## Question 10:

**States asymptotes:**

| Vertical: $x = 1$ and Horizontal: $y = 2$ | B1, B1 | 2 marks |

**Obtains quadratic form in $x$:**

| $yx^2 - 2yx + y = 2x^2 - 3x - 2 \Rightarrow (y-2)x^2 - (2y-3)x + (y+2) = 0$ | M1A1 | |

**Uses $B^2 - 4AC \geq 0$ for real roots:**

| For real $x$: $(2y-3)^2 - 4(y-2)(y+2) \geq 0$ | M1 | |

| $\Rightarrow 12y \leq 25 \Rightarrow y \leq \frac{25}{12}$ | A1 | 4 marks |

**Finds condition for $y' = 0$:**

| $y' = 0 \Rightarrow (x^2 - 2x + 1)(4x - 3) - (2x^2 - 3x - 2)(2x - 2) = 0$ | M1 | |

**Solves:**

| $\Rightarrow x^2 - 8x + 7 = 0 \Rightarrow (x-7)(x-1) = 0$ | A1 | |

| $\Rightarrow x = 7$, (since $x = 1$ is vertical asymptote) | | |

**Obtains stationary point:**

| Stationary point is $\left(7, \frac{25}{12}\right)$ | A1 | 3 marks |

**Sketch showing:**

| Axes and asymptotes | B1 | |
| $(-0.5, 0)$, $(2, 0)$, $(0, -2)$ and $(4, 2)$ | B1 | |
| Left hand branch | B1 | |
| Right hand branch | B1 | 4 marks; **[13 total]** |

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## Question 11E:

**Differentiation:**

| $y = 2\sec x \Rightarrow y' = 2\sec x \tan x$ | M1 | |

**Use of $\sec^2 x = 1 + \tan^2 x$:**

| $1 + (y')^2 = 1 + 4\sec^2 x(\sec^2 x - 1) = 4\sec^4 x - 4\sec^2 x + 1 = (2\sec^2 x - 1)^2$ | M1, A1 | |

**Substitute in arc length formula:**

| $s = \int_0^{\frac{\pi}{4}}(2\sec^2 x - 1)\,dx$ | A1 | 4 marks |

**Integrate:**

| $= \left[2\tan x - x\right]_0^{\frac{\pi}{4}}$ | M1 | |

**Substitute limits:**

| $= \left[2 - \frac{1}{4}\pi\right]$ | A1 | 2 marks |

**(i) Use surface area formula:**

| $S = 2\pi\int_0^{\frac{\pi}{4}} 2\sec x(2\sec^2 x - 1)\,dx$ | M1A1 | |

| $= 4\pi\int_0^{\frac{\pi}{4}}(2\sec^3 x - \sec x)\,dx$ (AG) | A1 | 3 marks |

**(ii) Differentiates:**

| $\frac{d}{dx}(\sec x \tan x) = \sec x \tan^2 x + \sec^3 x = \sec x(\sec^2 x - 1) + \sec^3 x$ | M1A1 | |

**And obtains printed result:**

| $= 2\sec^3 x - \sec x$ (AG) | A1 | 3 marks |

**Uses result to deduce surface area:**

| $S = 4\pi\left[\sec x \tan x\right]_0^{\frac{\pi}{4}}$ | M1 | 2 marks |

| $= 4\pi\sqrt{2}$ | A1 | **[14 total]** |

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## Question 11O:

**Vector perpendicular to $\Pi_1$:**

| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & 0 & 1\\1 & -3 & 1\end{vmatrix} = \begin{pmatrix}3\\-1\\-6\end{pmatrix}$ | M1A1 | |

**Obtains Cartesian equation:**

| $3\times2 + 1\times(-1) + (-2)\times(-6) = 17 \Rightarrow 3x - y - 6z = 17$ | M1, A1 | 4 marks |

**Obtains area of triangle $ABC$:**

| $\frac{1}{2}\sqrt{3^2 + 1^2 + 6^2} = \frac{1}{2}\sqrt{46}\ (= 3.39)$ | M1 A1 | |

**Obtains length of perpendicular from $D$ to triangle $ABC$:**

| $\left|\frac{9 - 6 - 12 - 17}{\sqrt{3^2 + 1^2 + 6^2}}\right| = \frac{26}{\sqrt{46}}$ | M1A1 | |

**Uses $\frac{1}{3} \times$ Base area $\times$ Height:**

| $\frac{1}{3} \times \frac{1}{2}\sqrt{46} \times \frac{26}{\sqrt{46}} = \frac{13}{3}$ | M1A1 | |

**Or triple scalar product method:**

| Or e.g. $\frac{1}{6}\begin{pmatrix}1\\5\\4\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\-6\end{pmatrix} = \frac{26}{6} = \frac{13}{3}$ | | 6 marks |

**Obtains normal to $ABD$:**

| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 5 & 4\\2 & 0 & 1\end{vmatrix} = \begin{pmatrix}5\\7\\-10\end{pmatrix}$ | M1A1 | |

**Uses scalar product:**

| $\sqrt{3^2+1^2+6^2}\sqrt{5^2+7^2+10^2}\cos\theta = \begin{pmatrix}3\\-1\\-6\end{pmatrix}\cdot\begin{pmatrix}5\\7\\-10\end{pmatrix}$ | M1 | |

**To find angle between normals and hence angle between $\Pi_1$ and $\Pi_2$:**

| $\Rightarrow \cos\theta = \frac{68}{\sqrt{46}\sqrt{174}} \Rightarrow \theta = 40.5°$ | A1 | 4 marks; **[14 total]** |