Find the domain of the following ` y = cos^(-1) ( x^(2)/( 1 + x^(2)))`

Options

A. `[-1,1]
B. R
C. `[0,1]`
D. `[-1,5]`

Answer

To find the domain of the function \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \), we need to ensure that the expression inside the inverse cosine function falls within the valid range of the cosine inverse function, which is from -1 to 1.

### Step-by-step Solution:

1. **Identify the Function**:
We start with the function:
\[ y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \]

2. **Determine the Range of the Argument**:
The argument of the cosine inverse function, \( \frac{x^2}{1 + x^2} \), must lie within the interval \([-1, 1]\):
\[ -1 \leq \frac{x^2}{1 + x^2} \leq 1 \]

3. **Analyze the Lower Bound**:
Since \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)), the fraction \( \frac{x^2}{1 + x^2} \) is also non-negative. Therefore, we can ignore the left part of the inequality:
\[ \frac{x^2}{1 + x^2} \geq 0 \]
This is always true for all real numbers \( x \).

4. **Analyze the Upper Bound**:
Now, we focus on the upper bound:
\[ \frac{x^2}{1 + x^2} \leq 1 \]
To solve this inequality, we can multiply both sides by \( 1 + x^2 \) (which is always positive):
\[ x^2 \leq 1 + x^2 \]
This simplifies to:
\[ 0 \leq 1 \]
This inequality is always true.

5. **Conclusion on the Range**:
Since both conditions are satisfied, we conclude that:
\[ 0 \leq \frac{x^2}{1 + x^2} \leq 1 \]
holds for all real numbers \( x \).

6. **Final Domain**:
Therefore, the domain of the function \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \) is:
\[ x \in \mathbb{R} \]

### Final Answer:
The domain of \( y = \cos^{-1} \left( \frac{x^2}{1 + x^2} \right) \) is \( x \in \mathbb{R} \).