Higher November 2018 Paper 1 Q25
25. A solution of the equation \(x^2 - 0{\cdot}5 = 0\) lies between 0 and 1.
Jane uses the method of decimal search to find the solution correct to one decimal place.
(a) Complete Jane’s table. [1]
| \(x\) | 0 | 0·1 | 0·2 | 0·3 | 0·4 | 0·5 | 0·6 | 0·7 | 0·8 | 0·9 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(x^2 - 0{\cdot}5\) | −0·5 | −0·41 | −0·34 | −0·25 | −0·14 | −0·01 | 0·14 | 0·31 | 0·5 |
(b) Explain why Jane’s solution lies between 0·7 and 0·8. [1]
(c) You are given that \(74^2 = 5476\).
Show that Jane’s solution is 0·7 correct to one decimal place. [2]
Show that Jane’s solution is 0·7 correct to one decimal place. [2]
| Scheme | Marks | Comments |
|---|---|---|
| \(-0.49,\ -0.46\) | B1 |
| Scheme | Marks | Comments |
|---|---|---|
| Valid explanation e.g. ‘There is a change of sign (so the value will be in this range).’ | E1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(0.74^2 - 0.5\) or \(0.74^2 = 0.5476\) | B1 | |
| \((0.5476 - 0.5 =)\ 0.0476\) AND valid conclusion e.g. ‘The answer is between 0.7 and 0.74’ or ‘\(0.0476 \gt 0\)’ | B1 | |
| (4) |