Higher June 2025 Paper 2 Q25
25 \(\mathrm{f}(x) = ax + b \quad\) and \(\quad \mathrm{g}(x) = \dfrac{x + b}{a} \quad\) where \(a\) and \(b\) are positive integers.
Prove that \(\quad \mathrm{fg}(x) - a\mathrm{f}^{-1}(x) \quad\) is always a multiple of 3 [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(a\left(\dfrac{x + b}{a}\right) + b\) | B1 | oe expression for \(\mathrm{fg}(x)\) eg \(x + b + b\) or \(x + 2b\) implied by 3rd B1 |
| \(\dfrac{x - b}{a}\) | B1 | oe expression for \(\mathrm{f}^{-1}(x)\) eg \(\dfrac{x}{a} - \dfrac{b}{a}\) implied by \(a\left(\dfrac{x - b}{a}\right)\) or \(-a\left(\dfrac{x - b}{a}\right)\) implied by 3rd B1 |
| \(a\left(\dfrac{x + b}{a}\right) + b - a\left(\dfrac{x - b}{a}\right)\) | B1 | oe correct full expression in terms of \(x\) eg \(x + b + b - (x - b)\) or \(x + 2b - x + b\) implies B3 |
| B3 awarded and \(3b\) with no errors | B1 |
Additional guidance
| \(a\left(\dfrac{x + b}{a}\right) + b = x + 2b\) \(x + 2b - a\left(\dfrac{x - b}{a}\right)\) \(= 3b \qquad\) (no errors) | B1 B1B1 B1 |
| \(x + b + b - (x - b)\) \(x + b + b - x - b \qquad\) (error) \(= 3b\) | B1B1B1 B0 |
| Ignore explanation about why \(3b\) is a multiple of 3 | |
| Substitution of values for letters with no further correct work | Zero |
| Accept eg \(a\left(\dfrac{x + b}{a}\right)\) written as \(a\dfrac{x + b}{a}\) | |
| \(x + 2b\) \(\dfrac{x - b}{a}\) \(x + 2b - x - b \qquad\) (do not allow recovery of missing brackets) \(= 3b\) | B1 B1 B0 B0 |