## Question 6:

### Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| Correct shape: curve with correct curvature and straight line with negative gradient | B1 | Must start and end on the $x$-axis |
| $-1,\;1,\;3$ and $4k$ (or 0.6) labelled in correct place | B1 | |

### Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\int_{-1}^{1}k(x+1)^2\,dx + \int_{1}^{3}k(6-2x)\,dx = 1$ | M1 | Adding two areas and putting equal to 1 |
| $\int_{-1}^{1}k(x^2+2x+1)\,dx + \int_{1}^{3}k(6-2x)\,dx = 1$ | M1A1 | M1: attempting to integrate (at least one $x^n\to x^{n+1}$) or finding area of triangle; A1: correct integration $k\!\left(\frac{x^3}{3}+x^2+x\right)$ and $k(6x-x^2)$ or equivalents |
| $k\!\left[\frac{x^3}{3}+x^2+x\right]_{-1}^{1} + k\!\left[6x-x^2\right]_{1}^{3} = 1$ | | |
| $k\!\left[2\tfrac{1}{3}+\tfrac{1}{3}\right] + k[9-5] = 1$ | dM1 | Dependent on previous two M marks; using correct limits |
| $6\tfrac{2}{3}k = 1$ | | |
| $k = \frac{3}{20}$ | A1cso | **AG** |

### Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\int_{-1}^{x}k(x^2+2x+1)\,dx = k\!\left[\frac{x^3}{3}+x^2+x\right]_{-1}^{x}$ or $\left[\frac{k}{3}(x+1)^3\right]_{-1}^{x}$ | M1 | Attempt to integrate line 1 of $f(x)$ with correct limits, or with $+c$ substituting $-1$ and setting $=0$ |
| $= \frac{3}{20}\!\left(\frac{x^3}{3}+x^2+x+\frac{1}{3}\right)$ or $\frac{1}{20}(x+1)^3$ | | |
| $\int_{1}^{x}k(6-2x)\,dx + \int_{-1}^{1}k(x^2+2x+1)\,dx = k\!\left[6x-x^2\right]_{1}^{x}+\frac{2}{5}$ | M1 | Attempt to integrate line 2 of $f(x)$ with correct limits and adding $\frac{2}{5}$ or their $F(1)$; or with $+c$ substituting $3$ and setting $=1$ |
| $= \frac{3}{20}(6x-x^2-5)+\frac{2}{5}$ | | |
| $= \frac{9}{10}x - \frac{3}{20}x^2 - \frac{7}{20}$ | | |
| $F(x) = \begin{cases} 0 & x < -1 \\ \dfrac{3}{20}\!\left(\dfrac{x^3}{3}+x^2+x+\dfrac{1}{3}\right) & -1\leq x \leq 1 \\[6pt] \dfrac{9}{10}x - \dfrac{3}{20}x^2 - \dfrac{7}{20} & 1 < x \leq 3 \\ 1 & x>3 \end{cases}$ | B1, A1, A1 | B1: top and bottom rows correct; 1st A1: 2nd line of $F(x)$ with correct range; 2nd A1: 3rd line of $F(x)$ with correct range. Do not penalise $\leq$ instead of $<$ or $\geq$ instead of $>$ |

### Part (d):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{9}{10}x - \frac{3}{20}x^2 - \frac{7}{20} = 0.5$ | M1 | Setting their 2nd or 3rd line of $F(x) = 0.5$ |
| $3x^2 - 18x + 17 = 0$ | | |
| $x = \frac{18 \pm \sqrt{18^2 - 4\times3\times17}}{6}$ | dM1 | Solving a 3-term quadratic dependent on first M1; must be using their 3rd line of $F(x)$ |
| $x = 1.17$ only | A1 | condone awrt 1.17; must reject other solution (4.825....) |