Explanation
consider the given function ,
$$f\left( x \right)=\sin {{\log }_{e}}\left( \dfrac{\sqrt{4-{{x}^{2}}}}{1-x} \right)$$
Now ,
for $$\log_e\dfrac{\sqrt{4x^2}}{1-x}$$ to be defined $$ \dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0 $$
$$ \dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0 $$
$$ \sqrt{4-{{x}^{2}}}>0 $$, $$x\not=1$$ and $$1-x>0$$
$$ 4-{{x}^{2}}>0 $$ $$1>x$$
$$ 4>{{x}^{2}} $$ $$x<1$$
$$ x^2-4<0$$
$$ (x+2)(x-2)<0 $$
$$-2<x<2$$
combining the two we get
$$-2<x<1$$
$$x\epsilon(-2,1)$$
Hence , the domain of the function is $$(-2,1)$$
Hence, this is the range of the function $$[-1, 1]$$
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