Higher January 2022 Paper 2R Q15
15 Make \(t\) the subject of \(n^2 = \dfrac{4d + t^3}{t^3}\)
(4)
| Scheme | Marks |
|---|---|
\(n^2t^3 = 4d + t^3\) or \(n^2 = \dfrac{4d}{t^3} + 1\) | M1 |
\(t^3\left(n^2 - 1\right) = 4d\) oe or \(n^2 - 1 = \dfrac{4d}{t^3}\) | M1 |
\(t^3 = \dfrac{4d}{\left(n^2 - 1\right)}\) oe or \(t^3 = \dfrac{4d}{\left(n^2 - 1\right)}\) | M1 |
| \(t = \sqrt[3]{\dfrac{4d}{\left(n^2 - 1\right)}}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for multiplying by the denominator
or for dividing the RHS by \(t^3\)
M1: for isolating terms in \(t^3\) and factorising the correct expression of the equation
or for isolating the \(\dfrac{4d}{t^3}\) term
M1: for making \(t^3\) the subject
A1: oe eg. \(t = \sqrt[3]{\dfrac{-4d}{\left(1 - n^2\right)}}\) or \(t = \left(\dfrac{4d}{\left(n^2 - 1\right)}\right)^{\frac{1}{3}}\)
SC B2 for \(t = \sqrt[3]{\dfrac{4d}{\left(n^2 + 1\right)}}\)