Higher November 2017 Paper 3 Q15
15
(a) Show that the equation \(x^3 + 7x - 5 = 0\) has a solution between \(x = 0\) and \(x = 1\) (2)
(b) Show that the equation \(x^3 + 7x - 5 = 0\) can be arranged to give \(x = \dfrac{5}{x^2 + 7}\) (2)
(c) Starting with \(x_0 = 1\), use the iteration formula \(x_{n+1} = \dfrac{5}{x_n^{\,2} + 7}\) three times to find an estimate for the solution of \(x^3 + 7x - 5 = 0\) (3)
(d) By substituting your answer to part (c) into \(x^3 + 7x - 5\), comment on the accuracy of your estimate for the solution to \(x^3 + 7x - 5 = 0\) (2)
| Answer | Mark | Notes |
|---|---|---|
| Shown | M1 | for method to establish at least one root between \(x = 0\) and \(x = 1\), eg \(\mathrm{f}(0) = -5\) and \(\mathrm{f}(1) = 3\) |
| C1 | for correct values and a deduction about the roots eg as there is a sign change there must be at least one root between \(x = 0\) and \(x = 1\) (as f is continuous) |
| Answer | Mark | Notes |
|---|---|---|
| Shown | C1 | for a correct first step in rearrangement, eg \(x(x^2 + 7) - 5 = 0\) or \(x^3 + 7x = 5\) |
| C1 | for clear and correct steps showing complete rearrangement |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \({x_1 = 0.625}\) \({x_2 = 0.6765327696}\) \({x_3 = 0.6704483001}\) | 0.6704(483001) | M1 | for substitution of 1 into the formula (to get 0.625) |
| M1 | for substitution of \(\text{``}x_1 = 0.625\text{''}\) and \(\text{``}x_2 = 0.6765327696\text{''}\) to give \(x_2\) and \(x_3\) | ||
| A1 | 0.6704(483001) |
| Answer | Mark | Notes |
|---|---|---|
| Comment | M1 | substitutes answer to (c) into expression (to get \(-0.00549\ldots\)) |
| C1 | appropriate comment, eg accurate as answer is close to 0 |