## Question 14:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $8 - x = 14 + 3x - 2x^2$ or $y = 14 + 3(8-y) - 2(8-y)^2$ | M1 | Uses the given line and curve to obtain an equation in one variable |
| $2x^2 - 4x - 6 = 0 \Rightarrow x = \ldots$ or $2y^2 - 28y + 90 = 0 \Rightarrow y = \ldots$ | dM1 | Solves their 3TQ as far as $x = \ldots$ or $y = \ldots$ **Dependent on first method mark** |
| $x = -1,\ x = 3$ or $y = 5,\ y = 9$ | A1 | Correct $x$ values or correct $y$ values |
| $(-1, 9)\ \ (3, 5)$ | ddM1A1 | ddM1: Solves for $y$ or $x$ using at least one value; dependent on both previous method marks. A1: Correct coordinates (do not need to be paired) |
| Special case: Fully correct answers only with no working | M0M0A0M1A1 | |

**Total: (5)**

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### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = 0 \Rightarrow y = 8$ or $\int(8-x)\,dx = 8x - \dfrac{x^2}{2}$ or $\pm(\text{curve}-\text{line}) = \pm(14+3x-2x^2-(8-x))$ | B1 | Correct $y$-intercept (may be seen on diagram), or correct integration of $8-x$, or correct $\pm(\text{curve}-\text{line})$ |
| $14 + 3x - 2x^2 = 0 \Rightarrow x = 3.5$ | B1 | Correct value — may be seen on diagram |
| $\int(14 + 3x - 2x^2)\,dx = 14x + \dfrac{3x^2}{2} - \dfrac{2x^3}{3}\ (+c)$ | M1A1 | M1: $x^n \to x^{n+1}$ on at least two terms for the curve $C$ or their $\pm(\text{curve}-\text{line})$. A1: Correct integration (ignore $+c$) |

**Next two M marks — any one of the following combinations:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left[14x + \dfrac{3x^2}{2} - \dfrac{2x^3}{3}\right]_{``3"}^{``3.5"} = \dfrac{31}{24}$ | M1 1(i) | Correct use of limits "3" and "3.5" on integrated curve $C$ |
| $\left[8x - \dfrac{x^2}{2}\right]_0^{``3"} \text{ or } \tfrac{1}{2}\times``3"\times(``8"+``5")= \dfrac{39}{2}$ | M1 1(ii) | Correct method for trapezium/integration between $x=0$ and $x=3$ |
| $\left[14x + \dfrac{3x^2}{2} - \dfrac{2x^3}{3}\right]_0^{``3.5"} = \dfrac{931}{24}$ | M1 2(i) | Correct use of limits "3.5" and 0 on integrated curve |
| $\left[\pm(\text{curve}-\text{line})\right]_0^{``3"} = 18$ | M1 2(ii) | Correct use of limits "3" and 0 on integrated $\pm(\text{curve}-\text{line})$ |
| $\left[\pm(\text{line}-\text{curve})\right]_{``3"}^{``3.5"} = \dfrac{13}{12}$ | M1 3(i) | Correct use of limits "3" and "3.5" on integrated $\pm(\text{line}-\text{curve})$ |
| $\left[8x - \dfrac{x^2}{2}\right]_0^{``3.5"} \text{ or } \tfrac{1}{2}\times``3.5"\times(``8"+``4.5") = \dfrac{175}{8}$ | M1 3(ii) | Correct trapezium/integration between $x=0$ and $x=3.5$ |

| Working/Answer | Mark | Guidance |
|---|---|---|
| Area $R = \dfrac{39}{2} + \dfrac{31}{24} = \dfrac{499}{24}$ (Way 1) | dM1A1 | dM1: Correct combination — must be 1(i)+1(ii), or 2(i)$-$2(ii), or 3(ii)$-$3(i); dependent on **all** previous method marks. A1: $\dfrac{499}{24}$ or exact equivalent e.g. $20\dfrac{19}{24}$ |

**Total part (b): (8)**

**Total Question 14: (13)**