Let $f(x) = \frac{{a + b{{\text{e}}^x}}}{{a{{\text{e}}^x} + b}}$, where $0 < b < a$.

(a)     Show that $f'(x) = \frac{{({b^2} - {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}$.

(b)     Hence justify that the graph of f has no local maxima or minima.

(c)     Given that the graph of f has a point of inflexion, find its coordinates.

(d)     Show that the graph of f has exactly two asymptotes.

(e)     Let a = 4 and b =1. Consider the region R enclosed by the graph of $y = f(x)$, the y-axis and the line with equation $y = \frac{1}{2}$.

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

(a)     $f'(x) = \frac{{b{{\text{e}}^x}(a{{\text{e}}^x} + b) - a{{\text{e}}^x}(a + b{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^2}}}$     M1A1

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

$= \frac{{({b^2} - {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}$     AG

[3 marks]

 

(b)     EITHER

$f'(x) = 0 \Rightarrow ({b^2} - {a^2}){{\text{e}}^x} = 0 \Rightarrow b =  \pm a{\text{ or }}{{\text{e}}^x} = 0$     A1

which is impossible as $0 < b < a$ and ${{\text{e}}^x} > 0$ for all $x \in \mathbb{R}$     R1

OR

$f'(x) < 0$ for all $x \in \mathbb{R}$ since $0 < b < a$ and ${{\text{e}}^x} > 0$ for all $x \in \mathbb{R}$     A1R1

OR

$f'(x)$ cannot be equal to zero because ${{\text{e}}^x}$ is never equal to zero     A1R1

[2 marks]

 

(c)     EITHER

$f''(x) = \frac{{({b^2} - {a^2}){{\text{e}}^x}{{(a{{\text{e}}^x} + b)}^2} - 2a{{\text{e}}^x}(a{{\text{e}}^x} + b)({b^2} - {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^4}}}$     M1A1A1

Note: Award A1 for each term in the numerator.

 

$= \frac{{({b^2} - {a^2}){{\text{e}}^x}(a{{\text{e}}^x} + b - 2a{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^3}}}$

$= \frac{{({b^2} - {a^2})(b - a{{\text{e}}^x}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^3}}}$

OR

$f'(x) = ({b^2} - {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ - 2}}$

$f''(x) = ({b^2} - {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ - 2}} + ({b^2} - {a^2}){{\text{e}}^x}( - 2a{{\text{e}}^x}){(a{{\text{e}}^x} + b)^{ - 3}}$     M1A1A1

Note: Award A1 for each term.

 

$= ({b^2} - {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ - 3}}\left( {(a{{\text{e}}^x} + b) - 2a{{\text{e}}^x}} \right)$

$= ({b^2} - {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ - 3}}(b - a{{\text{e}}^x})$

THEN

$f''(x) = 0 \Rightarrow b - a{{\text{e}}^x} = 0 \Rightarrow x = \ln \frac{b}{a}$     M1A1

$f\left( {\ln \frac{b}{a}} \right) = \frac{{{a^2} + {b^2}}}{{2ab}}$     A1

coordinates are $\left( {\ln \frac{b}{a},\frac{{{a^2} + {b^2}}}{{2ab}}} \right)$

[6 marks]

 

(d)     $\mathop {\lim }\limits_{x - \infty } f(x) = \frac{a}{b} \Rightarrow y = \frac{a}{b}$ horizontal asymptote     A1

$\mathop {\lim }\limits_{x \to  + \infty } f(x) = \frac{b}{a} \Rightarrow y = \frac{b}{a}$ horizontal asymptote     A1

$0 < b < a \Rightarrow a{{\text{e}}^x} + b > 0$ for all $x \in \mathbb{R}$ (accept $a{{\text{e}}^x} + b \ne 0$)

so no vertical asymptotes     R1

Note: Statement on vertical asymptote must be seen for R1.

 

[3 marks]

 

(e)     $y = \frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}$

$y = \frac{1}{2} \Leftrightarrow x = \ln \frac{7}{2}$ (or 1.25 to 3 sf)     (M1)(A1)

$V = \pi \int_0^{\ln \frac{7}{2}} {\left( {{{\left( {\frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}} \right)}^2} - \frac{1}{4}} \right){\text{d}}x}$     (M1)A1

$= 1.09$ (3 sf)     A1     N4

[5 marks]

Total [19 marks]

This question was well attempted by many candidates. In some cases, candidates who skipped other questions still answered, with some success, parts of this question. Part (a) was in general well done but in (b) candidates found difficulty in justifying that f’(x) was non-zero. Performance in part (c) was mixed: it was pleasing to see good levels of algebraic ability of good candidates who successfully answered this question; weaker candidates found the simplification required difficult. There were very few good answers to part (d) which showed the weaknesses of most candidates in dealing with the concept of asymptotes. In part (e) there were a large number of good attempts, with many candidates evaluating correctly the limits of the integral and a smaller number scoring full marks in this part.