## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -5$ and $y = a$ | B1 B1 | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + (a+10)x + 5a + 26 = (x+5)(x+a+5)+1$ | M1 | By inspection or long division |
| Oblique asymptote is $y = x + a + 5$ | A1 | |

## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + 10x + 5a + 26 = 0$ | M1 | Puts $y$-values equal and forms quadratic equation |
| $10^2 - 4(5a+26) = -4 - 20a < 0$ so no intersection point | A1 | Correct discriminant and conclusion |

## Question 10(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+5)(2x+a+10) - x^2 - ax - 10x - 5a - 26 = 0$ | M1 | Differentiates and forms quadratic equation |
| $x^2 + 10x + 24 = 0$ | A1 | |
| Stationary points are $(-4, a+2)$ and $(-6, a-2)$ | A1 | Must have both points |

## Question 10(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph with asymptotes drawn, intersection correct | B1 | Asymptotes drawn, intersection correct |
| $C_1$ correct | B1 | $C_1$ correct |
| $C_2$ correct | B1 | $C_2$ correct |

## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \sqrt{2}\theta^{\frac{1}{2}}\cos\theta$ | **M1** | Uses $x = r\cos\theta$ |
| $\frac{d}{d\theta}\left(\sqrt{2}\theta^{\frac{1}{2}}\cos\theta\right) = \sqrt{2}\left(-\theta^{\frac{1}{2}}\sin\theta + \frac{1}{2}\theta^{-\frac{1}{2}}\cos\theta\right) = 0$ | **M1 A1** | Sets derivative of $r\cos\theta$ equal to zero |
| $-\theta^{\frac{1}{2}}\sin\theta + \frac{1}{2}\theta^{-\frac{1}{2}}\cos\theta = 0 \Rightarrow \cos\theta = 2\theta\sin\theta \Rightarrow 2\theta\tan\theta = 1$ | **A1** | AG |
| $2(0.6)\tan(0.6) - 1 = -0.179$ and $2(0.7)\tan(0.7) - 1 = 0.179$ | **B1** | Shows sign change |
| | **5** | |

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## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\theta = \theta\sec^2\theta \Rightarrow \theta^{\frac{1}{2}}\left(\sec\theta - \sqrt{2}\right) = 0 \Rightarrow \theta = \frac{\pi}{4}$ | **M1 A1** | Finds value of $\theta$ |
| | **2** | |

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## Question 11E(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| [Correct curve sketch] | **B1** | Correct shape |
| [Correct intersection shown] | **B1** | Intersection correct |
| | **2** | |

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## Question 11E(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\frac{\pi}{4}}2\theta\, d\theta - \frac{1}{2}\int_0^{\frac{\pi}{4}}\theta\sec^2\theta\, d\theta = \frac{1}{2}\int_0^{\frac{\pi}{4}}\theta\left(2 - \sec^2\theta\right)d\theta$ | **M1** | Forms correct integral |
| $\frac{1}{2}\left[\theta(2\theta - \tan\theta)\right]_0^{\frac{\pi}{4}} - \frac{1}{2}\int_0^{\frac{\pi}{4}}(2\theta - \tan\theta)\,d\theta$ | **M1 A1** | Integrates by parts |
| $\frac{1}{2}\left[\theta(2\theta-\tan\theta)\right]_0^{\frac{\pi}{4}} - \frac{1}{2}\left[\theta^2 + \ln\cos\theta\right]_0^{\frac{\pi}{4}}$ | **A1** | |
| $\frac{\pi}{8}\left(\frac{\pi}{2}-1\right) - \frac{1}{2}\left(\left(\frac{\pi}{4}\right)^2 + \frac{1}{2}\ln 2\right) = \frac{1}{4}\ln 2 + \frac{\pi}{8}\left(\frac{\pi}{4}-1\right)$ | **A1** | AEF, must be exact |
| | **5** | |

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## Question 11O(i)(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1&2&3&4\\1&0&1&-1\\1&-2&-3&a\\1&2&5&2\end{pmatrix} \rightarrow\cdots\rightarrow \begin{pmatrix}-1&2&3&4\\0&2&4&3\\0&0&0&a+4\\0&0&0&0\end{pmatrix}$ | **M1 A1** | Reduces **M** or $\mathbf{M}^T$ to echelon form |
| $\dim V = \text{rank} = 3$ | **A1** | |
| $c_1v_1' + c_2v_2' + c_3v_4' = 0 \Rightarrow c_1 = c_2 = c_3 = 0$ | **M1** | Shows $v_1', v_2', v_4'$ are linearly independent |
| Thus $v_1, v_2, v_4$ are linearly independent (and so form a basis for $V$) | **A1** | |
| | **5** | |

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## Question 11O(i)(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1&2&4&x\\1&0&-1&y\\1&-2&a&z\\1&2&2&t\end{pmatrix}\rightarrow\cdots\rightarrow\begin{pmatrix}-1&2&4&x\\0&2&3&y+x\\0&0&a+4&z+x\\0&4&6&t+x\end{pmatrix}$ | **M1 A1** | $x=-\alpha+2\beta+4\gamma$, $y=\alpha-\gamma$, uses row operations or $z=\alpha-2\beta+a\gamma$, $t=\alpha+2\beta+2\gamma$ |
| System is consistent when $t+x = 2(y+x) \Rightarrow x+2y=t$ or $(-\alpha+2\beta+4\gamma)+2(\alpha-\gamma)=\alpha+2\beta+2\gamma = t$ | **M1 A1** | AG |
| | **4** | |

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## Question 11O(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-x+2y+3z+4t=0$, $2y+4z+3t=0$ | **M1** | Finds basis for null space |
| $t=\mu,\ z=\lambda,\ y=-2\lambda-\frac{3}{2}\mu,\ x=-\lambda+\mu$ | **A1** | |
| A basis is $\left\{\begin{pmatrix}-1\\-2\\1\\0\end{pmatrix}, \begin{pmatrix}2\\-3\\0\\2\end{pmatrix}\right\} = \{\mathbf{e}_1, \mathbf{e}_2\}$ | **A1** | AEF |
| $\mathbf{M}\begin{pmatrix}1\\0\\0\\0\end{pmatrix} = \begin{pmatrix}-1\\1\\1\\1\end{pmatrix}$ so particular solution is $\mathbf{e} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}$ | **B1** | Finds particular solution |
| General solution is $\mathbf{x} = \mathbf{e} + \lambda\mathbf{e}_1 + \mu\mathbf{e}_2$ | **A1** | FT: accept their basis; must have correct particular solution |
| | **5** | |