## Question 10:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{6}{7}x^2 + \frac{2}{7}x - \frac{5}{2}$ | M1 A1 | Finds $\frac{dy}{dx}$; look for at least two terms correct; need not be simplified; ISW after correct answer |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x = -\frac{7}{2}$, $\frac{dy}{dx} = \frac{6}{7}\left(-\frac{7}{2}\right)^2 + \frac{2}{7}\left(-\frac{7}{2}\right) - \frac{5}{2} = ...(= 7)$ | M1 | Substitutes $-\frac{7}{2}$ into $\frac{dy}{dx}$ to find gradient of $C$ at $A$ |
| So at $B$, $\frac{dy}{dx} = -\frac{1}{7}$ | M1 | Applies perpendicular condition to find gradient at $B$ |
| $\frac{6}{7}x^2 + \frac{2}{7}x - \frac{5}{2} = -\frac{1}{7}$ | dM1 | Equates $\frac{dy}{dx}$ to gradient of normal at $B$; depends on first M and a changed gradient |
| $\Rightarrow 12x^2 + 4x - 35 = -2 \Rightarrow 12x^2 + 4x - 33 = 0$* | A1* | Reaches given equation; any correct intermediate line shown following $\frac{6}{7}x^2 + \frac{2}{7}x - \frac{5}{2} = -\frac{1}{7}$ |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $12x^2 + 4x - 33 = 0 \Rightarrow (2x-3)(6x+11) = 0 \Rightarrow x = ...$ | M1 | Solves the given quadratic |
| From graph $x$ coordinate is positive, so $x = \frac{3}{2}$ at $B$ | A1 | |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of $l$ is $y = -\frac{1}{7}x - 1$ | M1 | |
| Finds coordinates of $A$: $x = -\frac{7}{2} \Rightarrow y = -\frac{1}{7}\times-\frac{7}{2}-1 = \left(-\frac{1}{2}\right)$ | dM1 | |
| Substitutes $x = -\frac{7}{2}$, $y = -\frac{1}{2}$ into $y = \frac{2}{7}x^3 + \frac{1}{7}x^2 - \frac{5}{2}x + k \Rightarrow k = ...$ | ddM1 | |
| $k = \frac{5}{4}$ CSO | A1 | |

## Question (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| Any valid method to solve the quadratic (factorisation, completing square, formula or calculator) | **M1** | May be awarded for work in (b) |
| Correct coordinate $x_B = \frac{3}{2}$ given with reason referencing the sketch | **A1** | e.g. cannot be $-\frac{11}{6}$ as that is negative; condone reasons like "because $B$ is positive" |

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## Question (d):

**Main Method:** Find equation for $l$, then find coordinates for $A$ or $B$, then sub coordinates into equation for $C$ to find $k$

| Answer | Mark | Guidance |
|--------|------|----------|
| Uses gradient of $l$ and intercept $-1$ to form equation of $l$ | **M1** | Gradient must be result of a changed $\frac{dy}{dx}$ at $x = -\frac{7}{2}$; cannot be just made up |
| Finds coordinates of either $A$ or $B$ using equation for $l$ and either $x_A = -\frac{7}{2}$ or $x_B = \frac{3}{2}$ | **dM1** | FYI: $x=-\frac{7}{2} \Rightarrow y = -\frac{1}{7} \times -\frac{7}{2} - 1 = \left(-\frac{1}{2}\right)$; $x=\frac{3}{2} \Rightarrow y = -\frac{1}{7} \times \frac{3}{2} - 1 = \left(-\frac{17}{14}\right)$ |
| Full method to solve for $k$ by substituting coordinates of $A$ or $B$ into equation for $C$ | **ddM1** | e.g. $x=\frac{3}{2}$, $y=-\frac{17}{14}$ into $y = \frac{2}{7}x^3 + \frac{1}{7}x^2 - \frac{5}{2}x + k \Rightarrow k = \ldots$ |
| $k = \frac{5}{4}$ | **A1** | CSO |

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**Alt Method I (d):**

| Answer | Mark | Guidance |
|--------|------|----------|
| At $A$: $y = \frac{2}{7}\left(-\frac{7}{2}\right)^3 + \frac{1}{7}\left(-\frac{7}{2}\right)^2 - \frac{5}{2}\left(-\frac{7}{2}\right) + k = \ldots\left(= k - \frac{7}{4}\right)$ | **M1** | Substitutes $x = -\frac{7}{2}$ or $x = \frac{3}{2}$ into equation for $C$; finds $y$ coordinate in terms of $k$ |
| Equation of $l$ is $y - \left(k - \frac{7}{4}\right) = -\frac{1}{7}\left(x + \frac{7}{2}\right)$ or $y$ intercept is $-\frac{1}{7} \times \frac{7}{2} + k - \frac{7}{4}$ | **dM1** | Uses gradient of normal at $A$ (or tangent at $B$) and their $y$ coordinate to find equation of $l$ or expression for intercept |
| $\Rightarrow y = -\frac{1}{7}x + k - \frac{9}{4} \Rightarrow k - \frac{9}{4} = -1 \Rightarrow k = \ldots$ | **ddM1** | Sets their intercept to $-1$ and solves for $k$ |
| $k = \frac{5}{4}$ | **A1** | |

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**Alt Method II (d):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt at equation for $l$; score for $y = -\frac{1}{7}x - 1$ | **M1** | |
| Equate equation for $l$ with equation for $C$; use fact that a root is known: $-\frac{1}{7}x - 1 = \frac{2}{7}x^3 + \frac{1}{7}x^2 - \frac{5}{2}x + k \Rightarrow 4x^3 + 2x^2 - 33x + 14k + 14 = 0$ | **dM1** | |
| Set $g\!\left(\pm\frac{3}{2}\right) = 0$ or $g\!\left(\pm\frac{7}{2}\right) = 0$ where $g(x) = 4x^3 + 2x^2 - 33x + 14k + 14$ | **ddM1** | Must lead to a value for $k$ |
| $k = \frac{5}{4}$ | **A1** | |