Higher November 2019 Unit 2 Q15
15. Make \(a\) the subject of the following formula. [5]
\[\frac{2a^2 - b}{a^2 b} = 1\]| Scheme | Marks | Comments |
|---|---|---|
| \(2a^2 - b = a^2 b\) | B1 | FT until 2nd error for equivalent level of difficulty. Allow sight of multiplication signs within expressions and allow multiplication by 1 at any stage. |
| \(2a^2 - a^2 b = b\) OR \(-b = a^2 b - 2a^2\) | B1 | FT a formula with three or more terms AND with at least two terms in \(a^2\). |
| \(a^2(2 - b) = b\) OR \(-b = a^2(b - 2)\) | B1 | |
| \(a^2 = \dfrac{b}{2-b}\) OR \(\dfrac{-b}{b-2} = a^2\) | B1 | |
| \(a = (\pm)\sqrt{\dfrac{b}{2-b}}\) OR \(a = (\pm)\sqrt{\dfrac{-b}{b-2}}\) | B1 |