Let ` f(x) = log _({x}) [x]`
`g (x) =log _({x})-{x}`
`h (x) = log _([x ]) {x}`
where `[], {}` denotes the greatest integer function and fractional part function respectively.
Domine of `h (x)` is :

Options

A. `[2,oo)`
B. `[1,oo)`
C. `[2,oo)-{I}`
D. `R ^(+) -{I}`

Answer

To find the domain of the function \( h(x) = \log_{[x]} \{x\} \), we need to analyze the components involved in this function.

### Step 1: Understanding the components of \( h(x) \)

1. **Greatest Integer Function**: The function \( [x] \) represents the greatest integer less than or equal to \( x \).
2. **Fractional Part Function**: The function \( \{x\} \) represents the fractional part of \( x \), which is defined as \( \{x\} = x - [x] \). This means \( \{x\} \) is always in the range \( [0, 1) \).

### Step 2: Conditions for the logarithm

For the logarithmic function \( \log_{[x]} \{x\} \) to be defined, two conditions must be satisfied:

1. The base of the logarithm, \( [x] \), must be greater than 0 and not equal to 1.
2. The argument of the logarithm, \( \{x\} \), must be greater than 0.

### Step 3: Analyzing the base \( [x] \)

- The base \( [x] \) is greater than 0 when \( x > 0 \). However, since \( [x] \) must also not equal 1, we have:
- \( [x] \neq 1 \) implies \( x < 1 \) or \( x \geq 2 \).

Thus, the valid intervals for \( x \) based on the base condition are \( (0, 1) \) and \( [2, \infty) \).

### Step 4: Analyzing the argument \( \{x\} \)

- The fractional part \( \{x\} \) is defined as \( \{x\} = x - [x] \). The fractional part is 0 when \( x \) is an integer. Therefore, \( x \) must not be an integer to ensure \( \{x\} > 0 \).

### Step 5: Combining the conditions

From the analysis:
- We have two intervals from the base condition: \( (0, 1) \) and \( [2, \infty) \).
- We need to exclude integers from these intervals.

1. **For \( (0, 1) \)**: The only integer in this interval is 0, which is not included anyway.
2. **For \( [2, \infty) \)**: We must exclude all integers starting from 2 onwards.

### Final Domain of \( h(x) \)

Thus, the domain of \( h(x) \) can be expressed as:
\[ (0, 1) \cup [2, \infty) \setminus \mathbb{Z} \]