Consider the following statements
Statement 1: The range of `log_(1)(1/(1+x^(2)))` is `(-oo,oo)`
Statement 2: If `0ltxle1`, then `log_(1)=xepsilon(-oo,0]`
Which of the following is correct

Options

A. Statement 1: is True, Statement 2 is True , Statement 2 is a correct explanation for statement 1
B. Statement 1 is True, Statement 2 is True Statement 2 is NOT a correct explanation for Statement 1
C. Statement 1 is True, Statement 2 is False
D. Statement 1 is False, Statement 2 is True

Explanation

To solve the given problem, we will analyze both statements one by one.

### Step 1: Analyze Statement 1
**Statement 1**: The range of \( \log_{1}\left(\frac{1}{1+x^{2}}\right) \) is \( (-\infty, \infty) \).

1. **Identify the function**: The function we are dealing with is \( \frac{1}{1+x^2} \).
2. **Determine the domain**: The expression \( 1+x^2 \) is always positive for all real \( x \), so \( \frac{1}{1+x^2} \) is defined for all \( x \in (-\infty, \infty) \).
3. **Evaluate the limits**:
- As \( x \to -\infty \) or \( x \to \infty \), \( \frac{1}{1+x^2} \to 0 \).
- At \( x = 0 \), \( \frac{1}{1+0^2} = 1 \).
4. **Range of the function**: The function \( \frac{1}{1+x^2} \) achieves values from \( 0 \) (not inclusive) to \( 1 \) (inclusive). Therefore, the range is \( (0, 1] \).
5. **Logarithm of the function**: Since the logarithm is defined only for positive values, the range of \( \log_{1}\left(\frac{1}{1+x^{2}}\right) \) will be \( (-\infty, 0] \) because:
- \( \log_{1}(1) = 0 \)
- As \( \frac{1}{1+x^2} \to 0 \), \( \log_{1}(0) \to -\infty \).

**Conclusion for Statement 1**: The statement is **false** because the range is \( (-\infty, 0] \), not \( (-\infty, \infty) \).

### Step 2: Analyze Statement 2
**Statement 2**: If \( 0 < x \leq 1 \), then \( \log_{1}(x) \in (-\infty, 0] \).

1. **Evaluate the logarithm**: The logarithm \( \log_{1}(x) \) is defined for \( x > 0 \).
2. **Check the range**:
- When \( x = 1 \), \( \log_{1}(1) = 0 \).
- As \( x \) approaches \( 0 \) from the right, \( \log_{1}(x) \to -\infty \).
3. **Range of \( \log_{1}(x) \)**: For \( 0 < x < 1 \), the logarithm will yield negative values, hence the range is indeed \( (-\infty, 0) \).

**Conclusion for Statement 2**: The statement is **true** because \( \log_{1}(x) \) does indeed lie in \( (-\infty, 0] \) for \( 0 < x \leq 1 \).

### Final Conclusion
- Statement 1 is **false**.
- Statement 2 is **true**.

Thus, the correct option is that Statement 1 is false and Statement 2 is true.

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