Find the set of values of x for which $\left| {x - 1} \right| > \left| {2x - 1} \right|$.

EITHER

$\left| {x - 1} \right| > \left| {2x - 1} \right| \Rightarrow {(x - 1)^2} > {(2x - 1)^2}$     M1

${x^2} - 2x + 1 > 4{x^2} - 4x + 1$

$3{x^2} - 2x < 0$     A1

$0 < x < \frac{2}{3}$     A1A1     N2

Note: Award A1A0 for incorrect inequality signs.

OR

$\left| {x - 1} \right| > \left| {2x - 1} \right|$

$x - 1 = 2x - 1$     $x - 1 = 1 - 2x$     M1A1

$- x = 0$     $3x = 2$

$x = 0$     $x = \frac{2}{3}$

Note: Award M1 for any attempt to find a critical value. If graphical methods are used, award M1 for correct graphs, A1 for correct values of x.

$0 < x < \frac{2}{3}$     A1A1     N2

Note: Award A1A0 for incorrect inequality signs.

[4 marks]

This question turned out to be more difficult than expected. Candidates who squared both sides or drew a graph generally gave better solutions than those who relied on performing algebraic operations on terms involving modulus signs.