AS June 2025 Paper 1 Q9
9 Solve the inequality
\[\frac{2x + 1}{3x - 9} \geqslant x - 1\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Multiplies LHS and RHS by \((3x - 9)^2\) or Subtracts one side from the other and correctly combines the fractions. or Cross multiplies and solves for \(x\) PI by obtaining at least one correct region or a correct quadratic factor or both \(\frac{2}{3}\) and 4 | M1 | 3.1a |
| Obtains a correct quadratic factor eg \(3x^2 - 14x + 8\) PI by obtaining \(\frac{2}{3}\) and 4 or States \(x \neq 3\) PI by \(x \gt 3\) | M1 | 1.1a |
| Obtains at least one correct region. Condone \(3 \leqslant x \leqslant 4\) | M1 | 1.1a |
| Obtains \(x \leqslant \dfrac{2}{3}\), \(3 \lt x \leqslant 4\) | A1 | 1.1b |
| (4 marks) |
Typical solution
\[(2x + 1)(3x - 9) \geqslant (x - 1)(3x - 9)^2\]but \(x \neq 3\)
\[0 \geqslant (3x - 9)\big((x - 1)(3x - 9) - (2x + 1)\big)\]\[(3x - 9)(3x^2 - 14x + 8) \leqslant 0\]\[(3x - 9)(3x - 2)(x - 4) \leqslant 0\]Critical values are: \(\frac{2}{3}\), 3, 4
\[x \leqslant \frac{2}{3}, \quad 3 \lt x \leqslant 4\]