Higher November 2021 Paper 2 Q19
19. Solve the following equation.
\[\frac{7}{3x + 1} + \frac{4}{x + 2} = 1\]Give your answers correct to 2 decimal places. [7]
| Scheme | Marks | Comments |
|---|---|---|
| \(7(x + 2) + 4(3x + 1) = (3x + 1)(x + 2)\) | M2 | soi M1 for \(\dfrac{7(x + 2) + 4(3x + 1)}{(3x + 1)(x + 2)}\ (= 1)\) soi |
| Expands the brackets and collects all terms on one side: \(7x + 14 + 12x + 4 = 3x^2 + 6x + x + 2\) \(3x^2 - 12x - 16 = 0\) | M2 | FT for possible M2 or M1 here, only if of equivalent difficulty; For M2 and M1 allow one error in expansion or collection; M1 for expansion not equated to zero e.g. \(7x + 14 + 12x + 4 = 3x^2 + 6x + x + 2\) leading to ‘their \(19x + 18\)’ = ‘their \(3x^2 + 7x + 2\)’ |
| Applies the quadratic formula: \(\dfrac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(-16)}}{2 \times 3}\) or better | M1 | FT ‘their derived 3-term quadratic’ (even if it does not result in real roots); allow one slip in substitution; formula must not be clearly incorrect; allow attempt to complete the square with at most one slip |
| Simplifies: \(\dfrac{12 \pm \sqrt{336}}{6}\) or \(\dfrac{6 \pm 2\sqrt{21}}{3}\) or equivalent | A1 | FT ‘their derived 3-term quadratic’; may be implied by correct decimal values \(-1.05505…\), 5.05505… |
| \(-1.06\), 5.06 | A1 | CAO |
| (7) |