Standard +0.8 This requires systematic case analysis of two modulus expressions with different critical points (x = -3/2 and x = -2), solving multiple inequalities across different intervals, and careful algebraic manipulation. While the technique is standard for P3, the execution demands precision and is more involved than routine single-modulus problems, placing it moderately above average difficulty.
State or imply non-modular inequality \((2x+3)^2 > 3^2(x+2)^2\), or corresponding quadratic equation, or pair of linear equations
B1
Make a reasonable attempt at solving a 3-term quadratic, or solve two linear equations for \(x\)
M1
Quadratic formula or \((5x+9)(x+3)\)
Obtain critical values \(x = -3\) and \(x = -\dfrac{9}{5}\)
A1
OE
State final answer \(-3 < x < -\dfrac{9}{5}\) or \(x > -3\) and \(x < -\dfrac{9}{5}\)
A1
Do not condone \(\leqslant\) for \(<\) in the final answer. No ISW
Alternative Method:
Answer
Marks
Guidance
Answer
Marks
Guidance
Obtain critical value \(x = -3\) from a graphical method, or by solving a linear equation or linear inequality
B1
\(2x + 3 = 3(x+2) \Rightarrow x = -3\)
Obtain critical value \(x = -\dfrac{9}{5}\) similarly
B2
State final answer \(-3 < x < -\dfrac{9}{5}\) or \(x > -3\) and \(x < -\dfrac{9}{5}\)
B1
Do not condone \(\leqslant\) for \(<\) in the final answer. No ISW
Total: 4 marks
## Question 1:
**Main Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply non-modular inequality $(2x+3)^2 > 3^2(x+2)^2$, or corresponding quadratic equation, or pair of linear equations | B1 | |
| Make a reasonable attempt at solving a 3-term quadratic, or solve two linear equations for $x$ | M1 | Quadratic formula or $(5x+9)(x+3)$ |
| Obtain critical values $x = -3$ and $x = -\dfrac{9}{5}$ | A1 | OE |
| State final answer $-3 < x < -\dfrac{9}{5}$ or $x > -3$ and $x < -\dfrac{9}{5}$ | A1 | Do not condone $\leqslant$ for $<$ in the final answer. No ISW |
**Alternative Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain critical value $x = -3$ from a graphical method, or by solving a linear equation or linear inequality | B1 | $2x + 3 = 3(x+2) \Rightarrow x = -3$ |
| Obtain critical value $x = -\dfrac{9}{5}$ similarly | B2 | |
| State final answer $-3 < x < -\dfrac{9}{5}$ or $x > -3$ and $x < -\dfrac{9}{5}$ | B1 | Do not condone $\leqslant$ for $<$ in the final answer. No ISW |
**Total: 4 marks**