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]