1MA0 Higher March 2013 Paper 2 Q8
8 The equation
\[x^3 - 3x = 15\]has a solution between 2 and 3
Use a trial and improvement method to find this solution.
Give your answer correct to 1 decimal place.
You must show all your working. (4)
| Working | Answer | Mark | Notes | ||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2.9 | 4 | B2 for a trial \(2.8 \leqslant x \leqslant 2.9\) evaluated correctly (B1 for a trial evaluated correctly for \(2 \leqslant x \leqslant 3\)) B1 for a different trial evaluated correctly for \(2.85 \leqslant x \lt 2.9\) B1 (dep on at least one previous B1) for 2.9 |
NB
For trials where \(x\) has one decimal place:
\(x \leqslant 2.6\) trials must be evaluated to at least 1 sf truncated or rounded
\(2.6 \lt x \lt 2.85\) trials must be evaluated to at least 2 sf truncated or rounded
\(2.85 \leqslant x \leqslant 2.9\) trials must be evaluated to at least 3 sf truncated or rounded
NB. Accept 15 or 15.0 for trial at \(x = 2.87\)
No working scores 0 marks.
If candidate is clearly working with \(x^3 - 3x - 15 = 0\) then use same scheme as above but subtract 15 from all evaluated values in the table