## Question 8:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $q(x) = x+2$ | B1 | For correct $q(x)$ **soi oe** |
| $y = \dfrac{A}{x+2} + \frac{1}{2}x + 1$ | M1 | For expressing $y$ in this form. Allow $cx+d$ for $A$ |
| $\left(-1, \frac{17}{2}\right) \Rightarrow A = 8$ | A1 | For correct $A$ |
| $y = \dfrac{\frac{1}{2}x^2 + 2x + 10}{x+2} \Rightarrow p(x) = \tfrac{1}{2}x^2 + 2x + 10$ | A1 | For correct $p(x)$. Allow equal multiples of $p(x)$ and $q(x)$ |

**[4]**

**Alternative:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $q(x) = x+2$ | B1 | For correct $q(x)$ **soi oe** |
| $y = \dfrac{ax^2+bx+c}{q(x)} = ax + (b-2a) + \dfrac{c-2b+4a}{x+2}$ | M1 | For division by their $q(x)$; 1st line of division and 1st term in quotient should be seen for correct method |
| $y = \frac{1}{2}x+1 \Rightarrow a=\frac{1}{2},\ b=2$ | A1 | For correct $a$ and $b$ **oe** |
| $\left(-1,\frac{17}{2}\right) \Rightarrow c-2b+4a = 8 \Rightarrow c=10$ | A1 | For correct $c$ **oe** |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}x^2 + (2-y)x + 10 - 2y = 0$ | M1 | For attempt to rearrange as quadratic in $x$ |
| $b^2 - 4ac \geq 0 \Rightarrow (2-y)^2 \geq 2(10-2y)$ | M1 | For use of $b^2-4ac$ ($\leq$ or $\geq$ or $=$ or $<$ or $>$) |
| $\Rightarrow y^2 \geq 16 \Rightarrow \{y \leq -4,\ y \geq 4\}$ | A1 | For critical values $\pm 4$ |
| *(pto for alternative)* | A1 | For correct range. (Must be $\leq$ and $\geq$) **www** |

**[4]**

**Alternative to 8(ii):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = \dfrac{(x+2)(x+2) - \left(\frac{1}{2}x^2+2x+10\right)}{(x+2)^2}$ | M1 | Differentiation using quotient rule |
| $= 0$ when $(x+2)^2 = \frac{1}{2}x^2+2x+10$ | M1 | Attempt to find solution using $\frac{dy}{dx}=0$ |
| $\Rightarrow \frac{1}{2}x^2+2x-6=0 \Rightarrow x^2+4x-12=0$ | | |
| $\Rightarrow (x+6)(x-2)=0$ | | |
| $\Rightarrow x=2,\ y=4 \quad x=-6,\ y=-4$ | A1 | |
| $\{y \leq -4,\ y \geq 4\}$ | A1 | For correct range. (Must be $\leq$ and $\geq$) **www** |

**Alternatively:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{1}{2}x+1+\dfrac{8}{x+2}$ | | |
| $\Rightarrow \dfrac{dy}{dx} = \frac{1}{2} - \dfrac{8}{(x+2)^2}$ | M1 | Differentiation using chain rule |
| $= 0$ when $\frac{1}{2} - \dfrac{8}{(x+2)^2} \Rightarrow (x+2)^2 = 16$ | M1 | Attempt to find solution using $\frac{dy}{dx}=0$ |
| $\Rightarrow x+2=\pm4 \Rightarrow x=2$ or $-6$ | | |
| $\Rightarrow y=4$ or $-4$ | A1 | For correct range. (Must be $\leq$ and $\geq$) **www** |
| $\{y \leq -4,\ y \geq 4\}$ | A1 | |

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{1}{2}x+1\right)^2 = \dfrac{\frac{1}{2}x^2+2x+10}{x+2}$ **OR** $y^2 = \dfrac{4}{y}+y$ | B1ft | For a correct equation derived from intersection of $C_2$ with $y=\frac{1}{2}x+1$ FT from (i) |
| $\Rightarrow x^3+4x^2+4x-32=0$ **OR** $y^3-y^2-4=0$ | M1 | For obtaining a cubic |
| | A1 | Correct cubic |
| $\Rightarrow (2,\ 2)$ | A1 | Coordinates correct **www** |

**[4]**

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