| Date | May 2017 | Marks available | 4 | Reference code | 17M.1.hl.TZ0.7 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
The function \(f\) is defined by
\[f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} + 2\cos x}}{4},{\text{ }}x \in \mathbb{R}.\]
The random variable \(X\) has a Poisson distribution with mean \(\mu \).
Show that \({f^{(4)}}x = f(x)\);
By considering derivatives of \(f\), determine the first three non-zero terms of the Maclaurin series for \(f(x)\).
Write down a series in terms of \(\mu \) for the probability \(p = {\text{P}}[X \equiv 0(\bmod 4)]\).
Show that \(p = {{\text{e}}^{ - \mu }}f(\mu )\).
Determine the numerical value of \(p\) when \(\mu = 3\).
Markscheme
\(f’(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}} - 2\sin x}}{4}\) (A1)
\(f’’(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} - 2\cos x}}{4}\) (A1)
\(f’’’(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}} + 2\sin x}}{4}\) (A1)
\({f^{(4)}}(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} + 2\cos x}}{4} = f(x)\) AG
[4 marks]
therefore,
\(f(0) = 1\) and \({f^{(4)}}(0) = 1\) (A1)
\(f’(0) = f''(0) = f'''(0) = 0\) (A1)
the sequence of derivatives repeats itself so the next non-zero derivative is \({f^{(8)}}(0) = 1\) (A1)
the MacLaurin series is \(1 + \frac{{{x^4}}}{{4!}} + \frac{{{x^8}}}{{8!}}( + \ldots )\) (M1)A1
[4 marks]
\(p = {\text{P}}(X = 0) + {\text{P}}(X = 4) + {\text{P}}(X = 8) + \ldots \) (M1)
\( = \frac{{{{\text{e}}^{ - \mu }}{\mu ^0}}}{{0!}} + \frac{{{{\text{e}}^{ - \mu }}{\mu ^4}}}{{4!}} + \frac{{{{\text{e}}^{ - \mu }}{\mu ^8}}}{{8!}} + \ldots \) A1
[??? marks]
\(p = {{\text{e}}^{ - \mu }}\left( {1 + \frac{{{\mu ^4}}}{{4!}} + \frac{{{\mu ^8}}}{{8!}} + \ldots } \right)\) A1
\( = {{\text{e}}^{ - \mu }}f(\mu )\) AG
[??? marks]
\(p = {{\text{e}}^{ - 3}}\left( {\frac{{{{\text{e}}^3} + {{\text{e}}^{ - 3}} + 2\cos 3}}{4}} \right)\) (M1)
\( = 0.226\) A1
[??? marks]
