# Question 7:

## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^2 kx\,dx = \left[\dfrac{kx^2}{2}\right]_0^2 = 2k = 1$, so $k = \tfrac{1}{2}$ | M1, A1 **2** | Use $\int_0^2 kx\,dx = 1$, or area of triangle; Correctly obtain $k = \tfrac{1}{2}$ **AG** |

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Sketch: straight line, positive gradient, through origin, truncated at $x=2$ | B1, B1 **2** | Straight line, positive gradient, through origin; Correct, some evidence of truncation, no need for vertical |

## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^2 \tfrac{1}{2}x^2\,dx = \left[\tfrac{1}{6}x^3\right]_0^2 = \tfrac{4}{3}$ | M1, A1 | Use $\int_0^2 kx^2\,dx$; $\tfrac{4}{3}$ seen or implied |
| $\int_0^2 \tfrac{1}{2}x^3\,dx = \left[\tfrac{1}{8}x^4\right]_0^2 = 2$ | M1, M1 | Use $\int_0^2 kx^3\,dx$; subtract their mean$^2$ |
| $2 - \left(\tfrac{4}{3}\right)^2 = \tfrac{2}{9}$ | A1 **5** | Answer $\tfrac{2}{9}$ or a.r.t. 0.222, c.a.o. |

## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Graph translated one unit to right; 1 and 3 indicated on axis | M1, A1$\sqrt{}$ **2** | Translate horizontally, allow stated, or "1, 2" on axis; One unit to right, 1 and 3 indicated, nothing wrong seen, no need for vertical or emphasised zero bits. [If in doubt as to $\rightarrow$ or $\downarrow$, M0] |

## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \tfrac{7}{3}$, Variance $= \tfrac{2}{9}$ | B1$\sqrt{}$, B1$\sqrt{}$ **2** | Previous mean $+1$; Previous variance. [If in doubt as to $\rightarrow$ or $\downarrow$, B1B1 in this part] |

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