Standard +0.3 This is a straightforward modulus inequality requiring students to consider two cases (x ≥ 3 and x < 3), solve linear inequalities in each case, and combine the solutions. While it requires systematic case-work, the algebraic manipulation is routine and the question follows a standard template that students practice extensively.
Find \(x\)-coordinate of intersection with \(y = 2-5x\)
M1
Find point of intersection with \(y = 2\
Obtain \(x = -\dfrac{4}{3}\)
A1
State final answer \(x < -\dfrac{4}{3}\)
A1
Do not accept \(x < -1.33\); [Do not condone \(\leqslant\) for \(<\) in the final answer.]
Alternative Method:
Answer
Marks
Guidance
Answer
Mark
Guidance
State or imply non-modular inequality/equality \((2-5x)^2 >, \geqslant, =, 2^2(x-3)^2\), or corresponding quadratic equation, or pair of linear equations \((2-5x) >, \geqslant, =, \pm\, 2(x-3)\)
B1
Two correct linear equations only
Make reasonable attempt at solving a 3-term quadratic, or solve one linear equation, or linear inequality for \(x\)
M1
\(21x^2 + 4x - 32 = (3x+4)(7x-8) = 0\); \(2-5x\) or \(-(2-5x)\) with \(2(x-3)\) or \(-2(x-3)\)
Obtain critical value \(x = -\dfrac{4}{3}\)
A1
State final answer \(x < -\dfrac{4}{3}\)
A1
Do not accept \(x < -1.33\); [Do not condone \(\leqslant\) for \(<\) in the final answer.]
Total: 4 marks
## Question 1:
**Main Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Make a recognisable sketch graph of $y = 2\|x-3\|$ and the line $y = 2-5x$ | **B1** | Need to see correct V at $x = 3$, roughly symmetrical, $x = 3$ stated, domain at least $(-2, 5)$ |
| Find $x$-coordinate of intersection with $y = 2-5x$ | **M1** | Find point of intersection with $y = 2\|x-3\|$ or solve $2-5x$ with $2(x-3)$ or $-2(x-3)$ |
| Obtain $x = -\dfrac{4}{3}$ | **A1** | |
| State final answer $x < -\dfrac{4}{3}$ | **A1** | Do not accept $x < -1.33$; [Do not condone $\leqslant$ for $<$ in the final answer.] |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality/equality $(2-5x)^2 >, \geqslant, =, 2^2(x-3)^2$, or corresponding quadratic equation, or pair of linear equations $(2-5x) >, \geqslant, =, \pm\, 2(x-3)$ | **B1** | Two correct linear equations only |
| Make reasonable attempt at solving a 3-term quadratic, or solve one linear equation, or linear inequality for $x$ | **M1** | $21x^2 + 4x - 32 = (3x+4)(7x-8) = 0$; $2-5x$ or $-(2-5x)$ with $2(x-3)$ or $-2(x-3)$ |
| Obtain critical value $x = -\dfrac{4}{3}$ | **A1** | |
| State final answer $x < -\dfrac{4}{3}$ | **A1** | Do not accept $x < -1.33$; [Do not condone $\leqslant$ for $<$ in the final answer.] |
**Total: 4 marks**