Higher June 2019 Paper 2R Q16
16 The following table gives values of \(x\) and \(y\) where \(y\) is inversely proportional to the square of \(x\).
| \(x\) | 1.5 | 2 | 3 | 4 |
|---|---|---|---|---|
| \(y\) | 16 | 9 | 4 | 2.25 |
(a) Find a formula for \(y\) in terms of \(x\). (3)
Given that \(x \gt 0\)
(b) find the value of \(x\) when \(y = 144\) (2)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{k}{x^2}\) condone proportion symbol in place of = | M1 |
| \(16 = \dfrac{k}{1.5^2}\) or \(9 = \dfrac{k}{2^2}\) or \(4 = \dfrac{k}{3^2}\) or \(2.25 = \dfrac{k}{4^2}\) | M1 |
| \(y = \dfrac{36}{x^2}\) | A1 |
| (3) |
Notes
M1: Setting up a correct equation “\(k\)” ≠ 1
M1: Using the values from the table to find the value of the constant or “\(k\)” = 36
A1: \(\dfrac{36}{x^2}\) = M2 A0
| Scheme | Marks |
|---|---|
| \(x^2 = \dfrac{36}{144}\) or \(x = \sqrt{\left(\dfrac{36}{144}\right)}\) | M1 |
| 0.5 oe | A1 |
| (2) | |
| (5 marks) |
Notes
M1: Substituting \(y = 144\) into the correct equation and making \(x^2\) or \(x\) the subject.
A1: cao