Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = f(x)$;

Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = \frac{1}{{f(x)}}$;

Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = \left| {\frac{1}{{f(x)}}} \right|$.

Find $\int {f(x)\cos x{\text{d}}x}$.

By finding $g'(x)$ explain why $g$ is an increasing function.

A1 for correct shape

A1 for correct $x$ and $y$ intercepts and minimum point

[2 marks]

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

A1 for reflecting negative branch from (ii) in the $x$-axis

A1 for correctly labelled minimum point

[2 marks]

EITHER

attempt at integration by parts     (M1)

$\int {({x^2} - {a^2})\cos x{\text{d}}x = ({x^2} - {a^2})\sin x - \int {2x\sin x{\text{d}}x} }$     A1A1

$= ({x^2} - {a^2})\sin x - 2\left[ { - x\cos x + \int {\cos x{\text{d}}x} } \right]$     A1

$= ({x^2} - {a^2})\sin x + 2x\cos - 2\sin x + c$     A1

OR

$\int {({x^2} - {a^2})\cos x{\text{d}}x = \int {{x^2}\cos x{\text{d}}x - \int {{a^2}\cos x{\text{d}}x} } }$

attempt at integration by parts     (M1)

$\int {{x^2}\cos x{\text{d}}x = {x^2}\sin x - \int {2x\sin x{\text{d}}x} }$     A1A1

$= {x^2}\sin x - 2\left[ { - x\cos x + \int {\cos x{\text{d}}x} } \right]$     A1

$= {x^2}\sin x + 2x\cos x - 2\sin x$

$- \int {{a^2}\cos x{\text{d}}x = - {a^2}\sin x}$

$\int {({x^2} - {a^2})\cos x{\text{d}}x = ({x^2} - {a^2})\sin x + 2x\cos x - 2\sin x + c}$     A1

[5 marks]

$g(x) = x{({x^2} - {a^2})^{\frac{1}{2}}}$

$g'(x) = {({x^2} - {a^2})^{\frac{1}{2}}} + \frac{1}{2}x{({x^2} - {a^2})^{ - \frac{1}{2}}}(2x)$     M1A1A1

Note:     Method mark is for differentiating the product. Award A1 for each correct term.

$g'(x) = {({x^2} - {a^2})^{\frac{1}{2}}} + {x^2}{({x^2} - {a^2})^{ - \frac{1}{2}}}$

both parts of the expression are positive hence $g'(x)$ is positive     R1

and therefore $g$ is an increasing function (for $\left| x \right| > a$)     AG

[4 marks]