# Question 5(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| V shaped graph | B1 | |
| Touches $x$-axis at $\frac{3}{4}$ and cuts $y$-axis at $3$ | B1 | |

# Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $4x-3=2-2x$ or $3-4x=2-2x$ to give either value of $x$ | M1 | |
| Both $x=\frac{5}{6}$ and $x=\frac{1}{2}$, or $x>\frac{5}{6}$ or $x<\frac{1}{2}$ | A1 | |
| $x<\frac{1}{2}$ or $x>\frac{5}{6}$ | dM1 A1 | |

# Question 5(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Draws graph or solves $|4x-3|=1\frac{1}{2}-2x$ to give one solution $x=\frac{3}{4}$ | M1 | |
| All values of $x$ except $x=\frac{3}{4}$, i.e. $x\neq\frac{3}{4}$, or $x<\frac{3}{4}$, $x>\frac{3}{4}$ | A1 | |

# Question 5 (Absolute Value/Modulus):

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| V-shaped graph | B1 | Must be V-shape, not a curve. Accept V shape with dotted extension of $y = 4x-3$ under x-axis |
| Meets x-axis at $x = \frac{3}{4}$, crosses y-axis at $y = 3$ | B1 | $A = \left(\frac{3}{4}, 0\right)$ and $B = (0,3)$. Condone $\left(0, \frac{3}{4}\right)$ and $(3,0)$ marked on correct axis |

## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts to solve $|4x-3| \ldots 2-2x$, finding at least one solution | M1 | Accept $\pm4x \pm 3 = 2-2x \Rightarrow x = ..$ or $\pm4x \pm 3 > 2-2x \Rightarrow x > ..$ or $x < ..$ |
| Both critical values $x = \frac{5}{6}$ and $x = \frac{1}{2}$ | A1 | Accept $x = 0.83$ and $x = 0.5$. Or one inequality $x > \frac{5}{6}$ or $x < \frac{1}{2}$ |
| Selects outside region | dM1 | Dependent on previous M. E.g. if $x = 0.83$ and $x = -2.5$ found, correct application gives $x < -2.5, x > 0.83$ |
| $x < \frac{1}{2}$ or $x > \frac{5}{6}$ | A1 | Accept $x < 0.5, x > 0.83$ |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sketch both lines showing single intersection at $x = \frac{3}{4}$, **or** solves $|4x-3| = 1\frac{1}{2} - 2x$ using both $4x-3 = 1\frac{1}{2}-2x$ and $-4x+3 = 1\frac{1}{2}-2x$ giving one solution | M1 | If two values obtained it is M0A0 |
| Solution set is all values apart from $x = \frac{3}{4}$ | A1 | Accept: all values of $x$ except $x = \frac{3}{4}$, or $x \in \mathbb{R},\ x \neq \frac{3}{4}$, or $x < \frac{3}{4},\ x > \frac{3}{4}$ |

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