Date | May 2009 | Marks available | 4 | Reference code | 09M.1.sl.TZ2.8 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Let \(f(x) = {{\rm{e}}^{ - 3x}}\) and \(g(x) = \sin \left( {x - \frac{\pi }{3}} \right)\) .
Write down
(i) \(f'(x)\) ;
(ii) \(g'(x)\) .
Let \(h(x) = {{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right)\) . Find the exact value of \(h'\left( {\frac{\pi }{3}} \right)\) .
Markscheme
(i) \( - 3{{\rm{e}}^{ - 3x}}\) A1 N1
(ii) \(\cos \left( {x - \frac{\pi }{3}} \right)\) A1 N1
[4 marks]
evidence of choosing product rule (M1)
e.g. \(uv' + vu'\)
correct expression A1
e.g. \( - 3{{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3x}}\cos \left( {x - \frac{\pi }{3}} \right)\)
complete correct substitution of \(x = \frac{\pi }{3}\) (A1)
e.g. \( - 3{{\rm{e}}^{ - 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right)\)
\(h'\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ - \pi }}\) A1 N3
[4 marks]
Examiners report
A good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixed success.
Often the substitution of \({\frac{\pi }{3}}\) was incomplete or not done at all.