Date | November 2015 | Marks available | 6 | Reference code | 15N.3ca.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
Let f(x)=exsinx.
Show that f″.
By further differentiation of the result in part (a) , find the Maclaurin expansion of f(x), as far as the term in {x^5}.
Markscheme
f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x M1A1
f''(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x = 2{{\text{e}}^x}\cos x A1
= 2\left( {{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x} \right) M1
= 2\left( {f'(x) - f(x)} \right) AG
[4 marks]
f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = 2(1 - 0) = 2 (M1)A1
Note: Award M1 for attempt to find f(0), f'(0) and f''(0).
f'''(x) = 2\left( {f''(x) - f'(x)} \right) (M1)
f'''(0) = 2(2 - 1) = 2,{\text{ }}{f^{IV}}(0) = 2(2 - 2) = 0,{\text{ }}{f^V}(0) = 2(0 - 2) = - 4 A1
so f(x) = x + \frac{2}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} - \frac{4}{{5!}} + \ldots (M1)A1
= x + {x^2} + \frac{1}{3}{x^3} - \frac{1}{{30}}{x^5} + \ldots
[6 marks]
Total [10 marks]