Higher June 2022 Paper 2 Q22
22. A solution, \(x\), of the equation \(x^3 + x - 1 = 0\) lies in the interval ‘between 0·5 and 1’.
Kate uses the method of interval bisection to find the solution correct to one decimal place.
(a) Show that Kate’s second interval for \(x\) is ‘between 0·5 and 0·75’. [2]
(b) Complete Kate’s third interval for \(x\),
‘between ………… and 0·75’. [1]
(c) Continue Kate’s method of interval bisection to show that the solution is \(x = 0{\cdot}7\) correct to one decimal place. [2]
| Scheme | Marks | Comments |
|---|---|---|
| \(0.5^3 + 0.5 - 1\ (= -0.375) \lt 0\) AND \(0.75^3 + 0.75 - 1\ (= 0.171875) \gt 0\) OR \(1^3 + 1 - 1\ (= 1) \gt 0\) AND \(0.75^3 + 0.75 - 1\ (= 0.171875) \gt 0\) | B2 | Allow rounded or truncated values, but must show or indicate a sign change convincingly B1 for any one of: \(0.5^3 + 0.5 - 1 \lt 0\) \(0.75^3 + 0.75 - 1 \gt 0\) \(1^3 + 1 - 1 \gt 0\) |
| Scheme | Marks | Comments |
|---|---|---|
| 0.625 | B1 | 0.625 must be written in the answer space or the interval 0.625 to 0.75 clearly implied |
| Scheme | Marks | Comments |
|---|---|---|
| Uses 0.6875 leading to interval (0.625, 0.6875) | M1 | |
| Uses 0.65625 leading to interval (0.65625, 0.6875) and correct conclusion | A1 | Allow the A1 if they continue with more intervals or fail to conclude with x = 0.7 to 1d.p. |
| (5) |