Easy -1.2 This is a straightforward modulus inequality requiring only the standard technique of splitting into two cases (2x+1 > 4 or 2x+1 < -4) and solving two simple linear inequalities. It's a routine textbook exercise testing basic recall of the modulus inequality method with minimal algebraic manipulation.
\(x<-2\tfrac{1}{2}\) mark final ans; allow 3 for correct unsupported answers. By squaring: \(4x^2+4x-15>(or=)0\) M1; \(x>3/2\) A1, \(x<-2\tfrac{1}{2}\) A1; Penalise \(\geq\) and \(\leq\) once only; \(3/2
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|2x+1|>4$ | | |
| $\Rightarrow 2x+1>4,\; x>3/2$ | B1 | $x>3/2$ mark final ans; if from $|2x|>3$ B0 |
| or $2x+1<-4$ | M1 | o.e., e.g. $-(2x+1)>4$ (or $2x+1=-4$); if $|2x+1|<-4$, M0 |
| $x<-2\tfrac{1}{2}$ | A1 **[3]** | $x<-2\tfrac{1}{2}$ mark final ans; allow 3 for correct unsupported answers. By squaring: $4x^2+4x-15>(or=)0$ M1; $x>3/2$ A1, $x<-2\tfrac{1}{2}$ A1; Penalise $\geq$ and $\leq$ once only; $3/2<x<-2\tfrac{1}{2}$ SC2 |
---