Easy -1.2 This is a straightforward absolute value inequality requiring only the standard technique of splitting into two cases: -1 < 9-2x < 1, then solving linear inequalities to get 4 < x < 5. It's a routine procedural question with minimal steps, easier than average A-level content.
State or imply non-modular inequality \((9-2x)^2 < 1\), or a correct pair of linear inequalities, combined or separate, e.g. \(-1 < 9-2x < 1\)
B1
Obtain both critical values 4 and 5
B1
State correct answer \(4 < x < 5\); accept \(x > 4, x < 5\)
B1
OR: State a correct equation or pair of equations for both critical values e.g. \(9-2x = 1\) and \(9-2x = -1\), or \(9-2x = \pm 1\)
B1
Obtain critical values 4 and 5
B1
State correct answer \(4 < x < 5\); accept \(x > 4, x < 5\)
B1
OR: State one critical value (probably \(x = 4\)) from a graphical method or by inspection or by solving a linear inequality or equation
B1
State the other critical value correctly
B1
State correct answer \(4 < x < 5\); accept \(x > 4, x < 5\)
B1
Max 3 marks
| Content | Mark | Guidance |
|---------|------|----------|
| State or imply non-modular inequality $(9-2x)^2 < 1$, or a correct pair of linear inequalities, combined or separate, e.g. $-1 < 9-2x < 1$ | B1 | |
| Obtain both critical values 4 and 5 | B1 | |
| State correct answer $4 < x < 5$; accept $x > 4, x < 5$ | B1 | |
| **OR:** State a correct equation or pair of equations for both critical values e.g. $9-2x = 1$ and $9-2x = -1$, or $9-2x = \pm 1$ | B1 | |
| Obtain critical values 4 and 5 | B1 | |
| State correct answer $4 < x < 5$; accept $x > 4, x < 5$ | B1 | |
| **OR:** State one critical value (probably $x = 4$) from a graphical method or by inspection or by solving a linear inequality or equation | B1 | |
| State the other critical value correctly | B1 | |
| State correct answer $4 < x < 5$; accept $x > 4, x < 5$ | B1 | Max 3 marks |
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