Standard +0.8 This requires solving an inequality involving two modulus expressions, necessitating systematic case analysis across multiple critical points (x = 5/2 and x = -1/2), then solving quadratic inequalities in each region and combining solutions. More demanding than routine single-modulus problems but follows standard A-level technique.
State or imply non-modular inequality \((2x - 5)^2 > (3(2x + 1))^2\), or corresponding quadratic equation, or pair of linear equations \((2x - 5) = \pm 3(2x + 1)\)
B1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for \(x\)
M1
Obtain critical values \(-2\) and \(\frac{1}{4}\)
A1
State final answer \(-2 < x < \frac{1}{4}\)
A1
[4]
OR: Obtain critical value \(x = -2\) from a graphical method, or by inspection, or by solving a linear equation or inequality
B1
Obtain critical value \(x = \frac{1}{4}\) similarly
B2
State final answer \(-2 < x < \frac{1}{4}\)
B1
[Do not condone \(<\) for \(<\)]
State or imply non-modular inequality $(2x - 5)^2 > (3(2x + 1))^2$, or corresponding quadratic equation, or pair of linear equations $(2x - 5) = \pm 3(2x + 1)$ | B1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for $x$ | M1 |
Obtain critical values $-2$ and $\frac{1}{4}$ | A1 |
State final answer $-2 < x < \frac{1}{4}$ | A1 |
| [4] |
OR: Obtain critical value $x = -2$ from a graphical method, or by inspection, or by solving a linear equation or inequality | B1 |
Obtain critical value $x = \frac{1}{4}$ similarly | B2 |
State final answer $-2 < x < \frac{1}{4}$ | B1 |
[Do not condone $<$ for $<$] | |