Higher June 2024 Paper 1 Q24
24
(a) \(9k + 7 \quad\) and \(\quad 2k^2 + 3 \quad\) are consecutive integers.
\(9k + 7 \quad\) is the smaller integer.
Work out the value of the next consecutive integer. [5 marks]
(b) \(x\) is a square number.
Show that the next square number is \(\quad x + 2\sqrt{x} + 1\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2k^2 + 3 - (9k + 7)\ (= 1)\) or \(2k^2 - 9k - 4\ (= 1)\) | M1 | oe eg \(9k + 7 + 1 = 2k^2 + 3\) or \(9k + 8 = 2k^2 + 3\) |
| \(2k^2 - 9k - 5\ (= 0)\) | A1 | terms in any order implied by \(k = 5\) (and \(-\dfrac{1}{2}\)) or correct answer |
| \((2k + 1)(k - 5)\ (= 0)\) or \((k =)\ \dfrac{--9 \pm \sqrt{9^2 - 4 \times 2 \times -5}}{2 \times 2}\) or \((k =)\ \dfrac{9 \pm \sqrt{121}}{4}\) or \((k =)\ 2.25 \pm \sqrt{7.5625}\) | M1 | oe correct factorisation or correct use of quadratic formula or correct use of completing the square for their 3-term quadratic |
| \((k =)\ 5\) (or \(-\dfrac{1}{2}\)) | A1ft | ft at least one solution for their 3-term quadratic implied by correct answer |
| 54 | A1 |
Additional guidance
| Answer 54 not from incorrect working | 5 marks |
| Trial and improvement scores 0 or 5 | |
| Use of inequalities can score up to M0A0M1A1ftA0 | |
| Condone 52, 53, 54 on answer line | 5 marks |
| 54 and 4.5 | 4 marks |
| \(2k^2 + 3 - 9k + 7\ (= 1)\) \(2k^2 - 9k + 9\ (= 0)\) \((2k - 3)(k - 3)\ (= 0)\) \(k = 3\) (or \(\dfrac{3}{2}\)) 22 | M0 A0 M1 A1ft A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\left(\sqrt{x} + 1\right)^2\) or \(\left(\sqrt{x} + 1\right)\left(\sqrt{x} + 1\right)\) | M1 | |
| \(\left(\sqrt{x} + 1\right)^2\) or \(\left(\sqrt{x} + 1\right)\left(\sqrt{x} + 1\right)\) and \(x + \sqrt{x} + \sqrt{x} + 1\) \(= x + 2\sqrt{x} + 1\) | A1 | SC1 takes any square number and shows that \(x + 2\sqrt{x} + 1\) gives the next square number |
| Alternative method 2 | ||
| \(x = n^2\) | M1 | any letter for \(n\) except \(x\) |
| \((n + 1)^2 = n^2 + 2n + 1\) \(= x + 2\sqrt{x} + 1\) | A1 | SC1 takes any square number and shows that \(x + 2\sqrt{x} + 1\) gives the next square number |
| Alternative method 3 | ||
| \(x = n^2\) | M1 | any letter for \(n\) except \(x\) |
| \(n^2 + 2\sqrt{n^2} + 1 = n^2 + 2n + 1\) and \((n + 1)^2\) | A1 | SC1 takes any square number and shows that \(x + 2\sqrt{x} + 1\) gives the next square number |
Additional guidance
| Remember that the answer is given in the question | |
| eg for SC1 \(x = 9,\ 9 + 2 \times 3 + 1 = 16\) | SC1 |
| Allow \(x^{\frac{1}{2}}\) for \(\sqrt{x}\) throughout | |
| If only multiplication in a grid is seen then this is not sufficient for A1 |