Higher June 2023 Paper 3 Q18
18
\(\mathrm{f}(x) = x^2 + 6x\)
\(\mathrm{g}(x) = 2x + 4\)
(a) Show that \(\quad \mathrm{fg}(x) = 4x^2 + 28x + 40\) [3 marks]
(b) Solve \(\quad \mathrm{fg}(x) = -5\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((2x + 4)^2 + 6(2x + 4)\) | M1 | may be seen in a grid |
| \(4x^2 + 8x + 8x + 16 + 12x + 24\) or \(4x^2 + 16x + 16 + 12x + 24\) | M1dep | fully expanded expression with terms summed allow one omission or one arithmetic error |
| \(4x^2 + 8x + 8x + 16 + 12x + 24\) or \(4x^2 + 16x + 16 + 12x + 24\) and \(4x^2 + 28x + 40\) | A1 |
Additional guidance
\(4x^2 + 16 + 12x + 24\) is two errors
| Answer | Mark | Comments |
|---|---|---|
| \(4x^2 + 28x + 45\ (= 0)\) | M1 | must be correct |
| \((2x + 5)(2x + 9)\ (= 0)\) or \((2x + 7)^2 - 49 + 45\ (= 0)\) or \(\dfrac{-28 \pm \sqrt{28^2 - 4 \times 4 \times 45}}{2 \times 4}\) or \(\dfrac{-28 \pm \sqrt{64}}{8}\) or \(\dfrac{-28 \pm 8}{8}\) or \(\dfrac{-7 \pm \sqrt{4}}{2}\) | M1dep | oe implies first M1 |
| \((x =)\ {-2.5}\) and \((x =)\ {-4.5}\) | A1 | oe fraction or decimal SC2 \((x =)\) [\(-1.63\), \(-1.629\)] and \((x =)\) [\(-5.371\), \(-5.37\)] |
Additional guidance
| SC2 from using \(4x^2 + 28x + 35\ (= 0)\) | |
| Trial and improvement with both answers correct and chosen from any list | M1M1A1 |
| Trial and improvement with one answer correct | M0M0A0 |