4MA0 Higher June 2015 Paper 4R Q9
9
(a) Expand \(5(2p - 3)\) (1)
(b) Solve the inequality \(9 - 2x \lt 3\) (2)
(c) \(y = x^3 - kx + 5\)
Work out the value of \(k\) when \(y = 6\) and \(x = -2\) (3)
Work out the value of \(k\) when \(y = 6\) and \(x = -2\) (3)
(d) Solve \(\dfrac{1}{f + 2} = 3\) (2)
| Scheme | Marks |
|---|---|
| \(10p - 15\) | B1 |
| (1) |
Notes
B1: Accept \(10 \times p - 15\)
| Scheme | Marks |
|---|---|
| \(9 - 3 \lt 2x\) or \(6 \lt 2x\) or \(-2x \lt 3 - 9\) or \(-2x \lt -6\) | M1 |
| \(x \gt 3\) | A1 |
| (2) |
Notes
M1: Allow equals sign (Eg \(6 = 2x\)) or condone incorrect inequality sign (eg \(6 \gt 2x\))
A1: Allow \(3 \lt x\)
NB: Final answer must be an inequality
SCB1 for \(x \geqslant 3\) or \(x \lt 3\) or \(x = 3\) or 3 as an answer
| Scheme | Marks |
|---|---|
| \(6 = (-2)^3 - k(-2) + 5\) or \(6 = -8 + 2k + 5\) | M1 |
Eg \(6 + 8 - 5 = 2k\) or \(-2k = -8 + 5 - 6\) or \(9 = 2k\) or \(-9 = -2k\) or \(k = \dfrac{(-2)^3 - 6 + 5}{-2}\) or \(-k = \dfrac{6 - (-2)^3 - 5}{-2}\) or \(-k = -4.5\) | M1 |
| 4.5 | A1 |
| (3) |
Notes
M1: For correct substitution
Allow omission of brackets.
M1: For correctly isolating \(2k\) or \(-2k\) or \(k\) or \(-k\) in a correct equation.
A1: Accept \(4\dfrac{1}{2}\), \(\dfrac{9}{2}\)
| Scheme | Marks |
|---|---|
| \(1 = 3(f + 2)\) or \(1 = 3f + 6\) or \(\dfrac{1}{3} = f + 2\) | M1 |
| \(-1\frac{2}{3}\) oe | A1 |
| (2) |
Notes
A1: Accept −1.66(666...) correctly rounded or truncated to at least 3 significant figures.