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 below average difficulty as it involves direct application of a basic method with minimal algebraic manipulation.
allow M1 for \(1\frac{1}{2}\) seen; Penalise both \(>\) and \(<\) once only
or \(2x+1 \leq -4 \Rightarrow x \leq -2\frac{1}{2}\)
M1 A1
allow M1 for \(-2\frac{1}{2}\) seen; \(-1\) if both correct but final answer expressed incorrectly, e.g. \(-2\frac{1}{2} \geq x \geq 1\frac{1}{2}\) or \(1\frac{1}{2} \leq x \leq -2\frac{1}{2}\) (or even \(-2\frac{1}{2} \leq x \leq 1\frac{1}{2}\) from previously correct work) e.g. SC3
[4]
## Question 4:
| $|2x+1| \geq 4$ | | Same scheme for other methods, e.g. squaring, graphing |
|---|---|---|
| $2x+1 \geq 4 \Rightarrow x \geq 1\frac{1}{2}$ | M1 A1 | allow M1 for $1\frac{1}{2}$ seen; Penalise both $>$ and $<$ once only |
| or $2x+1 \leq -4 \Rightarrow x \leq -2\frac{1}{2}$ | M1 A1 | allow M1 for $-2\frac{1}{2}$ seen; $-1$ if both correct but final answer expressed incorrectly, e.g. $-2\frac{1}{2} \geq x \geq 1\frac{1}{2}$ or $1\frac{1}{2} \leq x \leq -2\frac{1}{2}$ (or even $-2\frac{1}{2} \leq x \leq 1\frac{1}{2}$ from previously correct work) e.g. SC3 |
| **[4]** | | |
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