Higher June 2025 Paper 2R Q17
17 Make \(k\) the subject of \(p = \dfrac{8k^2 + 5}{7 - 3k^2}\)
(4)
| Scheme | Marks |
|---|---|
| \(7p - 3k^2p = 8k^2 + 5\) | M1 |
| \(7p - 5 = 8k^2 + 3k^2p\) or \(-3k^2p - 8k^2 = 5 - 7p\) | M1ft |
| eg \(7p - 5 = k^2(8 + 3p)\) or \(k^2(-3p - 8) = 5 - 7p\) | M1ft |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(k = (\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for correctly multiplying both sides by the denominator and expanding the brackets
M1ft: dep on 2 terms in \(k^2\) and 2 other terms
for correctly collecting their \(k^2\) terms on one side and their other terms on the other side
Note: eg \(8k^2 + 3k^2\) does not count as 2 terms in \(k^2\)
M1ft: dep on previous M1
for correctly factorising for \(k^2\) or \(-k^2\) in their equation
A1: oe eg \(k = (\pm)\sqrt{\dfrac{5 - 7p}{-3p - 8}}\) or \(k = (\pm)\left(\dfrac{7p - 5}{8 + 3p}\right)^{\frac{1}{2}}\)
or \(k = (\pm)\left(\dfrac{7p - 5}{8 + 3p}\right)^{0.5}\) (condone omission of \(\pm\))
NB: to award A1 we must see \(k = (\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}\) in working if \((\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}\) alone is given as an answer