Question:
The domain of the real valued function
\[
f(x)=\sin\left(\log\left(\frac{\sqrt{4-x^2}}{1-x}\right)\right)
\]
is:
The domain of the real valued function
\[
f(x)=\sin\left(\log\left(\frac{\sqrt{4-x^2}}{1-x}\right)\right)
\]
is:
Show 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.
Updated On: Jun 22, 2026
- \((1,4)\)
- \((-1,1)\)
- \((-2,1)\)
- \((-2,4)\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Check the square root condition.
For the expression \[ \sqrt{4-x^2} \] to be real, we need \[ 4-x^2 \geq 0 \] So, \[ x^2 \leq 4 \] Hence, \[ -2 \leq x \leq 2 \]
Step 2: Check the logarithm condition.
For \[ \log\left(\frac{\sqrt{4-x^2}}{1-x}\right) \] to be defined, the argument of logarithm must be positive.
Therefore, \[ \frac{\sqrt{4-x^2}}{1-x}\gt 0 \]
Step 3: Analyze the numerator.
Since \[ \sqrt{4-x^2}\geq 0 \] For the fraction to be strictly positive, the numerator must be positive.
So, \[ \sqrt{4-x^2}\gt 0 \] This gives \[ 4-x^2\gt 0 \] Hence, \[ -2\lt x\lt 2 \]
Step 4: Analyze the denominator.
The denominator is \[ 1-x \] For the fraction to be positive, and since the numerator is positive, we need \[ 1-x\gt 0 \] Thus, \[ x\lt 1 \]
Step 5: Find the common interval.
From Step 3, \[ -2\lt x\lt 2 \] From Step 4, \[ x\lt 1 \] Taking the intersection, we get \[ -2\lt x\lt 1 \] Therefore, the domain is \[ (-2,1) \]
Step 6: Final conclusion.
Hence, \[ \boxed{(-2,1)} \]
For the expression \[ \sqrt{4-x^2} \] to be real, we need \[ 4-x^2 \geq 0 \] So, \[ x^2 \leq 4 \] Hence, \[ -2 \leq x \leq 2 \]
Step 2: Check the logarithm condition.
For \[ \log\left(\frac{\sqrt{4-x^2}}{1-x}\right) \] to be defined, the argument of logarithm must be positive.
Therefore, \[ \frac{\sqrt{4-x^2}}{1-x}\gt 0 \]
Step 3: Analyze the numerator.
Since \[ \sqrt{4-x^2}\geq 0 \] For the fraction to be strictly positive, the numerator must be positive.
So, \[ \sqrt{4-x^2}\gt 0 \] This gives \[ 4-x^2\gt 0 \] Hence, \[ -2\lt x\lt 2 \]
Step 4: Analyze the denominator.
The denominator is \[ 1-x \] For the fraction to be positive, and since the numerator is positive, we need \[ 1-x\gt 0 \] Thus, \[ x\lt 1 \]
Step 5: Find the common interval.
From Step 3, \[ -2\lt x\lt 2 \] From Step 4, \[ x\lt 1 \] Taking the intersection, we get \[ -2\lt x\lt 1 \] Therefore, the domain is \[ (-2,1) \]
Step 6: Final conclusion.
Hence, \[ \boxed{(-2,1)} \]
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