| Date | November 2011 | Marks available | 3 | Reference code | 11N.1.sl.TZ0.14 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 14 | Adapted from | N/A |
Question
\[f(x) = \frac{1}{3}{x^3} + 2{x^2} - 12x + 3\]
Find \(f'(x)\) .
Find the interval of \(x\) for which \(f(x)\) is decreasing.
Markscheme
\(f'(x) = {x^2} + 4x - 12\) (A1)(A1)(A1) (C3)
Notes: Award (A1) for each term. Award at most (A1)(A1)(A0) if other terms are seen.
[3 marks]
\( - 6 \leqslant x \leqslant 2\) OR \( - 6 < x < 2\) (A1)(ft)(A1)(ft)(A1) (C3)
Notes: Award (A1)(ft) for \( - 6\), (A1)(ft) for \(2\), (A1) for consistent use of strict (\(<\)) or weak (\(\leqslant\)) inequalities. Final (A1) for correct interval notation (accept alternative forms). This can only be awarded when the left hand side of the inequality is less than the right hand side of the inequality. Follow through from their solutions to their \(f'(x) = 0\) only if working seen.
[3 marks]
Examiners report
This question was quite a good differentiator with many able to score at least one mark in part (a). Part (b) proved however to be quite a challenge as many candidates did not seem to understand what was required and were unable to use their answer to part (a) to help them to meet the demands of this question part. The top quartile scored well with virtually everyone scoring at least three marks. The picture was somewhat reversed with the lower quartile with the majority of candidates scoring 2 or fewer marks.
This question was quite a good differentiator with many able to score at least one mark in part (a). Part (b) proved however to be quite a challenge as many candidates did not seem to understand what was required and were unable to use their answer to part (a) to help them to meet the demands of this question part. The top quartile scored well with virtually everyone scoring at least three marks. The picture was somewhat reversed with the lower quartile with the majority of candidates scoring 2 or fewer marks.
