Standard +0.3 This is a standard modulus inequality requiring case-by-case analysis based on critical points x = 2 and x = -1/3. While it involves multiple cases and careful 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 \(2(x-2)^2 > (3x+1)^2\), or corresponding quadratic equation, or pair of linear equations \(2(x-2) = \pm(3x+1)\)
B1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for \(x\)
M1
Obtain critical values \(x = -5\) and \(x = \frac{2}{5}\)
A1
State final answer \(-5 < x < \frac{2}{5}\)
A1
[4]
OR:
Answer
Marks
Guidance
Obtain critical value \(x = -5\) from a graphical method, or by inspection, or by solving a linear equation or inequality
B1
Obtain critical value \(x = \frac{2}{5}\) similarly
B2
State final answer \(-5 < x < \frac{2}{5}\)
B1
[4]
*[Do not condone \(\leq\) for \(<\)]*
State or imply non-modular inequality $2(x-2)^2 > (3x+1)^2$, or corresponding quadratic equation, or pair of linear equations $2(x-2) = \pm(3x+1)$ | B1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for $x$ | M1 |
Obtain critical values $x = -5$ and $x = \frac{2}{5}$ | A1 |
State final answer $-5 < x < \frac{2}{5}$ | A1 | [4]
**OR:**
Obtain critical value $x = -5$ from a graphical method, or by inspection, or by solving a linear equation or inequality | B1 |
Obtain critical value $x = \frac{2}{5}$ similarly | B2 |
State final answer $-5 < x < \frac{2}{5}$ | B1 | [4]
*[Do not condone $\leq$ for $<$]*
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