## Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\frac{1}{2} + \frac{(3x-1)^2}{2(x^2+1)} = \frac{-x^2-1+9x^2-6x+1}{2(x^2+1)} = \frac{8x^2-6x}{2(x^2+1)} = \frac{4x^2-3x}{x^2+1}$ | M1A1 | |
| $\frac{9}{2} - \frac{(x+3)^2}{2(x^2+1)} = \frac{9x^2+9-x^2-6x-9}{2(x^2+1)} = \frac{8x^2-6x}{2(x^2+1)} = \frac{4x^2-3x}{x^2+1}$ | A1 (3) | |
| $\frac{(3x-1)^2}{2(x^2+1)} \geq 0 \Rightarrow y \geq -\frac{1}{2}$ | B1 | |
| $\frac{(x+3)^2}{2(x^2+1)} \geq 0 \Rightarrow y \leq \frac{9}{2} \Rightarrow -\frac{1}{2} \leq y \leq \frac{9}{2}$ | B1 (2) | Not hence gets B1 only |
| Turning points at equality: $\left(-3,\ \frac{9}{2}\right)$ and $\left(\frac{1}{3},\ -\frac{1}{2}\right)$ | B1B1 (2) | |
| Asymptote is $y = 4$ | B1 (1) | |
| Intersection with axes at $(0, 0)$ and $\left(\frac{3}{4},\ 0\right)$; intersection with asymptote at $\left(-\frac{4}{3},\ 4\right)$ | B1B1 | |
| Correct graph | B1 (3) Total 11 | |

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## Question 11E(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&2&3&4\\1&-1&2&3\\1&-3&3&5\\1&4&2&2\end{pmatrix} \rightarrow \cdots \rightarrow \begin{pmatrix}1&2&3&4\\0&3&1&1\\0&0&5&8\\0&0&0&0\end{pmatrix}$ | M1A1 | |
| $\Rightarrow \text{Dim}(V) = 3$ | A1 (3) | |

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## Question 11E(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $a\begin{pmatrix}1\\1\\1\\1\end{pmatrix} + b\begin{pmatrix}2\\-1\\-3\\4\end{pmatrix} + c\begin{pmatrix}3\\2\\3\\2\end{pmatrix} = \begin{pmatrix}0\\0\\0\\0\end{pmatrix}$ | M1 | |
| i: $a+2b+3c=0$; ii: $a-b+2c=0$; iii: $a-3b+3c=0$ | | |
| i $-$ iii $\Rightarrow b = 0 \Rightarrow a+3c=0$ and $a+2c=0 \Rightarrow c=0$ and $a=0$. So vectors are linearly independent. | M1A1, A1 | |
| Since the set has three linearly independent vectors, it forms a basis for $V$. | B1 (5) | |

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## Question 11E(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $W$ is not a vector space as it does not contain the zero vector. | B1 (1) | |

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## Question 11E(iv):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Row reduction giving final row: $0\ 0\ 0\ {-x-y+z+t}$ | M1A1 | |
| **Alternatively:** $\begin{pmatrix}x\\y\\z\\t\end{pmatrix} = \alpha\begin{pmatrix}1\\1\\1\\1\end{pmatrix} + \beta\begin{pmatrix}2\\-1\\-3\\4\end{pmatrix} + \gamma\begin{pmatrix}3\\2\\3\\2\end{pmatrix}$ | (M1A1) | i.e. in $V$ |
| $x+y = 2\alpha+\beta+5\gamma$; $z+t = 2\alpha+\beta+5\gamma \Rightarrow (x+y)-(z+t)=0$ | A1 | |
| Vector belongs to $V$ iff $x+y = z+t$ (OE); belongs to $W$ iff $x+y \neq z+t$ | M1, A1 (5) Total 14 | AG |

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## Question 11O:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}12&-4&2\\-4&\alpha+9&6\\2&6&7\end{vmatrix} = 0 \Rightarrow 80\alpha + 80 = 0 \Rightarrow \alpha = -1$ | M1A1, A1 (3) | |

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## Question 11O(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}3-\lambda&-4&2\\-4&-1-\lambda&6\\2&6&-2-\lambda\end{vmatrix} = 0$ | M1 | |
| $\Rightarrow \lambda^3 - 63\lambda + 162 = 0$ | M1A1 | |
| $\Rightarrow (\lambda+9)(\lambda-3)(\lambda-6) = 0 \Rightarrow \lambda_1 = 6$ and $\lambda_2 = 3$ | M1A1 (5) | |

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## Question 11O(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = -9 \Rightarrow \mathbf{e} = \begin{pmatrix}1\\2\\-2\end{pmatrix}$ | M1A1 | Award M1A1 for any one |
| $\lambda_1 = 6 \Rightarrow \mathbf{e}_1 = \begin{pmatrix}-2\\2\\1\end{pmatrix}$ and $\lambda_2 = 3 \Rightarrow \mathbf{e}_2 = \begin{pmatrix}2\\1\\2\end{pmatrix}$ | A1 (3) | A1 for the other two |
| $\mathbf{Mx} = \mathbf{M}(a\mathbf{e}_1 + b\mathbf{e}_2) = a\mathbf{Me}_1 + b\mathbf{Me}_2$ | B1 | |
| $= a\lambda_1\mathbf{e}_1 + b\lambda_2\mathbf{e}_2$ | B1 | |
| $= 6a\mathbf{e}_1 + 3b\mathbf{e}_2$ | B1 (3) Total 14 | |