-u)^{\frac{1}{2}}\right]_{0}^{2} = 4a$ | M1A1 |

**Total: 4**

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## Question 12E(i):

$\int y\,dx = \frac{1}{2}\int_{0}^{4}(e^x + e^{-x})\,dx = \frac{1}{2}\left[e^x - e^{-x}\right]_{0}^{4} = \frac{1}{2}(e^4 - e^{-4})$ | M1A1 |

$\int xy\,dx = \frac{1}{2}\int_{0}^{4}x(e^x + e^{-x})\,dx$ | M1 |

$\frac{1}{2}\left[e^x(x-1) - e^{-x}(x+1)\right]_{0}^{4} = \frac{1}{2}(3e^4 + 2 - 5e^{-4})$ | A1A1 |

$\frac{1}{2}\int y^2\,dx = \frac{1}{8}\int_{0}^{4}(e^{2x} + 2 + e^{-2x})\,dx$ | M1 |

$\frac{1}{8}\left[\frac{1}{2}e^{2x} + 2x - \frac{1}{2}e^{-2x}\right]_{0}^{4} = \frac{1}{16}(e^8 + 16 - e^{-8})$ | A1A1 |

$\bar{x} = \frac{3e^4 + 2 - 5e^{-4}}{e^4 - e^{-4}}\left(= \frac{3e^2 + 5e^{-2}}{e^2 + e^{-2}}\right)$; $\bar{y} = \frac{1}{8}\left(\frac{e^8 + 16 - e^{-8}}{e^4 - e^{-4}}\right)$ (OE) | M1A1 | Uses correct formulae for $\bar{x}$ and $\bar{y}$

**Total: 10**

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## Question 12E(ii):

$\frac{ds}{dx} = \sqrt{1+(y')^2} = \sqrt{\frac{1}{4}(e^{2x}+2+e^{-2x})} = \sqrt{\frac{1}{4}(e^x+e^{-x})^2} = \frac{1}{2}(e^x+e^{-x})$ AG | B1 |

$S = 2\pi\int_{0}^{4}\frac{1}{4}(e^x+e^{-x})^2\,dx = \frac{\pi}{2}\int_{0}^{4}(e^{2x}+2+e^{-2x})\,dx$ | M1A1 |

$= \frac{\pi}{2}\left[\frac{e^{2x}}{2} + 2x - \frac{e^{-2x}}{2}\right]_{0}^{4} = \frac{\pi}{2}\left(\frac{e^8}{2} + 8 - \frac{e^{-8}}{2}\right)$ | A1 |

**Total: 4**

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## Question 12O(i):

$\overrightarrow{AC} = \begin{pmatrix}2\\-2\\-2\end{pmatrix}$, $\overrightarrow{AB} = \begin{pmatrix}2\\-2\\2\end{pmatrix}$, $\overrightarrow{CD} = \begin{pmatrix}2\\-4\\\alpha-1\end{pmatrix}$ | B1 | May find $\overrightarrow{AD}$, $\overrightarrow{BC}$ or $\overrightarrow{BD}$ instead of AC

$\overrightarrow{AB}\times\overrightarrow{CD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\2&-4&\alpha-1\end{vmatrix} = \begin{pmatrix}5-\alpha\\3-\alpha\\-2\end{pmatrix} \sim \begin{pmatrix}\alpha-5\\\alpha-3\\2\end{pmatrix}$ | M1A1 |

$\frac{\begin{pmatrix}2\\\-2\\-2\end{pmatrix}\cdot\begin{pmatrix}\alpha-5\\\alpha-3\\2\end{pmatrix}}{\sqrt{(\alpha-5)^2+(\alpha-3)^2+4}} = 2\sqrt{2} \Rightarrow 2\sqrt{2}\sqrt{2\alpha^2-16\alpha+38} = 8$ | M1A1 | Substitutes vectors into correct formula

$\Rightarrow 2\alpha^2 - 16\alpha + 38 = 8 \Rightarrow \alpha^2 - 8\alpha + 15 = 0$ | A1 |

$\Rightarrow (\alpha-3)(\alpha-5) = 0 \Rightarrow \alpha = 3$ or $5$ AG | A1 |

**Total: 7**

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## Question 12O(ii):

$\overrightarrow{AD} = \begin{pmatrix}4\\-6\\0\end{pmatrix}$ or $\overrightarrow{CD} = \begin{pmatrix}2\\-4\\2\end{pmatrix}$ | B1 | Alt method: Let P be point on AC with parameter $\lambda$ — M1

Distance of $D$ from $AC = \frac{1}{\sqrt{1+1+1}}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\4&-6&0\\1&-1&-1\end{vmatrix} = \sqrt{\frac{56}{3}} = 4.32$ (or with CD) | M1A1 | Use $DP\cdot AC = 0$ to find $\lambda\ (= -5/3)$ — M1; Find length — A1

**Total: 3**

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## Question 12O(iii):

$ABC$: $\mathbf{n}_1 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&-1\\1&-1&1\end{vmatrix} = \begin{pmatrix}-1\\-1\\0\end{pmatrix}$, $ABD$: $\mathbf{n}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\2&-3&0\end{vmatrix} = \begin{pmatrix}3\\2\\-1\end{pmatrix}$ | B1B1 |

$\cos\theta = \left|\frac{-3-2+0}{\sqrt{2}\sqrt{14}}\right| \Rightarrow \theta = 19.1°$ or $0.333$ rads | M1A1 |

**Total: 4**