Question

The domain of the real valued function \[ f(x)=\sin\left(\log\left(\frac{\sqrt{4-x^2}}{1-x}\right)\right) \] is:

• A. \((1,4)\)
• B. \((-1,1)\)
• C. \((-2,1)\)
• D. \((-2,4)\)

Hint:

For logarithmic functions, the argument of the logarithm must always be strictly positive. Also, for square root expressions, the quantity inside the root must be non-negative.

Answer 1

The Correct Option is C

Step 1: List the conditions for the domain.
For $f(x)=\sin\!\left(\log\!\left(\dfrac{\sqrt{4-x^2}}{1-x}\right)\right)$, the sine is fine everywhere, so the domain is decided by the inside log and the square root.
Step 2: Square root condition.
We need $4-x^2\ge 0$, so $x^2\le 4$, giving $-2\le x\le 2$.
Step 3: Log argument must be positive.
We need $\dfrac{\sqrt{4-x^2}}{1-x}>0$. The numerator $\sqrt{4-x^2}\ge 0$.
Step 4: Keep the numerator strictly positive.
If $\sqrt{4-x^2}=0$ the fraction is $0$ and the log is undefined, so we need $4-x^2>0$, i.e. $-2<x<2$.
Step 5: Fix the sign of the denominator.
With a positive numerator, the fraction is positive only when $1-x>0$, i.e. $x<1$.
Step 6: Intersect everything.
Combining $-2<x<2$ with $x<1$ gives $-2<x<1$.
\[ \boxed{(-2,\,1)} \]