`bb"Statement I"` The range of `log(1/(1+x^(2))) " is " (-infty,infty)`.
`bb"Statement II" " when " 0 lt x le 1, log x in (-infty,0].`

Answer

To solve the problem, we need to analyze the two statements provided regarding the function \( \log\left(\frac{1}{1+x^2}\right) \) and the behavior of \( \log x \) for \( 0 < x \leq 1 \).

### Step-by-step Solution:

**Step 1: Analyze Statement I**
- We need to find the range of the function \( f(x) = \frac{1}{1+x^2} \).
- The denominator \( 1 + x^2 \) is always positive and has a minimum value of 1 (when \( x = 0 \)).
- As \( x \) approaches infinity, \( 1 + x^2 \) approaches infinity, making \( f(x) \) approach 0.
- Therefore, the range of \( f(x) \) is \( 0 < f(x) \leq 1 \).

**Hint for Step 1:** Consider the behavior of the function as \( x \) changes, particularly at critical points like \( x = 0 \) and as \( x \) approaches infinity.

**Step 2: Determine the Range of \( \log f(x) \)**
- Since \( 0 < f(x) \leq 1 \), we can analyze the logarithm:
- The logarithm function \( \log x \) is defined for \( x > 0 \).
- For \( 0 < f(x) < 1 \), \( \log f(x) \) will yield negative values.
- Specifically, as \( f(x) \) approaches 0, \( \log f(x) \) approaches \( -\infty \), and when \( f(x) = 1 \), \( \log f(x) = 0 \).
- Hence, the range of \( \log f(x) \) is \( (-\infty, 0] \).

**Hint for Step 2:** Remember that the logarithm of a number less than 1 is negative, and the logarithm of 1 is zero.

**Step 3: Analyze Statement II**
- Statement II claims that when \( 0 < x \leq 1 \), \( \log x \) is in the interval \( (-\infty, 0] \).
- For \( 0 < x < 1 \), \( \log x \) is indeed negative, approaching \( -\infty \) as \( x \) approaches 0.
- At \( x = 1 \), \( \log 1 = 0 \).
- Therefore, the range of \( \log x \) for \( 0 < x \leq 1 \) is also \( (-\infty, 0] \).

**Hint for Step 3:** Consider the properties of the logarithm function and how it behaves in the given interval.

### Conclusion:
- Statement I is false because the range of \( \log\left(\frac{1}{1+x^2}\right) \) is \( (-\infty, 0] \), not \( (-\infty, \infty) \).
- Statement II is true as the range of \( \log x \) for \( 0 < x \leq 1 \) is indeed \( (-\infty, 0] \).

Thus, the final conclusion is:
- Statement I: False
- Statement II: True