| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 3 |
| Topic | Inequalities |
| Type | Solve absolute value inequality |
| Difficulty | Easy -1.2 This is a straightforward absolute value inequality requiring only the standard technique of splitting into two cases: 2x-1 < 3 and 2x-1 > -3, then solving both linear inequalities. It's a routine procedural question with no conceptual challenges, making it easier than average for A-level. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities |
**METHOD 1**
$x < 2$ seen — B1
$2x - 1 < 3$ AND $-(2x-1) < 3$ seen — M1
Obtain $-1 < x < 2$ — A1 **[3]**
**METHOD 2**
$(2x-1)^2 < 3^2$ seen — B1
Expand and obtain a 3 term quadratic $(x^2 - x - 2 < 0)$ — M1
Obtain $-1 < x < 2$ — A13 Solve the inequality $| 2 x - 1 | < 3$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2014 Q3 [3]}}