Higher November 2024 Paper 4 Q20
20
(a) Show that the equation \(x^3 - 3x - 4 = 0\) has a solution between \(x = 2\) and \(x = 3\). [3]
(b) Use \(x = 2.5\) to find a smaller interval for the solution to \(x^3 - 3x - 4 = 0\).
You must show your working. [2]
You must show your working. [2]
(c) Find this solution correct to 1 decimal place.
You must show your working. [3]
You must show your working. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(2^3 - 3 \times 2 - 4 =\)] –2 [\(3^3 - 3 \times 3 - 4 =\)] 14 | M1 M1 | Must indicate their input and output | Accept other values of \(x\) used between 2 and 3, correct to 2 figs rot, (see table in part (c)). For full marks, the two values need to produce a sign change. |
| and any indication of a sign change [so solution lies between 2 and 3] | A1 | Dep. on at least M1 and different signs | Alternative method SC3 for using an iterative equation that converges and concluding statement that first two values lie between 2 and 3 oe |
Appendix: Exemplar responses for Q20(a)
| Response | Mark |
|---|---|
| Sign change ….. | A1 |
| −2 < 0 < 14 | A1(BOD) |
| answer lies between 2 and 3 | A0 |
| answer is in the middle is insufficient | A0 |
| \(x\) | fn |
|---|---|
| 2 | -2 |
| 2.1 | -1.039 |
| 2.2 | 0.048 |
| 2.3 | 1.267 |
| 2.4 | 2.624 |
| 2.5 | 4.125 |
| 2.6 | 5.776 |
| 2.7 | 7.583 |
| 2.8 | 9.552 |
| 2.9 | 11.689 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(x = 2.5\)] 4.125 | B1 | ||
| \(2 \lt x \lt 2.5\) | B1 | Condone equals signs and condone in words, allow a smaller correct interval | |
| Answer | Marks | Part marks and guidance | |||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Two correct evaluations in the range \(2 \lt x \lt 2.5\), one which gives a positive value and the other giving a negative value | M2 | M1 for one correct evaluation in the range \(2 \lt x \lt 2.5\) | Working for (c) may be seen in (b) Examples
| ||||||||||||||||||
| [\(x\) =] 2.2 | A1 | Dependent on achieving at least M1 Alternative method M1 rearranges to a correct iterative formula (converging or diverging) and M1 attempts first two iterations (either substitution seen or found to at least 2dp rot) and A1 for 2.2 OR If 0 scored SC1 for 2.2 with no worthwhile working | condone missing suffixes here e.g. \(x_{n+1} = \sqrt[3]{3x_n + 4}\) with \(x_0 = 2\), \(x_1 = 2.1544\ldots\), \(x_2 = 2.1872\ldots\) | ||||||||||||||||||