Show that $f''(x) = - \frac{1}{{(1 + \sin x)}}$ .

(i)     Find the Maclaurin series for $f(x)$ up to and including the term in ${x^4}$ .

(ii)     Explain briefly why your result shows that f is neither an even function nor an odd function.

Determine the value of $\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}}$.

$f'(x) = \frac{{\cos x}}{{1 + \sin x}}$     A1

$f''(x) = \frac{{ - \sin x(1 + \sin x) - \cos x\cos x}}{{{{(1 + \sin x)}^2}}}$     M1A1

$= \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}$     A1

$= - \frac{1}{{1 + \sin x}}$     AG

[4 marks]

(i)     $f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}$     A1

${f^{(4)}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}$     M1A1

$f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = - 1$     M1

$f'''(0) = 1,{\text{ }}{f^{(4)}}(0) = - 2$     A1

$f(x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots$     A1

(ii)     the series contains even and odd powers of x     R1

[7 marks]

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} +  \ldots  - x}}{{{x^2}}}$     M1

$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} + \frac{x}{6} +  \ldots }}{1}$     (A1)

$= - \frac{1}{2}$     A1

Note: Use of l’Hopital’s Rule is also acceptable.

[3 marks]