Show that \(f'(x) = g(x)\) and \(g'(x) = f(x)\).

Find the first three non-zero terms in the Maclaurin expansion of \(f(x)\).

Hence find the value of \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}}\).

Find the value of the improper integral \(\int_0^\infty  {\frac{{g(x)}}{{{{\left[ {f(x)} \right]}^2}}}{\text{d}}x} \).

any correct step before the given answer     A1AG

eg, \(f'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } + {{\left( {{{\text{e}}^{ - x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2} = g(x)\)

any correct step before the given answer     A1AG

eg, \(g'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } - {{\left( {{{\text{e}}^{ - x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2} = f(x)\)

[2 marks]

METHOD 1

statement and attempted use of the general Maclaurin expansion formula     (M1)

\(f(0) = 1;{\text{ }}g(0) = 0\) (or equivalent in terms of derivative values)   A1A1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

METHOD 2

\({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

\({{\text{e}}^{ - x}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

adding and dividing by 2     M1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

Notes: Accept 1, \(\frac{{{x^2}}}{2}\) and \(\frac{{{x^4}}}{{24}}\) or 1, \(\frac{{{x^2}}}{{2!}}\) and \(\frac{{{x^4}}}{{4!}}\).

     Award A1 if two correct terms are seen.

[5 marks]

METHOD 1

attempted use of the Maclaurin expansion from (b)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - \left( {1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} +  \ldots } \right)}}{{{x^2}}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \left( { - \frac{1}{2} - \frac{{{x^2}}}{{24}} -  \ldots } \right)\)     A1

\( =  - \frac{1}{2}\)     A1

METHOD 2

attempted use of L’Hôpital and result from (a)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 - f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - g(x)}}{{2x}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ - f(x)}}{2}\)     A1

\( =  - \frac{1}{2}\)     A1

[3 marks]

METHOD 1

use of the substitution \(u = f(x)\) and \(\left( {{\text{d}}u = g(x){\text{d}}x} \right)\)     (M1)(A1)

attempt to integrate \(\int_1^\infty  {\frac{{{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { - \frac{1}{u}} \right]_1^\infty \) or \(\left[ { - \frac{1}{{f(x)}}} \right]_0^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

METHOD 2

\(\int_0^\infty  {\frac{{\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}}}{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}} \right)}^2}}}{\text{d}}x = \int_0^\infty  {\frac{{2\left( {{{\text{e}}^x} - {{\text{e}}^{ - x}}} \right)}}{{{{\left( {{{\text{e}}^x} + {{\text{e}}^{ - x}}} \right)}^2}}}{\text{d}}x} } \)     (M1)

use of the substitution \(u = {{\text{e}}^x} + {{\text{e}}^{ - x}}\) and \(\left( {{\text{d}}u = {{\text{e}}^x} - {{\text{e}}^{ - x}}{\text{d}}x} \right)\)     (M1)

attempt to integrate \(\int_2^\infty  {\frac{{2{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { - \frac{2}{u}} \right]_2^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

[6 marks]