4MA0 Higher January 2016 Paper 4R Q14
14
(a) Complete the table of values for \(y = x^3 - 5x + 2\) (2)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| \(y\) | 4 | \(-2\) | 14 |
(b) On the grid, draw the graph of \(y = x^3 - 5x + 2\) for \(-3 \leqslant x \leqslant 3\) (2)
The equation \(x^3 - 6x + m = 0\), where \(m\) is an integer, has one negative solution and two positive solutions.
(c) Given that \(x = 1\) is one of the positive solutions, show that \(m = 5\) (1)
(d) By drawing a suitable straight line on the grid, find an estimate for the negative solution of \(x^3 - 6x + 5 = 0\)
Give your estimate to 1 decimal place. (2)
Give your estimate to 1 decimal place. (2)
| Scheme | Marks |
|---|---|
| −10, (4), 6, 2, (−2), 0, (14) | B2 |
| (2) |
Notes
B2: Award B1 for any 2 correct.
| Scheme | Marks |
|---|---|
| (−3, −10), (−2, 4), (−1, 6), (0, 2), (1, −2), (2, 0), (3, 14) Answer: Correct curve | B2 |
| (2) |
Notes
B2: For the correct smooth curve.
B1 for at least 5 points plotted correctly; ft from table for plotting only provided at least B1 scored in (a).
| Scheme | Marks |
|---|---|
| \(1^3 - 6 \times 1 + m = 0\) or \(1 - 6 + m = 0\) Answer: 5 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| M1 | |
| −2.7 to −2.9 | A1 |
| (2) |
Notes
M1: \(y = x - 3\) drawn
A1: ft from graph (dep on M1)