Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 0 and x = -2/3) and testing regions, but the solution is methodical rather than requiring insight. It's slightly above average difficulty because students must correctly handle multiple cases and combine intervals, but it's a routine technique taught in P2.
EITHER: State or imply non-modular inequality \((3x+2)^2 < x^2\), or corresponding quadratic equation, or pair of linear equations \(3x + 2 = \pm x\)
M1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = -1\) and \(x = -\frac{1}{2}\)
A1
State answer \(-1 < x < -\frac{1}{2}\)
A1
OR: Obtain the critical value \(x = -1\) from a graphical method or by inspection, or by solving a linear equation or inequality
B1
Obtain the critical value \(x = -\frac{1}{2}\) similarly
B2
State answer \(-1 < x < -\frac{1}{2}\)
B1
[4]
**EITHER:** State or imply non-modular inequality $(3x+2)^2 < x^2$, or corresponding quadratic equation, or pair of linear equations $3x + 2 = \pm x$ | M1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = -1$ and $x = -\frac{1}{2}$ | A1 |
State answer $-1 < x < -\frac{1}{2}$ | A1 |
**OR:** Obtain the critical value $x = -1$ from a graphical method or by inspection, or by solving a linear equation or inequality | B1 |
Obtain the critical value $x = -\frac{1}{2}$ similarly | B2 |
State answer $-1 < x < -\frac{1}{2}$ | B1 | [4]