Date | November 2009 | Marks available | 2 | Reference code | 09N.1.sl.TZ0.5 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Consider \(f(x) = {x^2} + \frac{p}{x}\) , \(x \ne 0\) , where p is a constant.
Find \(f'(x)\) .
There is a minimum value of \(f(x)\) when \(x = - 2\) . Find the value of \(p\) .
Markscheme
\(f'(x) = 2x - \frac{p}{{{x^2}}}\) A1A1 N2
Note: Award A1 for \(2x\) , A1 for \( - \frac{p}{{{x^2}}}\) .
[2 marks]
evidence of equating derivative to 0 (seen anywhere) (M1)
evidence of finding \(f'( - 2)\) (seen anywhere) (M1)
correct equation A1
e.g. \( - 4 - \frac{p}{4} = 0\) , \( - 16 - p = 0\)
\(p = - 16\) A1 N3
[4 marks]
Examiners report
Candidates did well on (a).
For (b), a great number of candidates substituted into the function instead of into the derivative.
The derivate of \({x^2}\) was calculated without difficulties, but there were numerous problems regarding the derivative of \(\frac{p}{x}\) . There were several candidates who considered both p and x as variables; some tried to use the quotient rule and had difficulties, others used negative exponents and were not successful.