Standard +0.8 This requires solving an inequality involving two modulus expressions, necessitating systematic case analysis across multiple critical points (x = 2 and x = -1/2), careful algebraic manipulation in each region, and combining solution sets. More demanding than routine single-modulus problems but follows standard technique for P3 level.
State or imply non-modular inequality \((x-2)^2 > (3(2x+1))^2\), or corresponding quadratic equation, or pair of linear equations \((x-2) = \pm(2x+1)\)
B1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = -1\) and \(x = -\frac{1}{7}\)
A1
State answer \(-1 < x < -\frac{1}{7}\)
A1
[4 marks]
OR
Answer
Marks
Guidance
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 \(x = -\frac{1}{7}\) similarly
B2
State answer \(-1 < x < -\frac{1}{7}\)
B1
[Do not condone \(\leq\) for \(<\); accept \(-\frac{5}{35}\) and \(-0.14\) for \(-\frac{1}{7}\).]
State or imply non-modular inequality $(x-2)^2 > (3(2x+1))^2$, or corresponding quadratic equation, or pair of linear equations $(x-2) = \pm(2x+1)$ | B1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = -1$ and $x = -\frac{1}{7}$ | A1 |
State answer $-1 < x < -\frac{1}{7}$ | A1 | [4 marks]
**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 $x = -\frac{1}{7}$ similarly | B2 |
State answer $-1 < x < -\frac{1}{7}$ | B1 | [Do not condone $\leq$ for $<$; accept $-\frac{5}{35}$ and $-0.14$ for $-\frac{1}{7}$.]
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