22M.1.AHL.TZ2.6

attempts to replace $x$ with $-x$        M1

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

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

 

Note: Award M1A1 for an attempt to calculate both $f\left(-x\right)$ and $-f\left(-x\right)$ independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by $180°$ about the origin or the graph is invariant after a reflection in the $y$-axis and then in the $x$-axis (or vice versa).

 

so $f$ is an odd function         AG

  

[2 marks]

attempts both product rule and chain rule differentiation to find $f\text{'}\left(x\right)$        M1

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

$=\frac{1-2x^2}{\sqrt{1-x^2}}$

sets their $f\text{'}\left(x\right)=0$        M1

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

attempts to find at least one of $f\left(\pm \frac{1}{\sqrt{2}}\right)$         (M1)

 

Note: Award M1 for an attempt to evaluate $f\left(x\right)$ at least at one of their $f\text{'}\left(x\right)=0$  roots.

 

$a=-\frac{1}{2}$  and $b=\frac{1}{2}$         A1

 

Note: Award A1 for $-\frac{1}{2}\le y\le \frac{1}{2}$.

  

[6 marks]

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