Moderate -0.5 This is a straightforward modulus equation requiring consideration of cases based on critical values (x=0 and x=1/2). While it requires systematic case analysis, the algebraic manipulation is simple and this is a standard textbook exercise with no conceptual surprises, making it slightly easier than average.
www, or \(2x-1=-x\) must be exact for A1 (e.g. not 0.33, but allow \(0.\overline{3}\)); condone doing both equalities in one line e.g. \(-x = 2x-1 = x\), etc
*or* squaring \(\Rightarrow 3x^2 - 4x + 1 = 0\)
M1
\(\Rightarrow (3x-1)(x-1) = 0\)
M1 factorising, formula or completing the square
\(\Rightarrow x = 1, \frac{1}{3}\)
A1 A1
allow M1 for sign errors in factorisation; \(-1\) if more than two solutions offered, but isw inequalities
[4]
## Question 5:
| $|2x-1| = |x|$ | | allow unsupported answers or from graph |
|---|---|---|
| $2x-1 = x, \quad x=1$ | M1 A1 | www |
| or $-(2x-1) = x, \quad x = \frac{1}{3}$ | M1 A1 | www, or $2x-1=-x$ must be exact for A1 (e.g. not 0.33, but allow $0.\overline{3}$); condone doing both equalities in one line e.g. $-x = 2x-1 = x$, etc |
| *or* squaring $\Rightarrow 3x^2 - 4x + 1 = 0$ | M1 | |
| $\Rightarrow (3x-1)(x-1) = 0$ | | M1 factorising, formula or completing the square |
| $\Rightarrow x = 1, \frac{1}{3}$ | A1 A1 | allow M1 for sign errors in factorisation; $-1$ if more than two solutions offered, but isw inequalities |
| **[4]** | | |
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