Higher June 2024 Paper 2R Q17
17 \(\mathrm{f}(x) = \dfrac{x}{2x - 4} \qquad \mathrm{g}(x) = 3x + 1\)
Given that \(\mathrm{fg}(k) = 2\)
work out the value of \(k\)
(3)
| Scheme | Marks |
|---|---|
\((\mathrm{fg}(k) =)\, \dfrac{3k + 1}{2(3k + 1) - 4}\) oe or \(\dfrac{3k + 1}{2(3k + 1) - 4} = 2\) oe or \((\mathrm{fg}(k) =)\, \dfrac{3k + 1}{6k - 2}\) oe or \(\dfrac{3k + 1}{6k - 2} = 2\) oe or \(x = 2(2x - 4)\) or \(x = 4x - 8\) or \(x = \dfrac{8}{3}\) oe | M1 |
\(3k + 1 = 2(6k - 2)\) oe or \(3k + 1 = 2\big(2(3k + 1) - 4\big)\) oe or \(3k + 1 = 12k - 4\) oe or \(3k + 1 = \dfrac{8}{3}\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{5}{9}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct expression for fg(\(k\)) or fg(\(x\)) or
for f(\(x\)) = 2
Allow \(x\) instead of \(k\) for all marks
M1: dep on M1 for correctly removing the denominator to form a correct equation
or
for g(\(k\)) = \(\dfrac{8}{3}\)
A1: oe eg 0.55(555…) rounded or truncated
or
\(0.\dot{5}\) (must show recurring)