June 2018 Paper 2 Q3
3 \(p = \dfrac{m + 2}{m^2 + 1}\)
(a) Work out the value of \(p\) when \(\ m = -5.5\) [1 mark]
(b) Work out the values of \(m\) when \(\ p = 2\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(-0.112\) or \(-\dfrac{14}{125}\) | B1 | oe fraction |
Additional guidance
| Ignore incorrect conversion between fraction and decimal if correct value seen | |
| Ignore rounding or truncation after correct value seen | |
| Answer \(-\dfrac{3.5}{31.25}\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(2(m^2 + 1) = m + 2\) or \(2m^2 + 2 = m + 2\) or \(2m^2 = m\) or \(2m^2 - m = 0\) | M1 | oe equation without fractions |
| \(m(2m - 1)\ (= 0)\) or \(m(1 - 2m)\ (= 0)\) or \(\dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 2 \times 0}}{2 \times 2}\) | M1dep | oe eg \(\dfrac{1}{4} \pm \sqrt{\dfrac{1}{16}}\) may be implied by both correct solutions |
| 0.5 \(\quad\) 0 or \(\dfrac{1}{2} \quad 0\) | A1 | oe |
Additional guidance
| 0.5 \(\quad\) 0 in working but only one of these on answer line | M2A0 |
| Equation left in terms of \(p\) and \(m\) | Zero |
| Answers only of 0.5 \(\quad\) 0 | M2A1 |
| Answer only of 0.5 | Zero |
| Answer only of 0 | Zero |
| If using formula with an error seen the maximum mark is M1M0A0 eg \(\ 2m^2 - m = 0 \quad \dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 1 \times 0}}{2 \times 2}\) Answers 0.5 \(\quad\) 0 | M1M0A0 |