Moderate -0.8 This is a straightforward modulus inequality requiring the standard technique of splitting into two cases (2x+1 ≥ 4 or 2x+1 ≤ -4), then solving two linear inequalities. It's a routine textbook exercise testing basic recall of the modulus definition with minimal problem-solving, making it easier than average but not trivial since students must remember to consider both cases correctly.
or \(2x + 1 \leq -4 \Rightarrow x \leq -2\frac{1}{2}\)
Same scheme for other methods, e.g. squaring, graphing. Penalise both \(>\) and \(<\) once only. \(-1\) if both correct but final ans 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
$|2x + 1| \geq 4$ | M1, A1, M1, A1 [4] | allow M1 for $1\frac{1}{2}$ seen allow M1 for $-2\frac{1}{2}$ seen
$\Rightarrow \quad 2x + 1 \geq 4 \Rightarrow x \geq 1\frac{1}{2}$ | |
or $2x + 1 \leq -4 \Rightarrow x \leq -2\frac{1}{2}$ | | Same scheme for other methods, e.g. squaring, graphing. Penalise both $>$ and $<$ once only. $-1$ if both correct but final ans 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
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