## Question 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $mx = x-x^2 \Rightarrow m=1-x \Rightarrow x=\ldots$ or $y=\frac{y}{m}-\frac{y^2}{m^2} \Rightarrow m^2=m-y \Rightarrow y=\ldots$ | M1 | Attempts to eliminate either $x$ or $y$ and factors/cancels to get a linear equation and solve |
| $x=1-m$ and $y=m(1-m)$ | A1 | Both correct |

## Question 9(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x-x^2(-mx)\,dx = \frac{x^2}{2}-\frac{x^3}{3}\left(-m\frac{x^2}{2}\right)$ | M1 | $x^n \to x^{n+1}$ in at least one term |
| Area of $R_1 = \int_0^{1-m}\{x-x^2(-mx)\}\,dx = \frac{(1-m)^2}{2}-\frac{(1-m)^3}{3}\left(-m\frac{(1-m)^2}{2}\right)-0$ | M1 | Uses limits $1-m$ and $0$; subtracts (condone omission of $-0$) |
| Correct expression in $m$ with/without area under line subtracted | A1 | |
| Area of $R_1 = \int_0^{1-m}\{x-x^2-mx\}\,dx = \frac{(1-m)^2}{2}(1-m)-\frac{(1-m)^3}{3}(-0)$ | dM1 | Correct strategy for area (may be scored for finding separate areas and subtracting) |
| $= \frac{(1-m)^3}{6}$ | A1* | Correct expression |

## Question 9(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of $(R_1+R_2)=\int_0^1(x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{6}$ | M1 | Correct method for finding area of $R_1+R_2$, or correct method for finding area of $R_2$ |
| $R_1=R_2 \Rightarrow \frac{(1-m)^3}{6}=\frac{1}{12} \Rightarrow m=\ldots$ | dM1 | Sets up correct equation using answer to part (b) and solves for $m$ |
| $m = 1-\frac{1}{\sqrt[3]{2}}$ | A1 | Correct exact value in any form |

# Question (from first page - part b/c of unnamed question):

**Part (b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to integrate equation for curve $C$ to find area | M1 | May be part of integrating a difference of curve and line, or done separately |
| Applies limits of 0 and $x$ from (a) (in terms of $m$) to integral, subtracts correct way round | M1 | May or may not include line at this point |
| $\text{Area} = \frac{(1-m)^2}{2}(1-m) - \frac{(1-m)^3}{3}(-0)$ | A1 | Correct expression in $m$ for area under $C$ between 0 and $1-m$ |
| Correct overall strategy for area; for line use $\frac{1}{2} \times x \times y$ from part (a) | dM1 | Depends on previous M; curve minus line, or separate areas minus triangle |
| Fully correct work reaching expression of form $\frac{(1-m)^2}{2}(1-m) - \frac{(1-m)^3}{3}$ with $(1-m)^3$ terms before final answer | A1 | Going from $= \frac{(1-m)^2}{2} - \frac{(1-m)^3}{3} - m\frac{(1-m)^2}{2}$ to given answer with no intermediate step is A0 |

**Part (c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method for finding area of $R_1 + R_2$; integral of $C$ with limits 0 to 1; allow $\frac{1}{6}$ from correctly set up integral if integration not shown | M1 | Alternatively: correct method for area of $R_2$, integral of $C$ from $1-m$ to 1 with triangle area added |
| Sets up correct equation using answer to part (b), reaches value for $m$ | dM1 | — |
| $m = 1 - \dfrac{1}{\sqrt[3]{2}}$ | A1 | Correct exact value in any form |

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