Higher November 2018 Unit 1 Q1
1.
(a) Expand \(3x(x^2 - 2)\). [2]
(b) Make \(g\) the subject of the formula \(f = 2 - 3g\). [2]
(c)
(i) Solve \(7x - 3 \lt 29\). [2]
(ii) What is the greatest integer value of \(x\) that satisfies the above inequality? [1]
| Scheme | Marks | Comments |
|---|---|---|
| \(3x^3 - 6x\) | B2 | Must be in an expression for B2. B1 for sight of (+)\(3x^3\) or \(-6x\). Mark final answer. |
| Scheme | Marks | Comments |
|---|---|---|
| \(3g = 2 - f\) or \(f - 2 = -3g\) | B1 | |
| \(g = \dfrac{2 - f}{3}\) or \(g = \dfrac{f - 2}{-3}\) or \(g = \dfrac{2}{3} - \dfrac{f}{3}\) | B1 | F.T only from (±)\(3g = \pm f \pm 2\). B1B0 for \(-g = \dfrac{f - 2}{3}\). B1B0 for \(g = 2 - f \div 3\). B1B0 for \(\dfrac{2 - f}{3}\) (‘\(g\)’ missing). Mark final answer. |
| Scheme | Marks | Comments |
|---|---|---|
| (i) \(7x \lt 32\) | B1 | |
| \(x \lt 32/7\) or \(x \lt 4\,{}^{4}\!/_{7}\) | B1 | Use of ‘=’ is B0B0 unless replaced for final answer. FT from \(7x \lt k\). Allow \(x \lt 4{\cdot}57\)(….). Do not allow \(x \lt 4{\cdot}6\) or \(x \lt 4{\cdot}5\) unless \(x \lt 4{\cdot}57\)(….) seen. Mark final answer. Penalise consistent use of ‘\(\leqslant\)’ by –1. |
| (ii) 4 | B1 | OR F.T. ‘their answer (inequality) in (c)(i)’ if \(x \lt a\). No FT from \(x \leqslant a\). \(4x\) is B0. |