A2 June 2020 Paper 2 Q5
5 Solve the inequality
\[\frac{2x + 3}{x - 1} \leqslant x + 5\][5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Selects a correct approach which would lead to solving the inequality eg Multiplies the inequality by \((x - 1)^2\) or Rearranges to an inequality with 0 on LHS or RHS or Replaces “\(\leqslant\)” with “\(=\)” and multiplies by \((x - 1)\) | M1 | 1.1a |
| Manipulates their equation/inequality to allow the critical values to be found | M1 | 1.1a |
| Obtains critical values of \(-4\), \(1\) and \(2\) | M1 | 1.1a |
| Gives one correct region from \(x \geqslant 2\), \(-4 \leqslant x \lt 1\) Condone \(-4 \leqslant x \leqslant 1\) Must have three critical values. | A1 | 1.1b |
| Obtains correct solution \(x \geqslant 2\), \(-4 \leqslant x \lt 1\) | A1 | 1.1b |
| (5 marks) |
Typical solution
\[\frac{(x - 1)^2(2x + 3)}{x - 1} \leqslant (x - 1)^2(x + 5)\]\[(x - 1)(2x + 3) \leqslant (x - 1)^2(x + 5)\]\[(x - 1)\{(x - 1)(x + 5) - (2x + 3)\} \geqslant 0\]\[(x - 1)\{x^2 + 2x - 8\} \geqslant 0\]\[(x - 1)(x + 4)(x - 2) \geqslant 0\]Considering cubic curve:
\(x \geqslant 2\) or between \(-4\) and \(1\)
But \(x \neq 1\)
So \(x \geqslant 2\) or \(-4 \leqslant x \lt 1\)