Higher November 2018 Paper 1 Q22
22. The functions \(g(x)\) and \(h(x)\) are defined for all real \(x\) by
\(g(x) = 4 - 3x,\)
\(h(x) = x^2 + 5.\)
(a) Solve \(g^{-1}(x) = x + 2\). [4]
(b) Find and simplify an expression for \(gg(x)\). [2]
(c) Explain why \(gh(-1) = gh(1)\). [1]
| Scheme | Marks | Comments |
|---|---|---|
| \(g^{-1}(x) = \dfrac{4 - x}{3}\) | B2 | B1 for \(x = \dfrac{4 - y}{3}\) or equivalent unless \(x\) and \(y\) interchanged later or SC1 for \(y\) or \(g^{-1}(x) = \dfrac{4 + y}{3}\) or equivalent |
| \(4 - x = 3x + 6\) or equivalent | M1 | |
| \(x = -\dfrac{1}{2}\) | A1 |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(x = g(x + 2)\) | B1 | |
| \(x = 4 - 3(x + 2)\) | B1 | |
| \(x = -2 - 3x\) or equivalent | M1 | |
| \(x = -\dfrac{1}{2}\) | A1 |
Possible misprint (note from Westie's Workshop)
- The special case “SC1 for \(y\) or \(g^{-1}(x) = \dfrac{4 + y}{3}\)” mixes the letters. It probably rewards the sign slip \(\dfrac{4 + x}{3}\) (or \(x = \dfrac{4 + y}{3}\) before swapping), but the board does not make clear which.
We have left the board's wording exactly as printed. Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| \(4 - 3(4 - 3x)\) | M1 | |
| \(9x - 8\) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| Valid explanation e.g. ‘\(h(-1) = h(1) = 6\) so both are equal to \(g(6)\).’ | E1 | Allow ‘\(1^2\) is the same as \((-1)^2\)’. Allow for working out \(gh(-1)\) and \(gh(1)\) and showing they are both e.g. \(-14\). |
| (7) |