(a)     Write down the value of the constant term in the Maclaurin series for \(f(x)\) .

(b)     Find the first three derivatives of \(f(x)\) and hence show that the Maclaurin series for \(f(x)\) up to and including the \({x^3}\) term is \(x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\).

(c)     Use this series to find an approximate value for ln 2 .

(d)     Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.

(e)     How good is this upper bound as an estimate for the actual error?

(a)     Constant term = 0     A1

[1 mark]

(b)     \(f'(x) = \frac{1}{{1 - x}}\)     A1

\(f''(x) = \frac{1}{{{{(1 - x)}^2}}}\)     A1

\(f'''(x) = \frac{2}{{{{(1 - x)}^3}}}\)     A1

\(f'(0) = 1;{\text{ }}f''(0) = 1;{\text{ }}f'''(0) = 2\)     A1

Note: Allow FT on their derivatives.

\(f(x) = 0 + \frac{{1 \times x}}{{1!}} + \frac{{1 \times {x^2}}}{{2!}} + \frac{{2 \times {x^3}}}{{3!}} + …\)     M1A1

\( = x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\)     AG

[6 marks]

(c)     \(\frac{1}{{1 - x}} = 2 \Rightarrow x = \frac{1}{2}\)     (A1)

\(\ln 2 \approx \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}}\)     M1

\( = \frac{2}{3}{\text{ (0.667)}}\)     A1

[3 marks]

(d)     Lagrange error \({\text{ = }}\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}} \times {\left( {\frac{1}{2}} \right)^{n + 1}}\)     (M1)

\( = \frac{6}{{{{(1 - c)}^4}}} \times \frac{1}{{24}} \times {\left( {\frac{1}{2}} \right)^4}\)     A1

\( < \frac{6}{{{{\left( {1 - \frac{1}{2}} \right)}^4}}} \times \frac{1}{{24}} \times \frac{1}{{16}}\)     A2

giving an upper bound of 0.25.     A1

[5 marks]

(e)     Actual error \( = \ln 2 - \frac{2}{3} = 0.0265\)     A1

The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate.     R1

[2 marks]

Total [17 marks]

In (a), some candidates appeared not to understand the term ‘constant term’. In (b), many candidates found the differentiation beyond them with only a handful realising that the best way to proceed was to rewrite the function as \(f(x) = - \ln (1 - x)\). In (d), many candidates were unable to use the Lagrange formula for the upper bound so that (e) became inaccessible.