Higher June 2019 Unit 1 Q19
19.
(a) Write the following expression as a single fraction.
Give your answer in its simplest form. [2]
\[\frac{1}{x - a} - \frac{1}{x}\]Give your answer in its simplest form. [2]
(b) Solve the following equation. [5]
\[\frac{x - 1}{x(4x + 3)} + 2 = 0\]From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{a}{x(x - a)}\) or \(\dfrac{a}{x^2 - ax}\) | B2 | B1 for correct numerator - not from incorrect work – use of brackets may be implied. B1 for correct denominator in a single fraction (accept equivalent) If B2, penalise -1 for incorrect subsequent work |
| Scheme | Marks | Comments |
|---|---|---|
| \(x - 1 + 2x(4x + 3)\) [= 0 ] or \(x - 1 + 8x^2 + 6x\) [= 0] or \(x - 1 = -\,2x(4x + 3)\) | M1 | Clearing fraction Allow e. g. \(\dfrac{x - 1 + 2x(4x + 3)}{x(4x + 3)} = 0\) Allow M1 for \(x - 1 = 2x(4x + 3)\) |
| \(8x^2 + 7x - 1\) [= 0] | A1 | Collecting terms and re-arranging quadratic equation Ignore presence of denominator (provided correct). |
| \((8x - 1)(x + 1)\) [= 0] | B2 | B1 for \((8x \;….\; 1)(x \;….\; 1)\) FT their quadratic equation, provided of equivalent difficulty. |
| \(x = \frac{1}{8}\) or \(x = -\,1\) | B1 | Both answers required. Strict FT ‘their derived brackets’. |
Using quadratic formula FT their quadratic equation, provided of equivalent difficulty.
| Scheme | Marks | Comments |
|---|---|---|
| \((x =)\ \ \dfrac{-7 \pm \sqrt{[7^2 - 4(8)(-1)]}}{2(8)}\) | M1 | For M1, allow one error, in sign or substitution, but not in formula. |
| \(x = \dfrac{-7 \pm \sqrt{81}}{16}\) | A1 | |
| \(x = \frac{1}{8}\) or \(x = -\,1\) (both answers required) | A1 | |
| No marks for a trial and improvement method. |