Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x=3/2 and x=4) and case analysis across intervals. While it involves multiple cases and algebraic manipulation, it follows a routine procedure taught in P2 with no novel insight required, making it slightly easier than average.
State or imply non-modulus inequality \((4-x)^2 \leqslant (3-2x)^2\) or corresponding equation, pair of linear equations or linear inequalities
M1
Attempt solution of 3-term quadratic equation, of two linear equations or of two linear inequalities
M1
Obtain critical values \(-1\) and \(\frac{7}{3}\)
A1
SR Allow B1 for \(x \leqslant -1\) only or \(x \geqslant \frac{7}{3}\) only if first M1 is not given
State answer \(x \leqslant -1,\ x \geqslant \frac{7}{3}\)
A1
Do not accept \(\frac{7}{3} \leqslant x \leqslant -1\) or \(-1 \geqslant x \geqslant \frac{7}{3}\) for A1
Total:
4
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply non-modulus inequality $(4-x)^2 \leqslant (3-2x)^2$ or corresponding equation, pair of linear equations or linear inequalities | M1 | |
| Attempt solution of 3-term quadratic equation, of two linear equations or of two linear inequalities | M1 | |
| Obtain critical values $-1$ and $\frac{7}{3}$ | A1 | SR Allow B1 for $x \leqslant -1$ only or $x \geqslant \frac{7}{3}$ only if first M1 is not given |
| State answer $x \leqslant -1,\ x \geqslant \frac{7}{3}$ | A1 | Do not accept $\frac{7}{3} \leqslant x \leqslant -1$ or $-1 \geqslant x \geqslant \frac{7}{3}$ for A1 |
| **Total:** | **4** | |