# Question 9:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k = -1$ | B1 | Accept $x=-1$ or $x>-1$ |

## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0) = |2-4\ln(0+1)| = 2-0 = 2$ | B1 | Accept as $y$ coordinate; may be identified as $y$ or $f(0)$ or $(0,2)$ |

## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = 2-4\ln(x+1) \Rightarrow \ln(x+1) = \dfrac{1}{2} \Rightarrow x = e^p + q$ | M1 | Rearranges to $x=\ldots$ of correct form $e^p+q,\ q\neq 0$; may be implied by awrt 0.65 |
| $x = e^{\frac{1}{2}}-1$ | A1 | Accept $\sqrt{e}-1$ or $e^{\frac{2}{4}}-1$; may be seen as coordinate pair |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2-4\ln(x+1)=3 \Rightarrow \ln(x+1)=\ldots$ or $-2+4\ln(x+1)=3 \Rightarrow \ln(x+1)=\ldots$ | M1 | Forms one valid equation with modulus removed; accept $>$ or $<$ instead of $=$ for M |
| Both equations attempted to achieve values via $\ln(x+1)=\ldots$ | dM1 | Both equations attempted; suitable attempt to undo ln; allow error with base |
| CVs $e^{-\frac{1}{4}}-1,\ e^{\frac{5}{4}}-1$ | A1 | Both exact; decimals are $-0.221$ and $2.49$ to 3 s.f. |
| $\text{"}{{-1}\text{"}} < x < e^{-\frac{1}{4}}-1$ or $x > e^{\frac{5}{4}}-1$ | ddM1A1ft | ddM1: Chooses outside region; dependent on both previous M marks; requires two distinct critical values. A1ft: Must be exact; allow "and" instead of "or"; use of $\cap$ in set notation is A0 |