Higher November 2023 Paper 2 Q17
17
(a) Show that the equation \(x^3 + 2x - 6 = 0\) has a solution between \(x = 1\) and \(x = 2\) (2)
(b) Show that the equation \(x^3 + 2x - 6 = 0\) can be rearranged to give \(x = \dfrac{6}{x^2 + 2}\) (1)
(c) Starting with \(x_0 = 1.45\)
use the iteration formula \(x_{n + 1} = \dfrac{6}{x_n^{\,2} + 2}\) twice to find an estimate for the solution of \(x^3 + 2x - 6 = 0\)
Give your answer correct to 4 decimal places. (3)
use the iteration formula \(x_{n + 1} = \dfrac{6}{x_n^{\,2} + 2}\) twice to find an estimate for the solution of \(x^3 + 2x - 6 = 0\)
Give your answer correct to 4 decimal places. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | For a start to establish at least one root in [1,2], eg substitutes \(x = 1\) or \(x = 2\) into LHS, eg \(1^3 + 2 \times 1 - 6\), or \(2^3 + 2 \times 2 - 6\) |
| C1 | For a complete argument, eg evaluates for both values of \(x\) and states since there is a sign change there must be at least one root in \(1 \lt x \lt 2\) (as f is continuous) |
Additional guidance
\(\text{f}(x) = x^3 + 2x - 6\)
\(\text{f}(1) = -3\)
\(\text{f}(2) = 6\)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shows re- arrangement | C1 | for showing re-arrangement; must see \(x^3 + 2x - 6 = 0\) leading to \(x(x^2 + 2) - 6 = 0\) or \(x(x^2 + 2) = 6\) or \(x = \dfrac{6}{x^2 + 2}\) leading to \(x^3 + 2x = 6\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.4496 | M1 | for \(x_1 = \dfrac{6}{1.45^2 + 2}\ (= 1.462(522851\ldots))\) |
| M1 | for \(x_2 = \dfrac{6}{\text{``}1.462(52\ldots)\text{''}^2 + 2}\ (= 1.449634937)\) | |
| A1 | for answer in the range 1.4496 to 1.4497 |
Additional guidance
If a correct value is given and then rounded or rounded incorrectly award full marks.