4MA0 Higher June 2013 Paper 3R Q17
17 Make \(n\) the subject of the formula
\[t = \sqrt{\dfrac{n + 3}{n}}\](4)
| Scheme | Marks |
|---|---|
\(t^2 = \dfrac{n + 3}{n}\) \(nt^2 = n + 3\) | M1 |
| \(nt^2 - n = 3\) | M1 |
| \(n(t^2 - 1) = 3\) | M1 |
| \(n = \dfrac{3}{t^2 - 1}\) | A1 |
| (4) |
Notes
M1: squaring both sides
M1: isolating terms in \(n\)
M1: factorising
A1: or \(n = \dfrac{3}{(t+1)(t-1)}\) or \(n = \dfrac{-3}{1 - t^2}\) or \(n = \dfrac{-3}{(1-t)(1+t)}\)
(corrected from the printed mark scheme: the fractions in “\(t^2 = \)” and in the answer “\(n = \)” are blank in the printed scheme; filled in here as \(\dfrac{n+3}{n}\) and \(\dfrac{3}{t^2-1}\))