Determine $D$, the largest possible domain of $f$.

Show that the graph of $f$ has three asymptotes and state their equations.

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

Use your answers from parts (b) and (c) to justify that $f$ has an inverse and state its domain.

Find an expression for ${f^{ - 1}}(x)$.

Consider the region $R$ enclosed by the graph of $y = f(x)$ and the axes.

Find the volume of the solid obtained when $R$ is rotated through $2\pi$ about the $y$-axis.

attempting to solve either $2{{\text{e}}^x} - 1 = 0$ or $2{{\text{e}}^x} - 1 \ne 0$ for $x$     (M1)

$D = \mathbb{R}\backslash \left\{ { - \ln 2} \right\}$ (or equivalent eg $x \ne  - \ln 2$)     A1

Note: Accept $D = \mathbb{R}\backslash \left\{ { - 0.693} \right\}$ or equivalent eg $x \ne  - 0.693$.

[2 marks]

considering $\mathop {\lim }\limits_{x \to  - \ln 2} f(x)$     (M1)

$x =  - \ln 2{\text{ }}(x =  - 0.693)$    A1

considering one of $\mathop {\lim }\limits_{x \to  - \infty } f(x)$ or $\mathop {\lim }\limits_{x \to  + \infty } f(x)$     M1

$\mathop {\lim }\limits_{x \to  - \infty } f(x) =  - 2 \Rightarrow y =  - 2$    A1

$\mathop {\lim }\limits_{x \to  + \infty } f(x) =  - \frac{1}{2} \Rightarrow y =  - \frac{1}{2}$    A1

Note: Award A0A0 for $y =  - 2$ and $y =  - \frac{1}{2}$ stated without any justification.

[5 marks]

$f'(x) = \frac{{ - {{\text{e}}^x}(2{{\text{e}}^x} - 1) - 2{{\text{e}}^x}(2 - {{\text{e}}^x})}}{{{{(2{{\text{e}}^x} - 1)}^2}}}$    M1A1A1

$=  - \frac{{3{{\text{e}}^x}}}{{{{(2{{\text{e}}^x} - 1)}^2}}}$    AG

[3 marks]

$f'(x) < 0{\text{ (for all }}x \in D) \Rightarrow f$ is (strictly) decreasing     R1

Note: Award R1 for a statement such as $f'(x) \ne 0$ and so the graph of $f$ has no turning points.

one branch is above the upper horizontal asymptote and the other branch is below the lower horizontal asymptote     R1

$f$ has an inverse     AG

$- \infty  < x <  - 2 \cup  - \frac{1}{2} < x < \infty$    A2

Note: Award A2 if the domain of the inverse is seen in either part (d) or in part (e).

[4 marks]

$x = \frac{{2 - {{\text{e}}^y}}}{{2{{\text{e}}^y} - 1}}$    M1

Note: Award M1 for interchanging $x$ and $y$ (can be done at a later stage).

$2x{{\text{e}}^y} - x = 2 - {{\text{e}}^y}$    M1

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

${f^{ - 1}}(x) = \ln \left( {\frac{{x + 2}}{{2x + 1}}} \right){\text{ }}\left( {{f^{ - 1}}(x) = \ln (x + 2) - \ln (2x + 1)} \right)$    A1

[4 marks]

use of $V = \pi \int_a^b {{x^2}{\text{d}}y}$     (M1)

$= \pi \int_0^1 {{{\left( {\ln \left( {\frac{{y + 2}}{{2y + 1}}} \right)} \right)}^2}{\text{d}}y}$    (A1)(A1)

Note: Award (A1) for the correct integrand and (A1) for the limits.

$= 0.331$    A1

[4 marks]