Find the first three terms of the Maclaurin series for \(\ln (1 + {{\text{e}}^x})\) .

Hence, or otherwise, determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) - x - \ln 4}}{{{x^2}}}\) .

METHOD 1

\(f(x) = \ln (1 + {{\text{e}}^x});{\text{ }}f(0) = \ln 2\)     A1

\(f'(x) = \frac{{{{\text{e}}^x}}}{{1 + {{\text{e}}^x}}};{\text{ }}f'(0) = \frac{1}{2}\)     A1

Note: Award A0 for \(f'(x) = \frac{1}{{1 + {{\text{e}}^x}}};{\text{ }}f'(0) = \frac{1}{2}\)

\(f''(x) = \frac{{{{\text{e}}^x}(1 + {{\text{e}}^x}) - {{\text{e}}^{2x}}}}{{{{(1 + {{\text{e}}^x})}^2}}};{\text{ }}f''(0) = \frac{1}{4}\)     M1A1

Note: Award M0A0 for \(f''(x){\text{ if }}f'(x) = \frac{1}{{1 + {{\text{e}}^x}}}\) is used

\(\ln (1 + {{\text{e}}^x}) = \ln 2 + \frac{1}{2}x + \frac{1}{8}{x^2} + …\)     M1A1

[6 marks]

METHOD 2

\(\ln (1 + {{\text{e}}^x}) = \ln (1 + 1 + x + \frac{1}{2}{x^2} + …)\)     M1A1

\( = \ln 2 + \ln (1 + \frac{1}{2}x + \frac{1}{4}{x^2} + …)\)     A1

\( = \ln 2 + \left( {\frac{1}{2}x + \frac{1}{4}{x^2} + ...} \right) - \frac{1}{2}{\left( {\frac{1}{2}x + \frac{1}{4}{x^2} + ...} \right)^2} + …\)     A1

\( = \ln 2 + \frac{1}{2}x + \frac{1}{4}{x^2} - \frac{1}{8}{x^2} + …\)     A1

\( = \ln 2 + \frac{1}{2}x + \frac{1}{8}{x^2} + …\)     A1

[6 marks]

METHOD 1

\(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) - x - \ln 4}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\ln 2 + x + \frac{{{x^2}}}{4} + {x^3}{\text{ terms &  above}} - x - \ln 4}}{{{x^2}}}\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{4} + {\text{powers of }}x} \right) = \frac{1}{4}\)     M1A1

Note: Accept + … as evidence of recognition of cubic and higher powers.

Note: Award M1AOM1A0 for a solution which omits the cubic and higher powers.

[4 marks]

METHOD 2

using l’Hôpital’s Rule

\(\mathop {\lim }\limits_{x \to 0} \frac{{2\ln (1 + {{\text{e}}^x}) - x - \ln 4}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x} \div (1 + {{\text{e}}^x}) - 1}}{{2x}}\)     M1A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\text{e}}^x} \div {{(1 + {{\text{e}}^x})}^2}}}{2} = \frac{1}{4}\)     M1A1

[4 marks]

In (a), candidates who found the series by successive differentiation were generally successful, the most common error being to state that the derivative of \(\ln (1 + {{\text{e}}^x})\) is \({(1 + {{\text{e}}^x})^{ - 1}}\). Some candidates assumed the series for \(\ln (1 + x)\) and \({{\text{e}}^x}\) attempted to combine them. This was accepted as an alternative solution but candidates using this method were often unable to obtain the required series.

In (b), candidates were equally split between using the series or using l’Hopital’s rule to find the limit. Both methods were fairly successful, but a number of candidates forgot that if a series was used, there had to be a recognition that it was not a finite series.