Higher June 2025 Paper 2 Q16
16 Make \(t\) the subject of the formula \(c = \dfrac{t^2 + 3}{7 - 8t^2}\)
(4)
| Scheme | Marks |
|---|---|
| \(7c - 8ct^2 = t^2 + 3\) oe | M1 |
| \(7c - 3 = t^2 + 8ct^2\) oe or \(-8ct^2 - t^2 = 3 - 7c\) oe | M1 |
| \(7c - 3 = t^2(1 + 8c)\) oe or \(t^2(-8c - 1) = 3 - 7c\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(t = (\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for multiplying both sides by denominator and expanding the brackets
M1: ft dep on 2 terms in \(t^2\) and 2 other terms for collecting \(t^2\) terms on one side and other terms on the other side
M1: ft dep on previous M1
for factorising for \(t^2\)
A1: oe eg \(t = (\pm)\sqrt{\dfrac{3 - 7c}{-8c - 1}}\) or
\(t = (\pm)\left(\dfrac{7c - 3}{1 + 8c}\right)^{\frac{1}{2}}\) or \(t = (\pm)\left(\dfrac{7c - 3}{1 + 8c}\right)^{0.5}\)
NB To award A1 we must see \(t = (\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}\) in working if \((\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}\) alone is given as an answer