Higher November 2024 Paper 2 Q24
24. \(y\) is inversely proportional to \(w^3\).
\(w\) is directly proportional to the square root of \(x\).
\(y = \dfrac{1}{9}\) and \(x = 4\) when \(w = 6\).
Find a formula for \(y\) in terms of \(x\) in the form \(y = \mathrm{m}x^{\mathrm{n}}\), where m and n are fractions in their simplest form. [7]
| Scheme | Marks | Comments |
|---|---|---|
| \(y \propto \dfrac{1}{w^3}\) or \(y = \dfrac{k}{w^3}\) OR \(w \propto \sqrt{x}\) or \(w = c\sqrt{x}\) | B1 | |
| \(\dfrac{1}{9} = \dfrac{k}{6^3}\) | M1 | FT \(y \propto \dfrac{1}{w}\) or \(y \propto w^3\) Note: k = 24 implies M1 |
| \(y = \dfrac{24}{w^3}\) oe | A1 | CAO. May be seen in later work |
| \(6 = c\sqrt{4}\) | M1 | FT \(w \propto x^2\) c = 3 implies M1 |
| \(w = 3\sqrt{x}\) | A1 | CAO May be seen in later work |
| \(y = \dfrac{24}{(3\sqrt{x})^3}\) or \(y = \dfrac{8}{9(\sqrt{x})^3}\) oe | M1 | FT provided M1 M1 previously awarded. |
| \(y = \dfrac{8}{9}x^{\frac{-3}{2}}\) | A1 | Mark final answer. A0 on FT if \(y = \mathrm{m}x^{\mathrm{n}}\) and m and n are not both fractions |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(y \propto \dfrac{1}{w^3}\) or \(y = \dfrac{k}{w^3}\) OR \(w \propto \sqrt{x}\) or \(w = c\sqrt{x}\) | B1 | |
| (Substitutes in for \(w\)) \(y = \dfrac{k}{(c\sqrt{x})^3}\) | M1 | FT \(y \propto \dfrac{1}{w}\) or \(y \propto w^3\) or \(w \propto x^2\) |
| (Changes constants) \(y = \dfrac{k}{c^3(\sqrt{x})^3}\) or \(y = \dfrac{P}{(\sqrt{x})^3}\) oe | A1 | CAO |
| \(\dfrac{1}{9} = \dfrac{P}{(\sqrt{4})^3}\) | M1 | |
| \(P = \dfrac{8}{9}\) oe | A1 | CAO |
| \(y = \dfrac{24}{(3\sqrt{x})^3}\) or \(y = \dfrac{8}{9(\sqrt{x})^3}\) oe | M1 | FT provided M1 M1 previously awarded. |
| \(y = \dfrac{8}{9}x^{\frac{-3}{2}}\) | A1 | Mark final answer. A0 on FT if \(y = \mathrm{m}x^{\mathrm{n}}\) and m and n are not both fractions |
| (7) |