Standard +0.3 This is a straightforward modulus inequality requiring case-by-case analysis based on critical points x = -1 and x = 3. Students must consider three intervals, square both sides (or use sign analysis), and solve resulting linear inequalities—a standard technique taught explicitly in P3 with no novel insight required beyond methodical application.
State or imply non-modular inequality \((x-3)^2 > (2x+1)^2\), or corresponding quadratic equation, or pair of linear equations \((x-3) = \pm 2(x+1)\)
B1
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Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
—
Obtain critical values \(-5\) and \(\frac{1}{3}\)
A1
—
State answer \(-5 < x < \frac{1}{3}\)
A1
[4]
OR:
Answer
Marks
Guidance
Obtain the critical value \(x = -5\) from a graphical method, or by inspection, or by solving a linear equation or inequality
B1
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Obtain the critical value \(x = \frac{1}{3}\) similarly
B2
—
State answer \(-5 < x < \frac{1}{3}\)
B1
[4]
[Do not condone \(\leq\) for \(<\); accept 0.33 for \(\frac{1}{3}\).]
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State or imply non-modular inequality $(x-3)^2 > (2x+1)^2$, or corresponding quadratic equation, or pair of linear equations $(x-3) = \pm 2(x+1)$ | B1 | —
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 | —
Obtain critical values $-5$ and $\frac{1}{3}$ | A1 | —
State answer $-5 < x < \frac{1}{3}$ | A1 | [4]
**OR:**
Obtain the critical value $x = -5$ from a graphical method, or by inspection, or by solving a linear equation or inequality | B1 | —
Obtain the critical value $x = \frac{1}{3}$ similarly | B2 | —
State answer $-5 < x < \frac{1}{3}$ | B1 | [4]
| [Do not condone $\leq$ for $<$; accept 0.33 for $\frac{1}{3}$.] | — | —
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