Standard +0.3 This is a straightforward modulus inequality requiring consideration of critical points (x = 0 and x = 1/2) and testing regions, which is a standard technique taught in P2. While it requires systematic case analysis, it's a routine application of modulus inequality methods with no novel insight needed, making it slightly easier than average.
State or imply non-modular inequality \((2x-1)^2 > x^4\) or corresponding quadratic equation or pair of linear equations \(2x - 1 = \pm x\)
M1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = 1\) and \(x = -\frac{1}{3}\)
A1
State answer \(x < -\frac{1}{3}, x > 1\)
A1
4 marks
OR: Obtain critical value \(x = 1\) from a graphical method, or by inspection, or by solving a linear inequality or linear equation
B1
Obtain the critical value \(x = -\frac{1}{3}\) similarly
B2
State answer \(x < -\frac{1}{3}, x > 1\)
B1
4 marks
State or imply non-modular inequality $(2x-1)^2 > x^4$ or corresponding quadratic equation or pair of linear equations $2x - 1 = \pm x$ | M1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = 1$ and $x = -\frac{1}{3}$ | A1 |
State answer $x < -\frac{1}{3}, x > 1$ | A1 | 4 marks |
OR: Obtain critical value $x = 1$ from a graphical method, or by inspection, or by solving a linear inequality or linear equation | B1 |
Obtain the critical value $x = -\frac{1}{3}$ similarly | B2 |
State answer $x < -\frac{1}{3}, x > 1$ | B1 | 4 marks |