# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2, -10)$ | B1 B1 | First B1 for one correct coordinate; second B1 for full correct pair. Allow $x=..., y=...$; do not accept e.g. 6/3 unless 2 identified |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{ff}(0) = \text{f}(-4) = ...$ | M1 | Full attempt at ff(0); may score for f(-4); allow slips with modulus if intent clear |
| $= 8$ | A1cso | ff(0) = 8 only; A0 if other values given |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $-3(x-2)-10 = 5x+10 \Rightarrow x = ...$ | M1 | Allow equality or any inequality; alternatively rearranges to $|x-2| = ax+b$, squares and solves |
| $x > -\frac{7}{4}$ only | A1 | oe only; if another inequality or value given and not rejected, withhold mark |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\left(\text{or }|x|\right) = \frac{16}{3}$ | B1 | Allow when seen even from incorrect working; allow also for $|x| = \frac{16}{3}$ |
| Attempts $3(|x|-2)-10=0 \Rightarrow |x|=k$, $k>0$ or $3(-x-2)-10=0 \Rightarrow x=-k$ or $3(x-2)-10=0 \Rightarrow x=k \Rightarrow x=-k$ | M1 | Correct method to find root on negative $x$-axis; allow missing brackets |
| $x = \frac{16}{3}$ and $-\frac{16}{3}$ with no other values | A1 | Must give negative value, not just $|x|=\frac{16}{3}$; may be on sketch if referred to in (d) |

---