Higher January 2019 Paper 1 Q12
12 The curve \(C\) has equation \(y = \dfrac{1}{3}x^3 - 9x + 1\)
(a) Find \(\dfrac{\text{d}y}{\text{d}x}\) (2)
(b) Find the range of values of \(x\) for which \(C\) has a negative gradient. (3)
| Scheme | Marks |
|---|---|
| \(3 \times \dfrac{1}{3}x^2 - 9\) | M1 |
| \(x^2 - 9\) oe | A1 |
| (2) |
Notes
M1: for \(3 \times \dfrac{1}{3}x^2\) oe or −9 oe
A1: or for \(1x^2 - 9\) oe
| Scheme | Marks |
|---|---|
| \(-3 \lt x \lt 3\) oe | B3 |
| (3) | |
| (5 marks) |
Notes
B3: may be seen as two separate inequalities
if not B3 then award B2 for \(x \lt 3\) or \(x \gt -3\) or \(-3 \leqslant x \leqslant 3\)
if not B2 then award B1 for \(x^2 - 9 \lt 0\) or \(x^2 \lt 9\) oe or for \((x - 3)(x + 3)\) or for (\(x\) =) \(\pm 3\) (values maybe seen in incorrect inequalities)
SC: If no marks awarded and M1 awarded in (a) then award B1 for “quadratic” < 0