Higher June 2018 Paper 3 Q18
18
(a) Show that the equation \(x^3 + x = 7\) has a solution between 1 and 2 (2)
(b) Show that the equation \(x^3 + x = 7\) can be rearranged to give \(x = \sqrt[3]{7 - x}\) (1)
(c) Starting with \(x_0 = 2\),
use the iteration formula \(x_{n+1} = \sqrt[3]{7 - x_n}\) three times to find an estimate for a solution of \(x^3 + x = 7\) (3)
use the iteration formula \(x_{n+1} = \sqrt[3]{7 - x_n}\) three times to find an estimate for a solution of \(x^3 + x = 7\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Correct statement | C1 | for substituting both 1 and 2 into \(x^3 + x\) or into \(x^3 + x - 7\) |
| C1 | for values 2 and 10 plus explanation that these are above and below 7, or for values \(-5\) and 3 plus explanation that there is a change of sign, thus implying a solution lies between 1 and 2 |
Additional guidance
All arithmetic shown must be correct.
Ignore any additional trials shown.
| Answer | Mark | Mark scheme |
|---|---|---|
| Correct rearrangement | C1 | for correct algebraic rearrangement |
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.74 | M1 | for substitution of 2 into the formula eg \(\sqrt[3]{7 - 2}\) (= 1.70997…) |
| M1 | for a substitution of \(x_1\) to give \(x_2\) (= 1.74241...) | |
| A1 | for answer in the range 1.738 to 1.74 |
Additional guidance
\(x_1 = 1.70997....\)
\(x_2 = 1.74241....\)
\(x_3 = 1.73884....\)
Accept an accuracy of 2 dp or more rounded or truncated for values of \(x_1\) and \(x_2\)
Award the marks for 1.7 on the answer line provided correct iterations are shown in the working space.