Standard +0.8 This requires squaring both sides to eliminate moduli, expanding and rearranging a quadratic inequality, then carefully checking critical points where expressions inside moduli change sign. More sophisticated than routine modulus inequalities which typically involve a single modulus compared to a constant, requiring systematic case analysis and algebraic manipulation beyond standard A-level fare.
State or imply non-modular inequality \((2x-3)^2 > 4^2(x+1)^2\), or corresponding quadratic equation, or pair of linear equations \((2x-3) = \pm4(x+1)\)
B1
\(12x^2 + 44x + 7 < 0\)
Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for \(x\)
M1
Correct method seen, or implied by correct answers
Obtain critical values \(x = -\frac{7}{2}\) and \(x = -\frac{1}{6}\)
A1
State final answer \(-\frac{7}{2} < x < -\frac{1}{6}\)
A1
Alternative method:
Obtain critical value \(x = -\frac{7}{2}\) from graphical method, inspection, or solving a linear equation or inequality
B1
Obtain critical value \(x = -\frac{1}{6}\) similarly
B2
State final answer \(-\frac{7}{2} < x < -\frac{1}{6}\)
B1
Total: 4
**Question 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality $(2x-3)^2 > 4^2(x+1)^2$, or corresponding quadratic equation, or pair of linear equations $(2x-3) = \pm4(x+1)$ | B1 | $12x^2 + 44x + 7 < 0$ |
| Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for $x$ | M1 | Correct method seen, or implied by correct answers |
| Obtain critical values $x = -\frac{7}{2}$ and $x = -\frac{1}{6}$ | A1 | |
| State final answer $-\frac{7}{2} < x < -\frac{1}{6}$ | A1 | |
| **Alternative method:** | | |
| Obtain critical value $x = -\frac{7}{2}$ from graphical method, inspection, or solving a linear equation or inequality | B1 | |
| Obtain critical value $x = -\frac{1}{6}$ similarly | B2 | |
| State final answer $-\frac{7}{2} < x < -\frac{1}{6}$ | B1 | |
| **Total: 4** | | |