Higher November 2023 Paper 2 Q20
20 Rearrange \(\quad p = \dfrac{2m + 1}{1 - m} \quad\) to make \(m\) the subject. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(p(1 - m) = 2m + 1\) | M1 | |
| \(p - pm = 2m + 1\) | M1dep | |
| \(p - 1 = 2m + pm\) or \(p - 1 = m(2 + p)\) or \(\dfrac{p - 1}{2 + p}\) | M1dep | oe collection of terms in \(m\) eg \(-pm - 2m = 1 - p\) oe eg \(\dfrac{1 - p}{-p - 2}\) or \(\dfrac{p}{2 + p} - \dfrac{1}{2 + p}\) |
| \(m = \dfrac{p - 1}{2 + p}\) or \(\dfrac{p - 1}{2 + p} = m\) | A1 | oe eg \(m = \dfrac{1 - p}{-p - 2}\) or \(m = \dfrac{p}{2 + p} - \dfrac{1}{2 + p}\) |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone \(m = \dfrac{p - 1}{2 + p}\) in working with \(\dfrac{p - 1}{2 + p}\) on answer line | M1M1M1A1 |
| \(m = \dfrac{p - 1}{2 + p}\) followed by incorrect further work | M1M1M1A0 |
| \(p(1 - m)^2 = 2m + 1\) | M0M0M0A0 |