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]