16N.1.AHL.TZ0.H_11

pestleMathematicsAAHLPaper 116N· sl-5-8-testing-for-max-and-min-optimisation-points-of-inflexionsource ↗

Consider the function $f$ defined by $f(x) = {{\text{e}}^x}\sin x,{\text{ }}0 \leqslant x \leqslant \pi$ .

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

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

[2]

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

[2]

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

[3]

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

[6]

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

[3]

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

[2]

Markscheme / solution

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

N16/5/MATHL/HP1/ENG/TZ0/11.e/M

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

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

Examiners’ report

[N/A]