Higher June 2023 Paper 4 Q18
18
(a) Show that the equation \(x^3 + x^2 - 5 = 0\) has a solution between \(x = 1\) and \(x = 2\). [3]
(b) Find this solution correct to 1 decimal place.
You must show calculations to support your answer. [4]
You must show calculations to support your answer. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(1^3 + 1^2 - 5\)] = \(-3\) | M1 | Must indicate their input and output | Accept other values of \(x\) used between 1 and 2 (see table in part (b)). For full marks, the two values need to produce a sign change. Acceptable answers for third mark are \(x = 1\) gives answer < 0 and \(x = 2\) gives answer >0 Note: so answer lies between 1 and 2 is not sufficient on its own or answer is in the middle is insufficient If within part (a) a candidate refers to their working in part (b) you must award the marks for this method |
| [\(2^3 + 2^2 - 5\)] = 7 | M1 | ||
| Sign change [so solution between \(x = 1\) and \(x = 2\)] or \(-3 \lt 0 \lt 7\) | A1 | Dep. on at least M1 and different signs Alternative method 1 for \(x^3 + x^2 = 5\) M2 for \(1^3 + 1^2 = 2\) and \(2^3 + 2^2 = 12\) or M1 for \(1^3 + 1^2 = 2\) or \(2^3 + 2^2 = 12\) may be implied by 2 or 12 and A1 for e.g. 2 < 5 < 12 dep. on at least M1 and 5 lies inbetween their two values Alternative method 2 SC3 for using an iterative equation that converges to a value between 1.35 and 1.45 and concluding statement such as 1 < 1.35 to 1.45 < 2 or SC2 for using an iterative equation that converges to a value between 1.35 and 1.45 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Two correct evaluations in the range \(1.35 \leqslant\) values \(\leqslant 1.49\), one which gives a positive value and the other giving a negative value | M3 | M2 for two correct evaluations 1 < values < 2, one which gives a positive value and the other giving a negative value or M1 for one correct evaluation between 1 < value < 2 | figures may be rot to at least 2 s.f. (table of values below) condone missing suffixes in formula here e.g. \(x_{n+1} = \sqrt{\frac{5}{x_n + 1}}\) converges and leads to values 1.5811388… , 1.39180797… , 1.44584536… If they refer to their working in part (a) which is relevant then award up to full marks in part (b) |
| 1.4 | A1 | Dependent on achieving at least M2 Alternative method 1 See appendix for values of \(x^3 + x^2\). Alternative method 2 M1 rearranges to a correct iterative formula (converging or diverging) M1 attempts first iteration (either substitution seen or found to at least 2dp rot) M1 continues iteration to reach \(x\) in the range 1.35 to 1.45 A1 for 1.4 Dep. on M2 OR If 0 scored SC1 for 1.4 with no worthwhile working | |
Table of values of \(x^3 + x^2 - 5\)
| \(x\) | \(x^3 + x^2 - 5\) | \(x\) | \(x^3 + x^2 - 5\) |
|---|---|---|---|
| 1.1 | −2.459 | 1.35 | −0.71712 |
| 1.2 | −1.832 | 1.36 | −0.63494 |
| 1.3 | −1.113 | 1.37 | −0.55175 |
| 1.4 | −0.296 | 1.38 | −0.46753 |
| 1.5 | 0.625 | 1.39 | −0.38228 |
| 1.6 | 1.656 | 1.40 | −0.29600 |
| 1.7 | 2.803 | 1.41 | −0.20868 |
| 1.8 | 4.072 | 1.42 | −0.12031 |
| 1.9 | 5.469 | 1.43 | −0.03089 |
| 1.44 | 0.05958 | ||
| 1.45 | 0.15113 | ||
| 1.46 | 0.24374 | ||
| 1.47 | 0.33742 | ||
| 1.48 | 0.43219 | ||
| 1.49 | 0.52805 |
Appendix: Question 18(b)
Table for \(x^3 + x^2\)
| \(x\) | \(x^3 + x^2\) | \(x\) | \(x^3 + x^2\) |
|---|---|---|---|
| 1.1 | 2.541 | 1.35 | 4.28288 |
| 1.2 | 3.168 | 1.36 | 4.36506 |
| 1.3 | 3.887 | 1.37 | 4.44825 |
| 1.4 | 4.704 | 1.38 | 4.53247 |
| 1.5 | 5.625 | 1.39 | 4.61772 |
| 1.6 | 6.656 | 1.40 | 4.70400 |
| 1.7 | 7.803 | 1.41 | 4.79132 |
| 1.8 | 9.072 | 1.42 | 4.87969 |
| 1.9 | 10.469 | 1.43 | 4.96911 |
| 1.44 | 5.05958 | ||
| 1.45 | 5.15113 | ||
| 1.46 | 5.24374 | ||
| 1.47 | 5.33742 | ||
| 1.48 | 5.43219 | ||
| 1.49 | 5.52805 |