| Date | None Specimen | Marks available | 4 | Reference code | SPNone.1.hl.TZ0.11 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 11 | Adapted from | N/A |
Question
The function \(f\) is defined by \(f(x) = {{\rm{e}}^x}\cos x\) .
Show that \(f''(x) = - 2{{\rm{e}}^x}\sin x\) .
Determine the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
By differentiating your series, determine the Maclaurin series for \({{\rm{e}}^x}\sin x\) up to the term in \({x^3}\) .
Markscheme
\(f'(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x\) A1
\(f''(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\cos x\) A1
\( = - 2{{\rm{e}}^x}\sin x\) AG
[2 marks]
\(f'''(x) = - 2{{\rm{e}}^x}\sin x - 2{{\rm{e}}^x}\cos x\) A1
\({f^{IV}}(x) = - 4{{\rm{e}}^x}\cos x\) A1
\(f(0) = 1\), \(f'(0) = 1\) ,\(f''(0) = 0\), \(f'''(0) = - 2\), \({f^{IV}}(0) = - 4\) (A1)
the Maclaurin series is
\({{\rm{e}}^x}\cos x = 1 + x - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6} + \ldots \) M1A1
Note: Accept multiplication of series method.
[5 marks]
differentiating,
\({{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x = 1 - {x^2} - \frac{{2{x^3}}}{3} + \ldots \) M1A1
\({{\rm{e}}^x}\sin x = 1 + {x^{}} - \frac{{{x^3}}}{3} + \ldots - 1 + {x^2} + \frac{{2{x^3}}}{3} + \ldots \) M1
\( = x + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
[4 marks]
