## Question 12:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = k(x+1)(x-2)$ | M1 | Uses product of linear terms in an equation. Condone $k=1$ used |
| When $x=0$, $f(x)=-4$ so $k=2$ | B1 | Allow if $2x^2$ or $k=2$ seen |
| So $f(x) = 2x^2 - 2x - 4$ | A1 | Correct expanded expression |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-1,0)$: $0 = a - b + c$; $(2,0)$: $0 = 4a + 2b + c$ | M1 | Setting up simultaneous equations for at least $a$ and $b$ |
| $c = -4$ | B1 | Allow for correct constant term or explicit value for $c$ |
| $a=2, b=-2$ gives $f(x) = 2x^2 - 2x - 4$ | A1 | Correct expanded expression |

**[3 marks total]**

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### Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \int (2x^2 - 2x - 4)\, dx$ | M1 | Attempt to integrate their $f(x)$ |
| $y = \frac{2}{3}x^3 - x^2 - 4x + c$ | A1FT | FT their $f(x)$. Condone missing $+c$ and missing $y=$ |
| When $x=0$, $y=8$ | M1 | Uses $(0,8)$ to evaluate $c$. May be implied by correct constant term |
| So $y = \frac{2}{3}x^3 - x^2 - 4x + 8$ | A1 | Cao $y=$ must be seen |

**[4 marks total]**

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### Part (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 0$ when $x = -1, 2$ | M1 | Sets $f(x)$ or the derivative of their $y$ to zero and attempts to solve |
| $x = -1 \Rightarrow y = \frac{31}{3}$, so $\left(-1, \frac{31}{3}\right)$ | A1 | cao |
| $x = 2 \Rightarrow y = \frac{4}{3}$, so $\left(2, \frac{4}{3}\right)$ | A1 | cao |

**[3 marks total]**

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## Appendix — Exemplar responses for Q3 last mark:

| Response | Mark |
|----------|------|
| $8\left(k^3 + \frac{1}{2}\right)$ so not a multiple of 8 | A1 |
| $8\left(k^3 + \frac{1}{2}\right)$ so 8 is not a factor | A0 |
| $8k^3 + 4$ so 8 is not a factor so not a multiple of 8 | A1 |
| $8k^3 + 4$ so 8 is not a factor | A0 |
| $8k^3 + 4$ cannot take a factor of 8 so it's not a multiple of 8 | A1 |
| $8k^3 + 4$ is not a multiple of 8, so not a multiple of 8 | A1 |
| $8k^3$ is a multiple of 4 and 8 but $+4$ means multiple of 4 and not 8 | A1 |
| $\frac{8k^3+4}{8} = k^3 + \frac{1}{2}$ so not a multiple of 8 | A1 |

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## Appendix — Exemplar responses for Q9(e):

| Response | Mark |
|----------|------|
| "The velocity might be 0 and it doesn't stop" | B0 |
| "The velocity would be negative" on its own | B0 |
| "The velocity would be negative which is impossible" | B0 |
| "The velocity would be negative and velocity can't be negative" | B0 |
| "Because the model would represent the train going backwards" | B1 |
| "The velocity would be negative which means going backwards" | B1 |
| "The velocity would be negative which means going backwards which it can't do" | B1 |
| "The velocity would be negative which means going backwards which it says it doesn't do in this question" | B1 |
| "It would give an ever increasing negative velocity" | B0 |
| "The model predicts travelling infinitely fast which is not possible" | B1 |
| "The model predicts travelling infinitely fast in the opposite direction which is not possible" | B1 |