4MA0 Higher January 2018 Paper 3R Q17
17
(a) Show that \(\dfrac{x + 1}{2x + 1} - \dfrac{1}{(2x + 1)(x + 1)} = \dfrac{x^2 + 2x}{(2x + 1)(x + 1)}\) (2)
(b) Hence, solve \(\dfrac{x + 1}{2x + 1} - \dfrac{1}{(2x + 1)(x + 1)} = \dfrac{1}{(2x + 1)(x + 1)}\)
Give your solutions correct to 3 significant figures.
Show your working clearly. (4)
Give your solutions correct to 3 significant figures.
Show your working clearly. (4)
| Scheme | Marks |
|---|---|
eg \(\dfrac{(x + 1)^2}{(2x + 1)(x + 1)} - \dfrac{1}{(2x + 1)(x + 1)}\) OR \(\dfrac{(x + 1)^2 - 1}{(2x + 1)(x + 1)}\) OR \(\dfrac{x^2 + 2x + 1 - 1}{(2x + 1)(x + 1)}\) | M1 |
| Shown | A1 |
| (2) |
Notes
M1: for correctly writing both fractions over a common denominator
A1: shown with fully correct working
| Scheme | Marks |
|---|---|
| \(x^2 + 2x = 1\) | M1 |
\(\dfrac{-2 \pm \sqrt{2^2 - 4 \times 1 \times -1}}{2}\) or \((x + 1)^2 - 1 = 1\) | M1 |
\(\dfrac{-2 \pm \sqrt{8}}{2}\) OR \(\dfrac{-2 \pm \sqrt{2^2 + 4}}{2}\) OR \(\dfrac{-2 \pm 2\sqrt{2}}{2}\) or \((x + 1) = \pm\sqrt{2}\) | M1 |
| 0.414, −2.41 | A1 |
| (4) |
Notes
M1: for \(x^2 + 2x = 1\) oe
M1: for substituting values from their quadratic equation into the formula (condone one sign error in substitution) or a correct first step for completing the square
M1: for method to solve their equation
A1: awrt dep on the second M mark (accept 0.41)