Higher January 2023 Paper 1R Q19
19 \(P\) is inversely proportional to \(y^2\)
When \(y = 4\), \(P = a\)
(a) Find a formula for \(P\) in terms of \(y\) and \(a\) (3)
Given also that \(y\) is directly proportional to \(\sqrt{x}\)
and when \(x = a\), \(P = 4a\)
(b) find a formula for \(P\) in terms of \(x\) and \(a\) (3)
| Scheme | Marks |
|---|---|
| \(P = \dfrac{k}{y^2}\) | M1 |
| eg \(a = \dfrac{k}{4^2}\) or \(k = 16a\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(P = \dfrac{16a}{y^2}\) | A1 |
| (3) |
Notes
M1: oe (the constant term, \(k\), can be any other letter apart from \(a\) or \(P\) or \(y\))
M1: oe
A1: oe eg \(P = 16ay^{-2}\) or \(P = \dfrac{4^2a}{y^2}\)
| Scheme | Marks |
|---|---|
\(\sqrt{\text{``}{\dfrac{16a}{4a}}\text{''}} = c\sqrt{a}\) oe eg \(\dfrac{16a}{4a} = c^2a\) or \(4a = \dfrac{16a}{c^2a}\) or \(4a \times c^2a = 16a\) oe or (when \(P = 4a\)) \(y^2 = \dfrac{16a}{4a}\) or \(y^2 = 4\) or \(y = \sqrt{\dfrac{16a}{4a}}\;(= 2)\) oe | M1 |
\(c = \sqrt{\dfrac{4}{a}}\) or \(c = \dfrac{\pm 2}{\sqrt{a}}\) or \(c = \dfrac{\pm 2\sqrt{a}}{a}\) oe allow the constant term squared eg \(c^2 = \dfrac{16a}{4a^2}\left(= \dfrac{4}{a}\right)\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(P = \dfrac{4a^2}{x}\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: ft a correct formula involving the constant term (\(c\) used here) and \(a\)
or
ft for an expression or value of \(y^2\) or \(y\) given for when \(P = 4a\)
M1: (implies previous M1) a correct value, in terms of \(a\), for the constant term or the constant term squared – need not be simplified
A1: oe eg \(P = \dfrac{16a}{\frac{4x}{a}}\) or \(P = \dfrac{16a^2}{4x}\)