4MA0 Higher January 2018 Paper 4R Q13
13 Here is the graph of \(y = x^3 - 4x^2 + 1\)
(a) Use the graph to solve the equation \(x^3 - 4x^2 + 1 = 2\) (1)
(b) For which values of \(k\) does the equation \(x^3 - 4x^2 + 1 = k\) have exactly two solutions. (2)
The equation \(x^3 - 4x^2 + x - 2 = 0\) can be solved by finding the point of intersection of the line L with the graph of \(y = x^3 - 4x^2 + 1\)
(c) Find an equation of L. (2)
| Scheme | Marks |
|---|---|
| 4.06 | B1 |
| (1) |
Notes
B1: Accept 4 – 4.1
| Scheme | Marks |
|---|---|
| B1 | |
| 1 or \(k = -8.5\) (1d.p.) | B1 |
| (2) |
Notes
B1: For \(k = 1\)
B1: \(k = -8.5\) (accept \(k = -8.6\) to −8.4)
| Scheme | Marks |
|---|---|
| \(y = ax + 3\) or \(y = -x + b\) or for \(3 - x\) oe | M1 |
| \(y = 3 - x\) | A1 |
| (2) | |
| (5 marks) |
Notes
M1: For \(y = ax + 3\) or \(y = -x + b\) or for \(3 - x\)