The range of the function $\mathrm{f}\left(\mathrm{x}\right)=\sin \left[\log \left(\frac{\sqrt{4-{\mathrm{x}}^2}}{1-\mathrm{x}}\right)\right]\text{ is}$ $\left[3a,b\right]$ then a+b is equal to

For f (x) to be defined,
$$\begin{gathered} \frac{\sqrt{4-{\mathrm{x}}^2}}{1-\mathrm{x}}>0,4-{\mathrm{x}}^2>0 \mathrm{and} 1-\mathrm{x}\ne 0 \\ \mathrm{Since} \sqrt{4-{\mathrm{x}}^2}\nless 0\mathrm{, } \end{gathered}$$
$\therefore$ We have 1 – x > 0 and 4 – x² > 0
⇒ x < 1 and (x – 2) (x + 2) < 0
⇒ x < 1 and – 2 < x < 2
⇒ –2 < x < 1
$\therefore$ Domain of f = (– 2, 1).
$$\begin{gathered} \text{ Since }-\mathrm{\infty }<\log \left(\frac{\sqrt{4-{\mathrm{x}}^2}}{1-\mathrm{x}}\right)<\mathrm{\infty } \\ \Rightarrow -1\le \sin \left[\log \left(\frac{\sqrt{4-{\mathrm{x}}^2}}{1-\mathrm{x}}\right)\right]\le 1 \end{gathered}$$
$\therefore$ Range of f = [–1, 1] =[3a,b]

 then $a=\frac{-1}{3},b=1$