Find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$.

Show that $\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^x}\cos x$.

Show that the function $f$ has a local maximum value when $x = \frac{{3\pi }}{4}$.

Find the $x$-coordinate of the point of inflexion of the graph of $f$.

Sketch the graph of $f$, clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.

Find the area of the region enclosed by the graph of $f$ and the $x$-axis.

The curvature at any point $(x,{\text{ }}y)$ on a graph is defined as $\kappa  = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}$.

Find the value of the curvature of the graph of $f$ at the local maximum point.

Find the value $\kappa$ for $x = \frac{\pi }{2}$ and comment on its meaning with respect to the shape of the graph.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x{\text{ }}\left( { = {{\text{e}}^x}(\sin x + \cos x)} \right)$    M1A1

[2 marks]

$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x - \sin x)$    M1A1

$= 2{{\text{e}}^x}\cos x$    AG

[2 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{\frac{{3\pi }}{4}}}\left( {\sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4}} \right) = 0$    R1

$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} < 0$    R1

hence maximum at $x = \frac{{3\pi }}{4}$     AG

[2 marks]

$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0 \Rightarrow 2{{\text{e}}^x}\cos x = 0$    M1

$\Rightarrow x = \frac{\pi }{2}$    A1

Note: Award M1A0 if extra zeros are seen.

[2 marks]

correct shape and correct domain     A1

max at $x = \frac{{3\pi }}{4}$, point of inflexion at $x = \frac{\pi }{2}$     A1

zeros at $x = 0$ and $x = \pi$     A1

Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.

[3 marks]

EITHER

$\int_0^x {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi  - \int_0^\pi  {{{\text{e}}^x}\cos x{\text{d}}x} }$    M1A1

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

OR

$\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = [ - {{\text{e}}^x}\cos x]_0^\pi  + \int_0^\pi  {{{\text{e}}^x}\cos x{\text{d}}x} }$    M1A1

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

THEN

$\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}\left( {[{{\text{e}}^x}\sin x]_0^x - [{{\text{e}}^x}\cos x]_0^x} \right)}$    M1A1

$\int_0^\pi  {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}({{\text{e}}^x} + 1)}$    A1

[6 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$    (A1)

 $\frac{{{d^2}y}}{{d{x^2}}} = 2{e^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} =  - \sqrt 2 {e^{\frac{{3\pi }}{4}}}$ (A1)

$\kappa  = \frac{{\left| { - \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}} \right|}}{1} = \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}$    A1

[3 marks]

$\kappa  = 0$    A1

the graph is approximated by a straight line     R1

[2 marks]