Higher June 2019 Paper 4 Q16
16
(a) The table shows values of \(x\) and \(y\).
| \(x\) | 4 | 16 | 36 |
|---|---|---|---|
| \(y\) | 6 | 3 | 2 |
Show that these values fit the relationship that \(y\) is inversely proportional to \(\sqrt{x}\). [2]
(b) \(a\) is inversely proportional to \(b^2\) and \(a = 3.75\) when \(b = 4\).
Find a formula linking \(a\) and \(b\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| any correct method e.g. two of [\(y \times \sqrt{x}\) =] \(6 \times \sqrt{4} = 12\), \(3 \times \sqrt{16} = 12\), \(2 \times \sqrt{36} = 12\) oe or use one pair to find \(y = \dfrac{12}{\sqrt{x}}\) and check with another pair | 2 | accept 6 for \(\sqrt{36}\) etc M1 for correct method with one error or omission or uses \(y = \frac{k}{\sqrt{x}}\) to find \(k = 12\) or one of [\(y \times \sqrt{x}\) =] \(6 \times \sqrt{4} = 12\), \(3 \times \sqrt{16} = 12\), \(2 \times \sqrt{36} = 12\) alternative method : show \(x\) is × 4 and × 9 and \(y\) is ÷ 2 and ÷ 3. | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a = \dfrac{60}{b^2}\) oe | 3 | condone answer of \(a \propto \frac{60}{b^2}\) for 2 marks or M1 for \(a = \dfrac{k}{b^2}\) oe implied by \(3.75 = \frac{k}{4^2}\) B1 for [\(k\) =] 60 | |