State the two zeros of f .

Sketch the graph of f .

The region bounded by the graph, the x-axis and the y-axis is denoted by A and the region bounded by the graph and the x-axis is denoted by B . Show that the ratio of the area of A to the area of B is

$$\frac{{{e^\pi }\left( {{e^{\frac{\pi }{2}}} + 1} \right)}}{{{e^\pi } + 1}}.$$

${e^{ - x}}\cos x = 0$

$\Rightarrow x = \frac{\pi }{2},{\text{ }}\frac{{3\pi }}{2}$     A1

[1 mark]

     A1

Note: Accept any form of concavity for $x \in \left[ {0,\frac{\pi }{2}} \right]$.

Note: Do not penalize unmarked zeros if given in part (a).

Note: Zeros written on diagram can be used to allow the mark in part (a) to be awarded retrospectively.

[1 mark]

attempt at integration by parts     M1

EITHER

$I = \int {{{\text{e}}^{ - x}}\cos x{\text{d}}x =  - {{\text{e}}^{ - x}}\cos x{\text{d}}x - \int {{{\text{e}}^{ - x}}\sin x{\text{d}}x} }$     A1

$\Rightarrow I =  - {{\text{e}}^{ - x}}\cos x{\text{d}}x - \left[ { - {{\text{e}}^{ - x}}\sin x + \int {{{\text{e}}^{ - x}}\cos x{\text{d}}x} } \right]$     A1

$\Rightarrow I = \frac{{{{\text{e}}^{ - x}}}}{2}(\sin x - \cos x) + C$     A1 

Note: Do not penalize absence of C.

OR

$I = \int {{{\text{e}}^{ - x}}\cos x{\text{d}}x = {{\text{e}}^{ - x}}\sin x + \int {{{\text{e}}^{ - x}}\sin x{\text{d}}x} }$     A1

$I = {{\text{e}}^{ - x}}\sin x - {{\text{e}}^{ - x}}\cos x - \int {{{\text{e}}^{ - x}}\cos x{\text{d}}x}$     A1

$\Rightarrow I = \frac{{{{\text{e}}^{ - x}}}}{2}(\sin x - \cos x) + C$     A1 

Note: Do not penalize absence of C.

THEN

$\int_0^{\frac{\pi }{2}} {{{\text{e}}^{ - x}}\cos x{\text{d}}x = \left[ {\frac{{{{\text{e}}^{ - x}}}}{2}(\sin x - \cos x)} \right]} _0^{\frac{\pi }{2}} = \frac{{{{\text{e}}^{ - \frac{\pi }{2}}}}}{2} + \frac{1}{2}$     A1

$\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {{{\text{e}}^{ - x}}\cos x{\text{d}}x = \left[ {\frac{{{{\text{e}}^{ - x}}}}{2}(\sin x - \cos x)} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} =  - \frac{{{{\text{e}}^{ - \frac{{3\pi }}{2}}}}}{2} - \frac{{{{\text{e}}^{ - \frac{\pi }{2}}}}}{2}}$     A1

ratio of A:B is $\frac{{\frac{{{{\text{e}}^{ - \frac{\pi }{2}}}}}{2} + \frac{1}{2}}}{{\frac{{{{\text{e}}^{ - \frac{{3\pi }}{2}}}}}{2} + \frac{{{{\text{e}}^{ - \frac{\pi }{2}}}}}{2}}}$

$= \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}\left( {{{\text{e}}^{ - \frac{\pi }{2}}} + 1} \right)}}{{{{\text{e}}^{\frac{{3\pi }}{2}}}\left( {{{\text{e}}^{ - \frac{{3\pi }}{2}}} + {{\text{e}}^{ - \frac{\pi }{2}}}} \right)}}$     M1

$= \frac{{{{\text{e}}^\pi }\left( {{{\text{e}}^{\frac{\pi }{2}}} + 1} \right)}}{{{{\text{e}}^\pi } + 1}}$     AG

[7 marks] 

Many candidates stated the two zeros of f correctly but the graph of f was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign.

Many candidates stated the two zeros of f correctly but the graph of f was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign.

Many candidates stated the two zeros of f correctly but the graph of f was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign.