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

Determine the Maclaurin series for $f(x)$ as far as the term in ${x^4}$ .

Deduce the Maclaurin series for $\ln (1 - \sin x)$ as far as the term in ${x^4}$ .

By combining your two series, show that $\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} +  \ldots$ .

Hence, or otherwise, find $\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}$ .

He makes 5 independent measurements of the concentration of a particular solution and correctly calculates the following confidence interval for $\mu$ .

[$22.7$ , $26.1$]

Determine the confidence level of this interval.

He is now given a different solution and is asked to determine a $95\%$ confidence interval for its concentration. The confidence interval is required to have a width less than $2$. Find the minimum number of independent measurements required.

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

$f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}$     M1

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

$= \frac{{ - 1}}{{1 + \sin x}}$     AG

[4 marks]

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

${f^{iv}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}$     A1

$f(0) = 0$ , $f'(0) = 1$ , $f''(0) =  - 1$ , $f'''(0) = 1$ , ${f^{iv}}(0) = - 2$     (A2)

Note: Award A1 for 2 errors and A0 for more than 2 errors.

$\ln (1 + \sin x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots$     M1A1

[6 marks]

$\ln (1 - \sin x) = \ln (1 + \sin ( - x)) =  - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots$     M1A1

[2 marks]

Adding,     M1

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

$= - {x^2} - \frac{{{x^4}}}{6} +  \ldots$     A1

$\ln \cos x = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}} +  \ldots$     A1

$\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} +  \ldots$    AG

[4 marks]

$\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} +  \ldots$     M1

Limit $= 0$     A1

[2 marks]

Interval width $= 26.1 - 22.7 = 3.4$

So $3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}$     M1A1

$z = 2.375 \ldots$     A1

Probability $= 0.9912$     A1

Confidence level $= 2 \times 0.4912 = 98.2\%$     A1

[5 marks]

$z$-value $= 1.96$     A1

We require

$2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2$     M1A1

Whence $n > 9.83$     A1

So we need $n = 10$     A1

Note: Accept $=$ signs throughout.

[5 marks]