Let `g(x) = (log (1 +x))^(-1) -x^(-1) , x gt 0` then

Options

A. `1 lt g(x) lt 2`
B. `-1 lt g(x) lt 0`
C. `0 lt g (x) lt 1`
D. none of these

Answer

To solve the problem, we need to analyze the function \( g(x) = \frac{1}{\log(1+x)} - \frac{1}{x} \) for \( x > 0 \) and find its range.

### Step 1: Rewrite the function
We start with the function:
\[ g(x) = \frac{1}{\log(1+x)} - \frac{1}{x} \]
To combine the fractions, we find a common denominator:
\[ g(x) = \frac{x - \log(1+x)}{x \log(1+x)} \]

### Step 2: Analyze the numerator
Next, we need to analyze the numerator \( x - \log(1+x) \). We will check if this expression is positive for \( x > 0 \).

### Step 3: Differentiate the numerator
Define \( f(x) = x - \log(1+x) \). We differentiate \( f(x) \):
\[ f'(x) = 1 - \frac{1}{1+x} \]
This simplifies to:
\[ f'(x) = \frac{x}{1+x} \]
Since \( f'(x) > 0 \) for \( x > 0 \), \( f(x) \) is an increasing function.

### Step 4: Evaluate \( f(x) \) at a specific point
Now we evaluate \( f(x) \) at \( x = 0 \):
\[ f(0) = 0 - \log(1+0) = 0 \]
As \( x \) increases from \( 0 \), \( f(x) \) will also increase from \( 0 \). Therefore, \( f(x) > 0 \) for \( x > 0 \).

### Step 5: Conclusion about the numerator
Since \( x - \log(1+x) > 0 \) for \( x > 0 \), the numerator of \( g(x) \) is positive for \( x > 0 \).

### Step 6: Analyze the denominator
Next, we analyze the denominator \( x \log(1+x) \):
- \( x > 0 \) and \( \log(1+x) > 0 \) for \( x > 0 \).
Thus, \( x \log(1+x) > 0 \) for \( x > 0 \).

### Step 7: Conclusion about \( g(x) \)
Since both the numerator and denominator of \( g(x) \) are positive for \( x > 0 \), we conclude that:
\[ g(x) > 0 \text{ for } x > 0 \]

### Step 8: Find the upper bound
Next, we need to find an upper bound for \( g(x) \). We know that \( \log(1+x) < x \) for \( x > 0 \) (since the logarithm grows slower than the linear function). Thus:
\[ \frac{1}{\log(1+x)} > \frac{1}{x} \]
This implies:
\[ g(x) < 1 \text{ for } x > 0 \]

### Final Result
Combining both results, we find:
\[ 0 < g(x) < 1 \text{ for } x > 0 \]
Thus, the range of \( g(x) \) is \( (0, 1) \).