Higher June 2023 Paper 6 Q7
7 The diagram shows the graph of \(y = kx - x^2 + 2\), where \(k\) is an integer.
(a) Show that \(k = 3\). [2]
(b) Use the graph to solve \(3x - x^2 + 2 = 3\).
Give your answers to 1 decimal place. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct substitution of \((x, y)\) from integer point on curve into equation leading to \(k = 3\) e.g. (2, 4) \(4 = 2k - 2^2 + 2\) or \(4 = 2k - 4 + 2\) leading to \(k = 3\) with at least one correct intermediate step | 2 | M1 for correct substitution of \((x, y)\) from integer point on curve into \(y = kx - x^2 + 2\) or \(y = 3x - x^2 + 2\) OR M1 for e.g. \(x = 2\) correctly substituted in \(y = 3x - x^2 + 2\) and finding \(y = 4\) Max M1 if \(k = 3\) substituted | (-1, -2) : \(-2 = -[1]k - (-1)^2 + 2\) (1, 4) : \(4 = [1]k - 1^2 + 2\) (2, 4) : \(4 = 2k - 2^2 + 2\) (3, 2): \(2 = 3k - 3^2 + 2\) (4, -2): \(-2 = 4k - 4^2 + 2\) Use of (0, 2) scores 0 but may be replaced with another point (ie do not treat as a choice) Examples of intermediate steps \(4 = 2k - 2^2 + 2\) then \(4 = 2k - 4 + 2\) is a sufficient int step or \(4 = 2k - 2\) is a sufficient int step or \(6 = 2k\) is a sufficient int step \(3 = k\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 0.4 and 2.6 | 2 | B1 for 0.4 or 2.6 or M1 for line \(y = 3\) drawn or for (0.4, 3) and (2.6, 3) indicated | Line to cut curve twice Treat \(x = 3\) drawn or multiple horizontal lines as choice unless \(y = 3\) clearly chosen Condone good freehand line eg circled or lines drawn down to \(x\)-axis |