Higher June 2024 Paper 2 Q20
20
\[(x - 9) = \dfrac{2(6 - x^2)}{x + 3} \qquad \text{and} \qquad x = \dfrac{d \pm \sqrt{e}}{f}\]Work out one set of possible values for \(d\), \(e\) and \(f\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 - 9x + 3x - 27\) or \(x^2 - 6x - 27\) | M1 | oe implied by eg \(\dfrac{1}{2}x^2 - \dfrac{9}{2}x + \dfrac{3}{2}x - \dfrac{27}{2}\) may be seen in a grid |
| their \((x^2 - 9x + 3x - 27) = 12 - 2x^2\) | M1dep | oe equation with brackets expanded eg their \(\left(\dfrac{1}{2}x^2 - \dfrac{9}{2}x + \dfrac{3}{2}x - \dfrac{27}{2}\right) = 6 - x^2\) |
| \(3x^2 - 6x - 39\ (= 0)\) or \(3x^2 - 6x = 39\) | M1dep | oe \(ax^2 + bx + c\ (= 0)\) or \(px^2 + qx = r\) eg \(x^2 - 2x - 13\ (= 0)\) or \(\dfrac{3}{2}x^2 - 3x - \dfrac{39}{2}\ (= 0)\) implied by eg \(\dfrac{2 \pm \sqrt{56}}{2}\) |
| \(d = k \quad e = 14k^2 \quad f = k\) where \(k\) is a non-zero constant | A1 | eg \(d = 1 \quad e = 14 \quad f = 1\) or \(d = 2 \quad e = 56 \quad f = 2\) or \(d = 6 \quad e = 504 \quad f = 6\) |
Additional guidance
| Take the values on the answer lines as the final answer eg \(\dfrac{2 \pm \sqrt{56}}{2}\) in working with \(d = 2 \quad e = \sqrt{56} \quad f = 2\) on answer lines | M3A0 |
| \(1 \pm \sqrt{14}\) in working with \(d = 1 \quad e = 14 \quad f =\) (blank) | M3A0 |
| For terms seen in a grid accept eg \(3x\) for \(+3x\) | |
| For up to M2 accept algebraic fractions but do not allow 3rd M1 unless recovered eg \(\dfrac{x^2 - 9x + 3x - 27}{x + 3} = \dfrac{12 - 2x^2}{x + 3}\) | M1M1 |