Standard +0.3 This is a straightforward modulus inequality requiring students to consider critical points at x = -1 and x = 4, then test regions or square both sides. It's slightly above average difficulty as it involves systematic case analysis, but the technique is standard and well-practiced in P2, with no conceptual surprises beyond applying the definition of modulus.
State or imply non-modular inequality \((x+1)^2 > (x-4)^2\), or corresponding equation or pair of linear equations
M1
Obtain critical value \(\frac{3}{2}\)
A1
State correct answer \(x > \frac{3}{2}\)
A1
[3]
OR:
Answer
Marks
Guidance
State a correct linear equation for the critical value, e.g. \(x+1 = -x+4\), or corresponding correct linear inequality, e.g. \(x+1 > -(x-4)\)
M1
Obtain critical value \(\frac{3}{2}\)
A1
State correct answer \(x > \frac{3}{2}\)
A1
[3]
State or imply non-modular inequality $(x+1)^2 > (x-4)^2$, or corresponding equation or pair of linear equations | M1 |
Obtain critical value $\frac{3}{2}$ | A1 |
State correct answer $x > \frac{3}{2}$ | A1 | [3]
OR:
State a correct linear equation for the critical value, e.g. $x+1 = -x+4$, or corresponding correct linear inequality, e.g. $x+1 > -(x-4)$ | M1 |
Obtain critical value $\frac{3}{2}$ | A1 |
State correct answer $x > \frac{3}{2}$ | A1 | [3]