# Question 3:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{f(x)\}$: a horizontal line drawn above the $x$-axis in the first quadrant | B1 | — |
| Labels of $c$, $2c$ and $\frac{1}{c}$ marked on the graph | dB1 | Dependent on first B mark; ignore $\{O\}$, $\{x\}$ and $\{f(x)\}$; allow label $\frac{1}{2c-c}$ as alternative to $\frac{1}{c}$ |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \frac{3c}{2}$ | B1 | Simplified or un-simplified; can be implied; B1 can be given for un-simplified $\left(\frac{(2c)^2}{c}\right) - \left(\frac{c^2}{c}\right)$ or $\frac{3c^2}{2c}$ or $2c - \frac{c}{2}$ etc.; $\int_c^{2c} \frac{1}{c}x\,dx$ or $\left[\frac{x^2}{2c}\right]_c^{2c}$ not sufficient |
| $\left\{E(X^2) =\right\} \int_c^{2c} \left(\frac{1}{2c-c}\right)x^2\,\{dx\}$ | M1 | $\int_c^{2c} x^2 f(x)\{dx\}$ where $f(x)$ equivalent to $\frac{1}{c}$; limits required; must have limits of $2c$ and $c$ |
| $= \left[\frac{1}{c}\left(\frac{x^3}{3}\right)\right]_{\{c\}}^{\{2c\}}$ | M1 | $\pm Ag(c)x^2 \to \pm Bg(c)x^3,\ A \neq 0,\ B \neq 0$; limits not required for this mark |
| $= \left(\frac{(2c)^3}{3c} - \frac{c^3}{3c}\right) \left\{= \frac{7c^2}{3}\right\}$ | dM1 | Dependent on first M; applies limits of $2c$ and $c$ to an integrated function in $x$ and subtracts correct way round |
| $\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{7c^2}{3} - \left(\frac{3c}{2}\right)^2$ | dM1 | Dependent on first M; applying variance formula correctly with their $E(X)$ |
| $= \frac{c^2}{12}$ * | A1 | Correct proof |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $X > 2(2c - X)$ | M1 | Correct un-simplified (or simplified) inequality statement; can be implied by $X > \frac{4c}{3}$; "P" not required for this mark |
| $\Rightarrow X > 4c - 2X \Rightarrow 3X > 4c \Rightarrow X > \frac{4c}{3}$ | dM1 | Dependent on first M; rearranges $X > 2(2c-X)$ to give $X>\ldots$ or $X<\ldots$; also allow $1-P\!\left(X < \pm\alpha c\right)$ or $1-P\!\left(X > \pm\alpha c\right),\ \alpha\neq 0$ |
| $\left\{P(X > 2(2c-X)) = P\!\left(X > \frac{4c}{3}\right)\right\} = \frac{2}{3}$ | A1 | $\frac{2}{3}$ or $\frac{4}{6}$ or $0.\dot{6}$; give M1M1A1 for final answer of $\frac{2}{3}$ from any working; give M2 for either $X>\frac{4c}{3}$ or $P\!\left(X>\frac{4c}{3}\right)$ or $1-P\!\left(X<\frac{4c}{3}\right)$ |

# Question 3:

## Part (b) – Alternative Method 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| Implied $E(X) = \frac{3c}{2}$ | B1 | |
| $\int_{c}^{2c} x^2 f(x)\{dx\}$ where $f(x)$ equivalent to $\frac{1}{c}$ | 1st M1 | Limits are required |
| $\int_{c}^{2c}\left(\frac{1}{2c-c}\left(x-\frac{3}{2}c\right)^2\right)\{dx\}$ | 4th dM1 | Applies $\int_{c}^{2c} f(x)\left(x-\frac{3c}{2}\right)^2\{dx\}$, $f(x)$ equivalent to $\frac{1}{c}$, limits required |
| $=\frac{1}{c}\left[\frac{1}{3}\left(x-\frac{3c}{2}\right)^3\right]_{\{c\}}^{\{2c\}}$ | 2nd M1 | $\pm Ag(c)(x-\delta)^2 \to \pm Bg(c)(x-\delta)^3$, $A,B,\delta \neq 0$, ignore limits |
| $=\frac{1}{3c}\left(\left(\frac{c}{2}\right)^3 - \left(-\frac{c}{2}\right)^3\right)$ | 3rd dM1 | Dependent on first M1. Applies limits of $2c$ and $c$, subtracts correct way round |
| $=\frac{1}{3c}\left(\frac{c^3}{4}\right) = \frac{c^2}{12}$ * | A1 | Correct proof |

**[6]**

## Part (b) – Alternative Method 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{c}^{2c}\left(\frac{1}{2c-c}\left(x-\frac{3}{2}c\right)^2\right)\{dx\}$ | B1, 1st M1, 4th dM1 | Award as in Alt. Method 1 |
| $=\frac{1}{c}\int_{c}^{2c}\left(x^2-3cx+\frac{9}{4}c^2\right)\{dx\}$ | | |
| $=\frac{1}{c}\left[\frac{1}{3}x^3-\frac{3}{2}cx^2+\frac{9}{4}c^2x\right]_{\{c\}}^{\{2c\}}$ | 2nd M1 | $\pm Ag(c)(x-\delta)^2 \to \pm Bg(c)(\pm\alpha x^3 \pm \beta x^2 \pm \delta x)^3$, $A,B,\alpha,\beta,\delta \neq 0$, ignore limits |
| $=\frac{1}{c}\left(\left(\frac{1}{3}(2c)^3-\frac{3}{2}c(2c)^2+\frac{9}{4}c^2(2c)\right)-\left(\frac{1}{3}(c)^3-\frac{3}{2}c(c)^2+\frac{9}{4}c^2(c)\right)\right)$ | 3rd dM1 | As earlier |
| $=\frac{1}{c}\left(\left(\frac{8}{3}c^3-6c^3+\frac{9}{2}c^3\right)-\left(\frac{1}{3}c^3-\frac{3}{2}c^3+\frac{9}{4}c^3\right)\right)$ | | |
| $=\frac{1}{c}\left(\left(\frac{7}{6}c^3\right)-\left(\frac{13}{12}c^3\right)\right)=\frac{1}{c}\left(\frac{c^3}{12}\right)$ | | |
| $=\frac{c^2}{12}$ * | A1 | Correct proof |

**[6]**

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