Moderate -0.3 This is a straightforward modulus inequality requiring students to consider cases based on critical points x = 3 and x = -2, then solve linear inequalities in each region. While it requires systematic case analysis, the algebraic manipulation is routine and the question is a standard textbook exercise with no novel insight needed, making it slightly easier than average.
Expand and solve a linear inequality, or equivalent
M1
Obtain critical value \(\frac{1}{2}\)
A1
State correct answer \(x < \frac{1}{2}\) (allow \(x \le \frac{1}{2}\))
A1
OR
Answer
Marks
State a correct linear equation for the critical value, e.g. \(3 - x = x + 2\), or corresponding correct inequality, e.g. \(-(x-3) > (x+2)\)
M1
Solve the linear equation, or inequality
M1
Obtain critical value \(\frac{1}{2}\)
A1
State correct answer \(x < \frac{1}{2}\)
A1
OR
Answer
Marks
Guidance
Make recognisable sketches of both \(y =
x-3
\) and \(y =
Obtain a critical value from the intersection of the graphs
M1
Obtain critical value \(\frac{1}{2}\)
A1
State final answer \(x < \frac{1}{2}\)
A1
[4 marks]
Expand and solve a linear inequality, or equivalent | M1 | |
Obtain critical value $\frac{1}{2}$ | A1 | |
State correct answer $x < \frac{1}{2}$ (allow $x \le \frac{1}{2}$) | A1 | |
**OR**
State a correct linear equation for the critical value, e.g. $3 - x = x + 2$, or corresponding correct inequality, e.g. $-(x-3) > (x+2)$ | M1 | |
Solve the linear equation, or inequality | M1 | |
Obtain critical value $\frac{1}{2}$ | A1 | |
State correct answer $x < \frac{1}{2}$ | A1 | |
**OR**
Make recognisable sketches of both $y = |x-3|$ and $y = |x+2|$ on a single diagram | B1 | |
Obtain a critical value from the intersection of the graphs | M1 | |
Obtain critical value $\frac{1}{2}$ | A1 | |
State final answer $x < \frac{1}{2}$ | A1 | [4 marks] |