Show that $f$ is an odd function.

Find $f'(x)$.

Hence find the $x$-coordinates of any local maximum or minimum points.

Find the range of $f$.

Sketch the graph of $y = f(x)$ indicating clearly the coordinates of the $x$-intercepts and any local maximum or minimum points.

Find the area of the region enclosed by the graph of $y = f(x)$ and the $x$-axis for $x \ge 0$.

Show that $\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > \left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right|}$.

$f( - x) = ( - x)\sqrt {1 - {{( - x)}^2}}$     M1

$=  - x\sqrt {1 - {x^2}}$

$=  - f(x)$     R1

hence $f$ is odd     AG

[2 marks]

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

[3 marks]

$f'(x) = \sqrt {1 - {x^2}}  - \frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}\;\;\;\left( { = \frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)$     A1

Note:     This may be seen in part (b).

Note:     Do not allow FT from part (b).

$f'(x) = 0 \Rightarrow 1 - 2{x^2} = 0$     M1

$x =  \pm \frac{1}{{\sqrt 2 }}$     A1

[3 marks]

$y$-coordinates of the Max Min Points are $y =  \pm \frac{1}{2}$     M1A1

so range of $f(x)$ is $\left[ { - \frac{1}{2},{\text{ }}\frac{1}{2}} \right]$     A1

Note:     Allow FT from (c) if values of $x$, within the domain, are used.

[3 marks]

Shape: The graph of an odd function, on the given domain, s-shaped,

where the max(min) is the right(left) of $0.5{\text{ }}( - 0.5)$     A1

$x$-intercepts     A1

turning points     A1

[3 marks]

${\text{area}} = \int_0^1 {x\sqrt {1 - {x^2}} {\text{d}}x}$     (M1)

attempt at “backwards chain rule” or substitution     M1

$=  - \frac{1}{2}\int_0^1 {( - 2x)\sqrt {1 - {x^2}} {\text{d}}x}$

Note:     Condone absence of limits for first two marks.

$= \left[ {\frac{2}{3}{{(1 - {x^2})}^{\frac{3}{2}}} \bullet  - \frac{1}{2}} \right]_0^1$     A1

$= \left[ { - \frac{1}{3}{{(1 - {x^2})}^{\frac{3}{2}}}} \right]_0^1$

$= 0 - \left( { - \frac{1}{3}} \right) = \frac{1}{3}$     A1

[4 marks]

$\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > 0}$     R1

$\left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right| = 0$     R1

so $\int_{ - 1}^1 {\left| {x\sqrt {1 - {x^2}} } \right|{\text{d}}x > \left| {\int_{ - 1}^1 {x\sqrt {1 - {x^2}} {\text{d}}x} } \right|}$     AG

[2 marks]

Total [20 marks]