Easy -1.2 This is a straightforward application of the basic modulus inequality rule, requiring only splitting into two cases (3x+1 > 8 or 3x+1 < -8) and solving two simple linear inequalities. It's a standard textbook exercise testing recall of the definition with minimal algebraic manipulation.
State or imply non-modular inequality \((3x + 1)^2 > 8^2\), or corresponding equation or pair of linear equations
M1
Obtain critical values \(\frac{7}{3}\) or \(-3\)
A1
State correct answer \(x < -3\) or \(x > \frac{7}{3}\)
A1
OR
Answer
Marks
Guidance
State one critical value, e.g. \(x = -3\), by solving a linear equation (or inequality) or from a graphical method or by inspection
B1
State the other critical value correctly
B1
State correct answer \(x < -3\) or \(x > \frac{7}{3}\)
B1
[3]
| State or imply non-modular inequality $(3x + 1)^2 > 8^2$, or corresponding equation or pair of linear equations | M1 |
| Obtain critical values $\frac{7}{3}$ or $-3$ | A1 |
| State correct answer $x < -3$ or $x > \frac{7}{3}$ | A1 |
**OR**
| State one critical value, e.g. $x = -3$, by solving a linear equation (or inequality) or from a graphical method or by inspection | B1 |
| State the other critical value correctly | B1 |
| State correct answer $x < -3$ or $x > \frac{7}{3}$ | B1 | [3] |