Challenging +1.2 This requires systematic case analysis of modulus inequalities with a parameter, squaring both sides to eliminate modulus signs, and careful algebraic manipulation. While the technique is standard for P3, the presence of the parameter 'a' and need to consider multiple cases elevates it above routine exercises, but it remains a recognizable problem type with well-established methods.
State or imply non-modular inequality \((x+3a)^2 > (2(x-2a))^2\), or corresponding quadratic equation, or pair of linear equations \((x+3a) = \pm 2(x-2a)\)
B1
EITHER path
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = \frac{1}{3}a\) and \(x = 7a\)
A1
State answer \(\frac{1}{3}a < x < 7a\)
A1
[4]
Obtain the critical value \(x = 7a\) from graphical method, by inspection, or by solving a linear equation or inequality
B1
OR path
Obtain the critical value \(x = \frac{1}{3}a\) similarly
B2
State answer \(\frac{1}{3}a < x < 7a\)
B1
[4]
*Note: Do not condone \(\leq\) for \(<\); accept 0.33 for \(\frac{1}{3}\)*
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply non-modular inequality $(x+3a)^2 > (2(x-2a))^2$, or corresponding quadratic equation, or pair of linear equations $(x+3a) = \pm 2(x-2a)$ | B1 | EITHER path |
| Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 | |
| Obtain critical values $x = \frac{1}{3}a$ and $x = 7a$ | A1 | |
| State answer $\frac{1}{3}a < x < 7a$ | A1 | [4] |
| Obtain the critical value $x = 7a$ from graphical method, by inspection, or by solving a linear equation or inequality | B1 | OR path |
| Obtain the critical value $x = \frac{1}{3}a$ similarly | B2 | |
| State answer $\frac{1}{3}a < x < 7a$ | B1 | [4] |
*Note: Do not condone $\leq$ for $<$; accept 0.33 for $\frac{1}{3}$*
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