Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = -2 and x = 2.5) and testing regions, but the algebraic manipulation is straightforward. It's slightly above average difficulty as it involves systematic case analysis rather than just routine manipulation, but remains a textbook exercise with well-established methods.
EITHER: State or imply non-modular inequality \((x+2)^2 < (5-2x)^2\), or corresponding equation
B1
Expand and make reasonable solution attempt at 2- or 3-term quadratic, or equivalent
M1
Obtain critical values 1 and 7
A1
State correct answer \(x < 1, x > 7\)
A1
4
OR: State one correct equation for a critical value e.g. \(x + 2 = 5 - 2x\)
M1
State two relevant equations separately e.g. \(x + 2 = 5 - 2x\) and \(x + 2 = -(5 - 2x)\)
A1
Obtain critical values 1 and 7
A1
State correct answer \(x < 1, x > 7\)
A1
OR: State one critical value (probably \(x = 1\)), from a graphical method or by inspection or by solving a linear inequality
B1
State the other critical value correctly
B2
State correct answer \(x < 1, x > 7\)
B1
[The answer \(7 < x < 1\) scores B0.]
4
**EITHER:** State or imply non-modular inequality $(x+2)^2 < (5-2x)^2$, or corresponding equation | B1 |
Expand and make reasonable solution attempt at 2- or 3-term quadratic, or equivalent | M1 |
Obtain critical values 1 and 7 | A1 |
State correct answer $x < 1, x > 7$ | A1 | 4
**OR:** State one correct equation for a critical value e.g. $x + 2 = 5 - 2x$ | M1 |
State two relevant equations separately e.g. $x + 2 = 5 - 2x$ and $x + 2 = -(5 - 2x)$ | A1 |
Obtain critical values 1 and 7 | A1 |
State correct answer $x < 1, x > 7$ | A1 |
**OR:** State one critical value (probably $x = 1$), from a graphical method or by inspection or by solving a linear inequality | B1 |
State the other critical value correctly | B2 |
State correct answer $x < 1, x > 7$ | B1 |
[The answer $7 < x < 1$ scores B0.] | | 4
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