Higher June 2025 Paper 1 Q17
17 \(\mathrm{g}(x) = 1 - 3x \qquad \mathrm{h}(x) = 2x^2 - 1\)
Show that \(3\mathrm{gh}(x) + \mathrm{hg}(x) = 0\) has just one solution for \(x\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for method to finding \(\mathrm{gh}(x)\), eg \(\mathrm{gh}(x) = 1 - 3(2x^2 - 1)\) |
| M1 | for method to find \(\mathrm{hg}(x)\), eg \(\mathrm{hg}(x) = 2(1 - 3x)^2 - 1\) | |
| M1 | (dep M2) for method to find \(3\mathrm{gh}(x) + \mathrm{hg}(x)\) eg \(3(1 - 3(2x^2 - 1)) + 2(1 - 3x)^2 - 1\ (= 0)\) | |
| M1 | for expanding all brackets as far as at least \(3 - 18x^2 + 9 + 2 - 12x + 18x^2 - 1\ (= 0)\) | |
| C1 | for reducing to a linear equation eg \(13 - 12x = 0\) and stating that this gives just one solution or stating \(x = \dfrac{13}{12}\) oe |
Additional guidance
\(= 1 - 6x^2 + 3\)
\(= 4 - 6x^2\)
\(= 2(1 - 3x - 3x + 9x^2) - 1\)
\(= 2 - 12x + 18x^2 - 1\)
\(= 1 - 12x + 18x^2\)
Expressions for \(\mathrm{gh}(x)\) and \(\mathrm{hg}(x)\) may have been incorrectly expanded and simplified
Need not be fully simplified but must be correct