The domain of f(x) = sin log left( frac{sqrt{ | vedclass

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(A) For $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} \right)$ to be defined,the argument of the logarithm must be strictly positive. First,the sq

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$(-2, 1)$

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(A) For $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} ight)$ to be defined,the argument of the logarithm must be strictly positive. First,the square root $\sqrt{4-x^2}$ is defined when $4-x^2 \geq 0$,which implies $x \in [-2, 2]$. Second,the expression inside the logarithm must be positive: $\frac{\sqrt{4-x^2}}{1-x} > 0$. Since $\sqrt{4-x^2} \geq 0$,we must have $1-x > 0$ (i.e.,$x Thus,$x Combining $x \in [-2, 2]$ with $x < 1$ and $x 
eq \pm 2$,we get $x \in (-2, 1)$.

The domain of $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} \right)$ is

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