Standard +0.8 This modulus inequality requires systematic case analysis (considering signs of expressions inside absolute values) and careful algebraic manipulation to solve for x in terms of parameter a. While the technique is standard for P3, the presence of a parameter and need to consider multiple cases elevates it above routine exercises, though it remains a straightforward application of learned methods without requiring novel insight.
Either: State or imply non-modular inequality \((x + 2a)^2 > (3(x - a))^2\), or corresponding quadratic equation, or pair of linear equations \((x + 2a) = \pm3(x - a)\)
B1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for \(x\)
M1
Obtain critical values \(x = \frac{1}{4}a\) and \(x = \frac{5}{2}a\)
A1
State answer \(\frac{1}{4}a < x < \frac{5}{2}a\)
A1
Or: Obtain critical value \(x = \frac{5}{2}a\) from a graphical method, or by inspection, or by solving a linear equation or inequality
B1
Obtain critical value \(x = \frac{1}{4}a\) similarly
B2
State answer \(\frac{1}{4}a < x < \frac{5}{2}a\)
B1
[Do not condone \(\leq\) for \(<\).]
Total: 4 marks
Either: State or imply non-modular inequality $(x + 2a)^2 > (3(x - a))^2$, or corresponding quadratic equation, or pair of linear equations $(x + 2a) = \pm3(x - a)$ | B1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for $x$ | M1 |
Obtain critical values $x = \frac{1}{4}a$ and $x = \frac{5}{2}a$ | A1 |
State answer $\frac{1}{4}a < x < \frac{5}{2}a$ | A1 |
Or: Obtain critical value $x = \frac{5}{2}a$ from a graphical method, or by inspection, or by solving a linear equation or inequality | B1 |
Obtain critical value $x = \frac{1}{4}a$ similarly | B2 |
State answer $\frac{1}{4}a < x < \frac{5}{2}a$ | B1 |
[Do not condone $\leq$ for $<$.] | Total: 4 marks |
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Find the set of values of $x$ satisfying the inequality
$$|x + 2a| > 3|x - a|,$$
where $a$ is a positive constant. [4]
\hfill \mbox{\textit{CAIE P3 2014 Q1 [4]}}