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]