Animated Solution for Mathematics - Functions: If $f(x)=\sin \ln \left(\frac{\sqrt{4-x^2}}{1-x}\right)$, then domain of $f(x)$ is .... and its range is .........

To find the domain of the function $f(x)=\sin \ln \left(\frac{\sqrt{4-x^2}}{1-x}\right)$, we must ensure every component is mathematically valid. We begin with the innermost layer, the square root.

For the expression $\sqrt{4-x^2}$ to yield a real number, the radicand must be non-negative: This restricts $x$ to the interval $[-2,2]$.

Next, we address the natural logarithm. The argument of $\ln (u)$ must be strictly positive:

Since the numerator $\sqrt{4-x^2}$ is always non-negative, the fraction is positive only if the numerator is non-zero and the denominator is positive. This implies: 1. $4 - x^2 eq 0 \implies x eq \pm 2$ 2. $1-x>0⟹x<1$

We now intersect our conditions: $x\in [-2,2]$, $x eq \pm 2$, and $x<1$.

Combining these constraints, we exclude the endpoints $-2$ and $2$ and restrict the upper bound to $1$. Thus, the domain is:

To determine the range, let $u=\ln \left(\frac{\sqrt{4-x^2}}{1-x}\right)$. We examine the behavior of $u$ as $x$ traverses the interval $(-2,1)$.

As $x\to 1^{-}$, the denominator $1-x$ approaches $0$ from the positive side. Consequently, the fraction $\frac{\sqrt{4-x^2}}{1-x}$ approaches $+\infty$, and $u=\ln (\text{fraction})\to \infty$.

As $x\to -2^{+}$, the numerator $\sqrt{4-x^2}$ approaches $0$, causing the fraction to approach $0^{+}$. The natural logarithm of a value approaching $0^{+}$ is $-\infty$.

Since the function is continuous on its domain, $u$ takes on all real values in the interval $(-\infty ,\infty )$.

Finally, we consider the outermost layer, $f(x)=\sin (u)$. Because $u$ spans the entire set of real numbers $\mathbb{R}$, the sine function oscillates through its complete cycle.

The range of $f(x)$ is the standard range of the sine function: