Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 0 and x = -1/2) and testing regions, which is a routine technique taught in P3. While it requires systematic case analysis and algebraic manipulation, it follows a well-established method with no novel insight needed, making it slightly easier than average.
EITHER: State or imply non-modular inequality \((2x+1)^2 < x^2\) or corresponding quadratic
equation or pair of linear equations \((2x + 1) = \pm x\)
B1
Expand and make reasonable solution attempt at 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = -1\) and \(x = -\frac{1}{3}\) only
A1
State answer \(-1 < x < -\frac{1}{3}\)
A1
OR: Obtain critical value \(x = -1\) from graphical method, or by inspection, or by solving a linear inequality or equation
B1
Obtain critical value \(x = -\frac{1}{3}\) (deduct B1 from B3 if extra values obtained)
B2
State answer \(-1 < x < -\frac{1}{3}\)
B1
Total: 4
[Condone \(\leq\) for \(<\); accept \(-0.33\) for \(-\frac{1}{3}\)]
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** State or imply non-modular inequality $(2x+1)^2 < x^2$ or corresponding quadratic | | |
| equation or pair of linear equations $(2x + 1) = \pm x$ | B1 | |
| Expand and make reasonable solution attempt at 3-term quadratic, or solve two linear equations | M1 | |
| Obtain critical values $x = -1$ and $x = -\frac{1}{3}$ only | A1 | |
| State answer $-1 < x < -\frac{1}{3}$ | A1 | |
| **OR:** Obtain critical value $x = -1$ from graphical method, or by inspection, or by solving a linear inequality or equation | B1 | |
| Obtain critical value $x = -\frac{1}{3}$ (deduct B1 from B3 if extra values obtained) | B2 | |
| State answer $-1 < x < -\frac{1}{3}$ | B1 | **Total: 4** |
| [Condone $\leq$ for $<$; accept $-0.33$ for $-\frac{1}{3}$] | | |
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