# Question 4:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^a k(a-x)^2\,dx = \left[k\left(a^2x - ax^2 + \frac{x^3}{3}\right)\right]_0^a$ or $\left[\frac{-k(a-x)^3}{3}\right]_0^a$ | M1 A1 | Integrating $f(x)$ at least 1 term correct. For M1 allow $\frac{\pm k(a-x)^3}{3}$; A1 correct integration (ignore limits) |
| $k\left(a^3 - a^3 + \frac{a^3}{3}\right) = 1$ or $\frac{ka^3}{3} = 1 \Rightarrow ka^3 = 3$ | A1 cso | Substitute limits and equate to 1 to form one expression in terms of $k$ and $a$ leading to $ka^3=3$ |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^a kx(a-x)^2\,dx = \left[k\left(\frac{a^2x^2}{2} - \frac{2ax^3}{3} + \frac{x^4}{4}\right)\right]_0^a$ or $\left[\frac{-kx(a-x)^3}{3} + \frac{k(a-x)^4}{12}\right]_0^a$ | M1 A1 | Indicating they are integrating $xf(x)$ with an attempt at integrating $x^n \to x^{n+1}$; A1 correct integration |
| $k\left(\frac{a^2 \cdot a^2}{2} - \frac{2a \cdot a^3}{3} + \frac{a^4}{4}\right) = 1.5$ or $\frac{ka^4}{12} = 1.5$ or $ka^4 = 18$ oe | dM1 | (dep on previous M1). Substitute limits and equate to 1.5 to form a 2nd expression in terms of $k$ and $a$ |
| $\frac{ka^4}{ka^3} = 6$ or $\frac{18}{3} = 6$ $[\therefore a=6]$ | A1cso | Correct method shown to solve their 2 equations to eliminate $k$ and show $a=6$ |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(x) = \frac{1}{72}\left(36x - 6x^2 + \frac{x^3}{3}\right)$, $\frac{1}{72}\left(36x - 6x^2 + \frac{x^3}{3}\right) = 0.5$ oe | M1 | Finding correct $F(x)$. Allow $F(x) = 1 - \frac{(6-x)^3}{216}$ but $F(x) = \frac{(6-x)^3}{216}$ is M0; allow in terms of $k$ |
| $F(1.15)(= 0.47\ldots)$ and $F(1.25)(= 0.5038\ldots)$, $1.2377\ldots$ | M1 | For attempting their $F(1.15)$ and their $F(1.25)$ or a suitable tighter interval or for 'solving' cubic leading to a value awrt 1.24 |
| $F(1.15) = \text{awrt } 0.47$, $F(1.25) = \text{awrt } 0.504$; $(0.47(18\ldots) < 0.5 < 0.503(8\ldots))$ therefore the median is **1.2** to 1 decimal place | A1 | Both correct values and correct conclusion (allow $x=1.2$) or awrt 1.24 and correct conclusion (allow $x=1.2$). Allow change of sign argument if they have subtracted 0.5 |

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