(i)     Show that $\frac{1}{{4f(x) - 2g(x)}} = \frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 3}}$.

(ii)     Use the substitution $u = {{\text{e}}^x}$ to find $\int_0^{\ln 3} {\frac{1}{{4f(x) - 2g(x)}}} {\text{d}}x$. Give your answer in the form $\frac{{\pi \sqrt a }}{b}$ where $a,{\text{ }}b \in {\mathbb{Z}^ + }$.

(i)     By forming a quadratic equation in ${{\text{e}}^x}$, solve the equation $h(x) = k$, where $k \in {\mathbb{R}^ + }$.

(ii)     Hence or otherwise show that the equation $h(x) = k$ has two real solutions provided that $k > \sqrt {{n^2} - 1}$ and $k \in {\mathbb{R}^ + }$.

(i)     Show that $t'(x) = \frac{{{{[f(x)]}^2} - {{[g(x)]}^2}}}{{{{[f(x)]}^2}}}$ for $x \in \mathbb{R}$.

(ii)     Hence show that $t'(x) > 0$ for $x \in \mathbb{R}$.

(i)     $\frac{1}{{4\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}} \right) - 2\left( {\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}} \right)}}$     (M1)

$= \frac{1}{{2({{\text{e}}^x} + {{\text{e}}^{ - x}}) - ({{\text{e}}^x} - {{\text{e}}^{ - x}})}}$    (A1)

$= \frac{1}{{{{\text{e}}^x} + 3{{\text{e}}^{ - x}}}}$    A1

$= \frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 3}}$    AG

(ii)     $u = {{\text{e}}^x} \Rightarrow {\text{d}}u = {{\text{e}}^x}{\text{d}}x$     A1

$\int {\frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 3}}{\text{d}}x = \int {\frac{1}{{{u^2} + 3}}{\text{d}}u} }$    M1

(when $x = 0,{\text{ }}u = 1$ and when $x = \ln 3,{\text{ }}u = 3$)

$\int_1^3 {\frac{1}{{{u^2} + 3}}{\text{d}}u\left[ {\frac{1}{{\sqrt 3 }}\arctan \left( {\frac{u}{{\sqrt 3 }}} \right)} \right]_1^3}$    M1A1

$\left( { = \left[ {\frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{{{\text{e}}^x}}}{{\sqrt 3 }}} \right)} \right]_0^{\ln 3}} \right)$

$= \frac{{\pi \sqrt 3 }}{9} - \frac{{\pi \sqrt 3 }}{{18}}$    (M1)

$= \frac{{\pi \sqrt 3 }}{{18}}$    A1

[9 marks]

(i)     $(n + 1){{\text{e}}^{2x}} - 2k{{\text{e}}^x} + (n - 1) = 0$     M1A1

${{\text{e}}^x} = \frac{{2k \pm \sqrt {4{k^2} - 4({n^2} - 1)} }}{{2(n + 1)}}$    M1

$x = \ln \left( {\frac{{k \pm \sqrt {{k^2} - {n^2} + 1} }}{{n + 1}}} \right)$    M1A1

(ii)     for two real solutions, we require $k > \sqrt {{k^2} - {n^2} + 1}$     R1

and we also require ${k^2} - {n^2} + 1 > 0$     R1

${k^2} > {n^2} - 1$    A1

$\Rightarrow k > \sqrt {{n^2} - 1} {\text{ }}({\text{ }}k \in {\mathbb{R}^ + })$    AG

[8 marks]

METHOD 1

$t(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}$

$t'(x) = \frac{{{{({{\text{e}}^x} + {{\text{e}}^{ - x}})}^2} - {{({{\text{e}}^x} - {{\text{e}}^{ - x}})}^2}}}{{{{({{\text{e}}^x} + {{\text{e}}^{ - x}})}^2}}}$    M1A1

$t'(x) = \frac{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}} \right)}^2} - {{\left( {\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}} \right)}^2}}}{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}} \right)}^2}}}$    A1

$= \frac{{{{\left[ {f(x)} \right]}^2} - {{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$    AG

METHOD 2

$t'(x) = \frac{{f(x)g'(x) = g(x)f'(x)}}{{{{\left[ {f(x)} \right]}^2}}}$    M1A1

$g'(x) = f(x)$ and $f'(x) = g(x)$     A1

$= \frac{{{{\left[ {f(x)} \right]}^2} - {{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$    AG

METHOD 3

$t(x) = ({{\text{e}}^x} - {{\text{e}}^{ - x}}){({{\text{e}}^x} + {{\text{e}}^{ - x}})^{ - 1}}$

$t'(x) = 1 - \frac{{{{({{\text{e}}^x} - {{\text{e}}^{ - x}})}^2}}}{{{{({{\text{e}}^x} + {{\text{e}}^{ - x}})}^2}}}$    M1A1

$= 1 - \frac{{{{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$    A1

$= \frac{{{{\left[ {f(x)} \right]}^2} - {{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$    AG

METHOD 4

$t'(x) = \frac{{g'(x)}}{{f(x)}} - \frac{{g(x)f'(x)}}{{{{\left[ {f(x)} \right]}^2}}}$    M1A1

$g'(x) = f(x)$ and $f'(x) = g(x)$ gives $t'(x) = 1 - \frac{{{{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$     A1

$= \frac{{{{\left[ {f(x)} \right]}^2} - {{\left[ {g(x)} \right]}^2}}}{{{{\left[ {f(x)} \right]}^2}}}$    AG

(ii)     METHOD 1

${\left[ {f(x)} \right]^2} > {\left[ {g(x)} \right]^2}$ (or equivalent)     M1A1

${\left[ {f(x)} \right]^2} > 0$    R1

hence $t'(x) > 0,{\text{ }}x \in \mathbb{R}$     AG

Note:     Award as above for use of either $f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}$ and $g(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}$ or ${{\text{e}}^x} + {{\text{e}}^{ - x}}$ and ${{\text{e}}^x} - {{\text{e}}^{ - x}}$.

METHOD 2

${\left[ {f(x)} \right]^2} - {\left[ {g(x)} \right]^2} = 1$ (or equivalent)     M1A1

${\left[ {f(x)} \right]^2} > 0$    R1

hence $t'(x) > 0,{\text{ }}x \in \mathbb{R}$     AG

Note:     Award as above for use of either $f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}}}}{2}$ and $g(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}$ or ${{\text{e}}^x} + {{\text{e}}^{ - x}}$ and ${{\text{e}}^x} - {{\text{e}}^{ - x}}$.

METHOD 3

$t'(x) = \frac{4}{{{{({{\text{e}}^x} + {{\text{e}}^{ - x}})}^2}}}$

${\left( {{{\text{e}}^x} + {{\text{e}}^{ - x}}} \right)^2} > 0$    M1A1

$\frac{4}{{{{\left( {{{\text{e}}^x} + {{\text{e}}^{ - x}}} \right)}^2}}} > 0$    R1

hence $t'(x) > 0,{\text{ }}x \in \mathbb{R}$     AG

[6 marks]

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

Part (a)(i) was reasonably well done with most candidates able to show that $\frac{1}{{4f(x) - 2g(x)}} = \frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 3}}$. In part (a)(ii), a number of candidates correctly used the required substitution to obtain $\int {\frac{{{{\text{e}}^x}}}{{{{\text{e}}^{2x}} + 3}}{\text{d}}x = \int {\frac{1}{{{u^2} + 3}}{\text{d}}u} }$ but then thought that the antiderivative involved natural log rather than arctan.

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

In part (b)(i), a reasonable number of candidates were able to form a quadratic in ${{\text{e}}^x}$ (involving parameters $n$ and $k$) and then make some progress towards solving for ${{\text{e}}^x}$ in terms of $n$ and $k$. Having got that far, a small number of candidates recognised to then take the natural logarithm of both sides and hence solve $h(x) = k$ for $\chi$. In part (b)(ii), a small number of candidates were able to show from their solutions to part (b)(i) or through the use of the discriminant that the equation $h(x) = k$ has two real solutions provided that $k > \sqrt {{k^2} - {n^2} + 1}$ and $k > \sqrt {{n^2} - 1}$.

Parts (a) and (c) were accessible to the large majority of candidates. Candidates found part (b) considerably more challenging.

It was pleasing to see the number of candidates who attempted part (c). In part (c)(i), a large number of candidates were able to correctly apply either the quotient rule or the product rule to find $t'(x)$. A smaller number of candidates were then able to show equivalence between the form of $t'(x)$ they had obtained and the form of $t'(x)$ required in the question. A pleasing number of candidates were able to exploit the property that $f'(x) = g(x)$ and $g'(x) = f(x)$. As with part (c)(i), part (c)(ii) could be successfully tackled in a number of ways. The best candidates offered concise logical reasoning to show that $t'(x) > 0$ for $x \in \mathbb{R}$.