Higher November 2022 Paper 1 Q19
19 Solve \(\dfrac{1}{x} - \dfrac{1}{x + 1} = 4\)
Give your answer in the form \(a \pm b\sqrt{2}\) where \(a\) and \(b\) are fractions. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-\dfrac{1}{2} \pm \dfrac{1}{2}\sqrt{2}\) | P1 | for using a common denominator, eg \(\dfrac{x + 1}{x(x + 1)} - \dfrac{x}{x(x + 1)}\ (= 4)\) or \(\dfrac{x + 1 - x}{x(x + 1)}\ (= 4)\) or \(x + 1 - x = 4x(x + 1)\) |
| P1 | for expanding and rearranging to get \(4x^2 + 4x - 1\ (= 0)\) | |
| P1 | (dep P1) ft for a method to solve their 3 term quadratic equation, eg \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times -1}}{2 \times 4}\) or \(4\left[\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2\right] - 1 = 0\) oe | |
| A1 | for values of \(x\), eg \(\dfrac{-4 \pm \sqrt{32}}{8}\) or \(\pm\sqrt{\dfrac{1}{2}} - \dfrac{1}{2}\) oe | |
| A1 | for \(-\dfrac{1}{2} \pm \dfrac{1}{2}\sqrt{2}\) oe in the form \(a \pm b\sqrt{2}\) where are \(a\) and \(b\) are fractions |
Additional guidance
Note we don’t need to see “= 0”; just the LHS is sufficient
Accept other forms of the 3 term quadratic, eg \(4x^2 + 4x = 1\)
Correct use of formula or completing the square
Accept \(a = -\dfrac{1}{2}\), \(b = \dfrac{1}{2}\)
or \(a = -\dfrac{1}{2}\), \(b = -\dfrac{1}{2}\)