Moderate -0.3 This is a straightforward modulus inequality requiring students to consider two cases (x ≥ 1 and x < 1), solve linear inequalities in each case, and combine the solutions. It's slightly easier than average as it involves only basic algebraic manipulation with no complex reasoning, though the case-work structure requires some care.
State or imply non-modular inequality \((2x)^2 > (x-1)^2\), or corresponding equation
B1
Expand and make a reasonable solution attempt at a 2- or 3-term quadratic
M1
Obtain critical value \(x = \frac{1}{3}\)
A1
State answer \(x > \frac{1}{3}\) only
A1
OR:
Answer
Marks
State the relevant critical linear equation, i.e. \(2x = 1 - x\)
B1
Obtain critical value \(x = \frac{1}{3}\)
B1
State answer \(x > \frac{1}{3}\)
B1
State or imply by omission that no other answer exists
B1
OR:
Answer
Marks
Guidance
Obtain the critical value \(x = \frac{1}{3}\) from a graphical method, or by inspection, or by solving a linear inequality
B2
State answer \(x > \frac{1}{3}\)
B1
State or imply by omission that no other answer exists
B1
4
**EITHER:**
| State or imply non-modular inequality $(2x)^2 > (x-1)^2$, or corresponding equation | B1 |
| Expand and make a reasonable solution attempt at a 2- or 3-term quadratic | M1 |
| Obtain critical value $x = \frac{1}{3}$ | A1 |
| State answer $x > \frac{1}{3}$ only | A1 |
**OR:**
| State the relevant critical linear equation, i.e. $2x = 1 - x$ | B1 |
| Obtain critical value $x = \frac{1}{3}$ | B1 |
| State answer $x > \frac{1}{3}$ | B1 |
| State or imply by omission that no other answer exists | B1 |
**OR:**
| Obtain the critical value $x = \frac{1}{3}$ from a graphical method, or by inspection, or by solving a linear inequality | B2 |
| State answer $x > \frac{1}{3}$ | B1 |
| State or imply by omission that no other answer exists | B1 | 4 |