(a)     Assuming the Maclaurin series for ${{\text{e}}^x}$, determine the first three non-zero terms in the Maclaurin expansion of $\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}$.

(b)     The random variable $X$ has a Poisson distribution with mean $\mu$. Show that ${\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = a + b{{\text{e}}^{c\mu }}$ where $a$, $b$ and $c$ are constants whose values are to be found.

(a)     ${{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \frac{{{x^5}}}{{5!}} +  \ldots$

${{\text{e}}^{ - x}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^5}}}{{5!}} +  \ldots$     A1

$\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2} = x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} +  \ldots$     (M1)A1

Note: Accept any valid (otherwise) method.

[3 marks]

(b)     ${\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = {\text{P}}(X = 1,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots )$     (M1)

$= {{\text{e}}^{ - \mu }}\left( {\mu  + \frac{{{\mu ^3}}}{{3!}} + \frac{{{\mu ^5}}}{{5!}} +  \ldots } \right)$     A1

$= \frac{{{{\text{e}}^{ - \mu }}({{\text{e}}^\mu } - {{\text{e}}^{ - \mu }})}}{2}$     A1

$= \frac{1}{2} - \frac{1}{2}{{\text{e}}^{ - 2\mu }}$     A1

$\left( {a = \frac{1}{2},{\text{ }}b =  - \frac{1}{2},{\text{ }}c =  - 2} \right)$

[4 marks]