Answer
The given function is an odd function and has symmetry about the origin only.
Work Step by Step
Step I:
To check if even or odd.
In the given equation, if $f\left( -x \right)=f\left( x \right)$ , then it is an even function, and if $f\left( -x \right)=-f\left( x \right)$ , then it is odd.
$\begin{align}
& f\left( -x \right)=2\left( -x \right)\sqrt{1-{{\left( -x \right)}^{2}}} \\
& =-2x\sqrt{1-{{x}^{2}}} \\
& =-f\left( x \right)
\end{align}$
It is an odd function.
Step II:
To check symmetry about the y-axis:
Now, putting $x=-x$ in the given equation, if the equation remains the same, then it has symmetry about the y-axis.
$\begin{align}
& f\left( x \right)=2\left( -x \right)\sqrt{1-{{\left( -x \right)}^{2}}} \\
& =-2x\sqrt{1-{{x}^{2}}}
\end{align}$
It is not the same as the provided equation above, hence it is not symmetric about the y-axis.
Step III:
To check symmetry about the x-axis:
Now, put $y=-y$ in the given equation; if the equation remains the same, then it has symmetry about the x-axis.
$\begin{align}
& y=2x\sqrt{1-{{x}^{2}}} \\
& -y=2x\sqrt{1-{{x}^{2}}} \\
& y=-2x\sqrt{1-{{x}^{2}}}
\end{align}$
It is not the same as the provided equation above, hence it is not symmetric about the x-axis.
Step IV:
To check symmetry about the origin:
Now, put $x=-x\text{ and }y=-y$ in the given equation; if the equation remains the same, then it has symmetry about the origin.
$\begin{align}
& y=2x\sqrt{1-{{x}^{2}}} \\
& -y=2\left( -x \right)\sqrt{1-{{\left( -x \right)}^{2}}} \\
& y=2x\sqrt{1-{{x}^{2}}}
\end{align}$
Thus, it is the same as the provided equation above, hence it has symmetry about the origin.
Therefore, the given function is an odd function and has symmetry about the origin only.
