June 2019 Paper 1 Q19
19 \(\mathrm{f}(x) = \dfrac{x - 3}{2x}\)
Solve \(\quad \mathrm{f}(x + 1) - \mathrm{f}(2x) = 0.5\)
You must show your working. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{x - 2}{2x + 2}\) or \(\dfrac{x + 1 - 3}{2(x + 1)}\) or \(\dfrac{2x - 3}{4x}\) | M1 | oe substituting correctly in at least one expression |
| \(4x(x - 2)\) and \((2x + 2)(2x - 3)\) or \(4x(x - 2) - (2x + 2)(2x - 3)\) or \(4x^2 - 8x - 4x^2 + 2x + 6\) or \(6 - 6x\) or \(2x(x - 2)\) and \((x + 1)(2x - 3)\) | M1dep | oe (could be from using a different denominator) correct numerators or an expression for both, which need not be simplified do not award any follow through marks from an error in first M mark this one comes from a denominator of \(4x(x + 1)\) |
| \(4x(x - 2) - (2x + 2)(2x - 3)\) \(= 0.5 \times 4x \times 2(x + 1)\) | M1dep | oe but needs to be the correct equation setting up the quadratic by multiplying the RHS by the product of the denominators could be scored by both sides of the equation still having the same denominator dep on both previous M marks |
| \(4x^2 + 10x - 6 = 0\) or \(2x^2 + 5x - 3 = 0\) | A1 | |
| \((4x - 2)(x + 3) = 0\) or \((2x - 1)(2x + 6) = 0\) or \((2x - 1)(x + 3) = 0\) | M1dep | correct factors or correct use of quadratic formula oe |
| 0.5 and \(-3\) | A1 | both answers needed |
Additional guidance
| Stop marking as soon as an error is made after first M mark | |
| Look out for correct answer from incorrect working … eg \(x + 1 - 2x = 0.5\) … gives \(x = 0.5\) or \(\mathrm{f}(2x) = 2 \times \dfrac{(x - 3)}{2x} = \dfrac{(2x - 3)}{4x}\) ie \(\mathrm{f}(2x)\) written as \(2\,\mathrm{f}(x)\) then incorrect multiplication | M0A0 |