Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 1/2 and x = -1) and testing intervals. While it involves multiple cases and some algebraic manipulation, the technique is routine for P3 students and follows a well-practiced method with no conceptual surprises.
State or imply non-modular inequality \((2x-1)^2 < 3^2(x+1)^2\), or corresponding quadratic equation
B1
e.g. \(5x^2 + 22x + 8 = 0\); Allow recovery from 'invisible brackets' on RHS
Form and solve a 3-term quadratic in \(x\)
M1
Obtain critical values \(x = -4\) and \(x = -\dfrac{2}{5}\)
A1
State final answer \(x < -4\), \(x > -\dfrac{2}{5}\)
A1
Do not condone \(\leqslant\) for \(<\), or \(\geqslant\) for \(>\) in the final answer. Allow 'or' but not 'and'. \(-\dfrac{2}{5} < x < -4\) scores A0. Accept equivalent forms using brackets e.g. \(x \in (-\infty, -4) \cup (-0.4, \infty)\)
Alternative method for Question 1:
Answer
Marks
Guidance
Answer
Marks
Guidance
Obtain critical value \(x = -4\) from a graphical method, or by solving a linear equation or linear inequality
B1
Obtain critical value \(x = -\dfrac{2}{5}\) similarly
B2
State final answer \(x < -4\), \(x > -\dfrac{2}{5}\)
B1
Do not condone \(\leqslant\) for \(<\), or \(\geqslant\) for \(>\) in the final answer. Allow 'or' but not 'and'. \(-\dfrac{2}{5} < x < -4\) scores A0. Accept equivalent forms e.g. \(x \in (-\infty,-4) \cup (-0.4,\infty)\)
Total
4
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply non-modular inequality $(2x-1)^2 < 3^2(x+1)^2$, or corresponding quadratic equation | **B1** | e.g. $5x^2 + 22x + 8 = 0$; Allow recovery from 'invisible brackets' on RHS |
| Form and solve a 3-term quadratic in $x$ | **M1** | |
| Obtain critical values $x = -4$ and $x = -\dfrac{2}{5}$ | **A1** | |
| State final answer $x < -4$, $x > -\dfrac{2}{5}$ | **A1** | Do not condone $\leqslant$ for $<$, or $\geqslant$ for $>$ in the final answer. Allow 'or' but not 'and'. $-\dfrac{2}{5} < x < -4$ scores A0. Accept equivalent forms using brackets e.g. $x \in (-\infty, -4) \cup (-0.4, \infty)$ |
**Alternative method for Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain critical value $x = -4$ from a graphical method, or by solving a linear equation or linear inequality | **B1** | |
| Obtain critical value $x = -\dfrac{2}{5}$ similarly | **B2** | |
| State final answer $x < -4$, $x > -\dfrac{2}{5}$ | **B1** | Do not condone $\leqslant$ for $<$, or $\geqslant$ for $>$ in the final answer. Allow 'or' but not 'and'. $-\dfrac{2}{5} < x < -4$ scores A0. Accept equivalent forms e.g. $x \in (-\infty,-4) \cup (-0.4,\infty)$ |
| **Total** | **4** | |