Question

If $f\left(x\right)=\sin \left\{\log \left(\frac{\sqrt{4-x^2}}{1-x}\right)\right\};x\in R$ then range of $f\left(x\right)$ is given by:

Answer 1

for $g\left(x\right)=\frac{\sqrt{{4-x}^2}}{1-x}$

as $xϵ[-2,2]-\left\{1\right\}$

so domain $=[-2,2]-\left\{1\right\}$

now $g^{\prime }\left(x\right)=\sqrt{{4-x}^2}\times \frac{1}{1-x}+\frac{1}{1-x}\times \frac{-2x}{2\sqrt{{4-x}^2}}$
for $g^{\prime }\left(x\right)=0$

$\frac{\sqrt{{4-x}^2}}{{1-x}^2}=\frac{1}{1-x}\times \frac{2x}{2\sqrt{{4-x}^2}}$

$(4-x^2)=xx(1-x)$

so, $x=4$

but $x$ can't be $4$

and for $xϵ[-2,2]-\left\{1\right\}$

$f\left(x\right)$ increases for $xϵ[-2,2]$ and decreases for $xϵ[1,2]$

and ${lim}_{x\to 1}f\left(x\right)=\infty$

but $\log 0$ is not defined

${lim}_{x\to 0}\log \left(x\right)=-\infty$

so range of $\log \left(g\right(x\left)\right)$ is $(-\infty ,\infty )$

and thus

range of $f\left(x\right)$ will be $[-1,1]$