## Question 7:

### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = -\frac{16}{3}(3x-k)^{-2}$ | M1A1 | M1: Differentiates to form $A(3x-k)^{-2}$. A1: $-\frac{16}{3}(3x-k)^{-2}$. Not $-432(27x-9k)^{-2}$ or $-\frac{432}{(27x-9k)^2}$ as common factor exists. |

### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $-\frac{16}{3}(3x-k)^{-2} = -12 \Rightarrow (3x-k)^2 = \ldots$ | M1 | Sets derivative equal to $-12$ (not 12) and rearranges. Condone poor squaring. |
| $3 - k = \pm\frac{2}{3} \Rightarrow k = \ldots$ | dM1 | Substitutes $x=1$ and solves for 2 values of $k$. **Depends on having obtained $A < 0$.** FYI correct 3TQ is $9k^2 - 54k + 77 = 0$. |
| $k = \frac{7}{3},\ \frac{11}{3}$ | A1 | Accept equivalent exact fractions or recurring decimals $2.\dot{3},\ 3.\dot{6}$ but not rounded decimals. |

### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \frac{16}{9\left(3(1)-\frac{7}{3}\right)}$ | M1 (B1 on ePEN) | Uses $k$ from (b) (or 'made up' $k$) and $x=1$ to find $y$ at $P$. |
| $y - \frac{8}{3} = \frac{1}{12}(x-1)$ | dM1 | Attempts normal equation using their $y$-coordinate with gradient $\frac{1}{12}$. If $y=mx+c$ must reach $c = \ldots$ Depends on previous M mark. |
| $12y - x - 31 = 0$ | A1 | Or any equivalent integer multiple. Note: if $k=\frac{11}{3}$ used, get $x - 12y - 33 = 0$, scores 110. |

### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\frac{16}{9(3x-k)}\,dx = \frac{16}{27}\left[\ln(3x-k)\right]$ | M1 | Integrates to form $B\ln(3x-k)$. |
| $= \frac{16}{27}\left[\ln\!\left(3x-\frac{7}{3}\right)\right]$ | A1ft | Follow through their $k$; allow letter $k$ if $k$ not less than 3. Ignore $+c$. Also correct: $\frac{16}{27}\left[\ln(27x-21)\right]$ for $k=\frac{7}{3}$. |
| $\frac{16}{27}\left[\ln\!\left(3x-\frac{7}{3}\right)\right]_1^3 = \frac{16}{27}\!\left(\ln\!\left(3(3)-\frac{7}{3}\right) - \ln\!\left(3(1)-\frac{7}{3}\right)\right)$ | dM1 | Substitutes limits 3 and 1 and subtracts either way. Must have numeric $k$. Depends on first M mark. |
| $= \frac{16}{27}\ln(10)$ | A1 | Or exact equivalent e.g. $\frac{32}{54}\ln(10)$. **Use of incorrect $k$ in (d) scores maximum M1A1ft dM1A0.** Note if $k=\frac{11}{3}$: get $\frac{16}{27}\ln(8)$, scores 1110. |

**Substitution note:** $u = 3x-k \Rightarrow \frac{du}{dx}=3 \Rightarrow \int\frac{16}{9(3x-k)}\,dx = \frac{16}{9}\int\frac{1}{u}\cdot\frac{1}{3}\,du = \frac{16}{27}\ln u$. Score M1 for integrating to correct form $k\ln u$, A1 for $\frac{16}{27}\ln u$, then dM1 for applying correct changed limits or reverting to $x$ and using 3 and 1.

## Question 7:

**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{16}{9(3x-k)} = \frac{16}{27x-9k} \Rightarrow \frac{dy}{dx} = -\frac{432}{(27x-9k)^2}$ | M1A0 | |

**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\frac{432}{(27x-9k)^2} = -12 \Rightarrow 12(27x-9k)^2 = 432$ | M1 | |
| $12(27x-9k)^2 = 432 \Rightarrow (27x-9k)^2 = 36 \Rightarrow 27-9k = \pm 6 \Rightarrow k = \ldots$ | dM1 | |
| $k = \frac{7}{3}, \frac{11}{3}$ | A1 | |

**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{16}{27-21}$ | M1 | B1 on ePEN |
| $y - ``\frac{8}{3}`` = \frac{1}{12}(x-1)$ | dM1 | |
| $12y - x - 31 = 0$ | A1 | |

**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{16}{27x-9k}\,dx = \frac{16}{27}[\ln(27x-9k)]$ | M1 | |
| $= \frac{16}{27}[\ln(27x-21)]$ | A1ft | |
| $\frac{16}{27}[\ln(27x-21)]_1^3 = \frac{16}{27}(\ln(27(3)-21) - \ln(27-21))$ | dM1 | |
| $= \frac{16}{27}\ln(10)$ | A1 | |

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