## Question 9:

| Answer | Mark | Guidance |
|--------|------|----------|
| For $C_1$: $\frac{dy}{dx} = 2x - 4 \rightarrow m = 2$ | B1 | |
| $y - \text{'their }4\text{'} = \text{'their m'}(x-3)$ or using $y = mx + c$ | M1 | Use of $\frac{dy}{dx}$ and $(3, \text{their }4)$ to find the tangent equation |
| $y - 4 = 2(x-3)$ or $y = 2x - 2$ | A1 | If using $y = mx+c$, getting $c = -2$ is enough |
| $2x - 2 = \sqrt{4x+k}\ (\rightarrow 4x^2 - 12x + 4 - k = 0)$ | *M1 | Forms an equation in one variable using tangent & $C_2$ |
| Use of $b^2 - 4ac = 0$ on a 3 term quadratic set to 0 | *DM1 | Uses 'discriminant = 0' |
| $144 = 16(4-k) \rightarrow k = -5$ | A1 | |
| $4x^2 - 12x + 4 - k = 0 \rightarrow 4x^2 - 12x + 9 = 0$ | DM1 | Uses $k$ to form a 3 term quadratic in $x$ |
| $x = \frac{3}{2}\ \left(or\ \frac{1}{2}\right),\ y = 1\ (\text{or } -1)$ | A1 | Condone 'correct' extra solution |
| **Alternative:** For $C_2$: $\frac{dy}{dx} = A(4x+k)^{-\frac{1}{2}}$ | *M1 | Finds $\frac{dy}{dx}$ for $C_2$ in the form $A(4x+k)^{-\frac{1}{2}}$ |
| At P: 'their $2$' $= A(4x+k)^{-\frac{1}{2}}$ $\rightarrow \left(x = \frac{1-k}{4}\ or\ 4x+k=1\right)$ | *DM1 | Equating 'their 2' to 'their $\frac{dy}{dx}$', simplify to form a linear equation linking $4x+k$ and a constant |
| $(2x-2)^2 = 4x+k \rightarrow (2x-2)^2 = 1 \rightarrow (4x^2 - 8x + 3 = 0)$ | DM1 | Using $y = 2x-2$, $y^2 = 4x+k$ and $4x+k=1$ (but not =0) to form 3 term quadratic in $x$ |
| $x = \frac{3}{2}\ \left(or\ \frac{1}{2}\right)$ and from $k = -5\ (or\ -1)$ | A1 | Needs correct values for $x$ and $k$ |
| from $y^2 = 4x+k,\ y = 1\ (\text{or } -1)$ | A1 | Condone 'correct' extra solution |

## Question 9:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 1$ (or $-1$) and $x = \frac{3}{2}$ (or $\frac{1}{2}$) | A1 | Needs correct values for $y$ and $x$ |
| From $4x + k = 1$, $k = -5$ (or $-1$) | A1 | Condone 'correct' extra solution |
| | **8** | |

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