Moderate -0.8 This is a straightforward modulus equation requiring case-by-case analysis. Students need to consider when expressions are positive/negative and solve resulting linear equations, which is a standard technique with no conceptual difficulty beyond basic algebraic manipulation.
\) or \(2x - 1 = x, x = 1\) or \(-(2x-1) = x, x = 1/3\)
$2x - 1 = |x|$ or $2x - 1 = x, x = 1$ or $-(2x-1) = x, x = 1/3$ | M1A1, M1A1 | www, www, or $2x - 1 = -x$ must be exact for A1 (e.g. not 0.33, but allow 0.3); condone doing both equalities in one line e.g. $-x = 2x - 1 = x$, etc | allow unsupported answers or from graph or squaring $\Rightarrow 3x^2 - 4x + 1 = 0$ M1 $\Rightarrow (3x-1)(x-1) = 0$ M1 factorising, formula or comp. square $\Rightarrow x = 1, 1/3$ A1 A1 allow M1 for sign errors in factorisation $-1$ if more than two solutions offered, but isw inequalities. Doing fg: 2ln(e') = 2x SC1. Allow $x^2$ (but not 2x) unsupported