Range of `f(x) =log _((x))(9-x ^(2)),` where `[.]` denotes G.I.E.` is :

Options

A. `{1,2}`
B. `(-oo,2)`
C. `(-oo, log _(2)5]`
D. `[log )_(2)5,3]`

Answer

To find the range of the function \( f(x) = \log_{x}(9 - x^2) \) where \([.]\) denotes the greatest integer function, we will first determine the domain of the function and then derive the range based on that domain.

### Step 1: Determine the Domain
For the logarithmic function \( \log_{x}(9 - x^2) \) to be valid, two conditions must be satisfied:
1. The argument of the logarithm must be positive:
\[ 9 - x^2 > 0 \]
2. The base \( x \) must be positive and not equal to 1:
\[ x > 0 \quad \text{and} \quad x \neq 1 \]

#### Solving \( 9 - x^2 > 0 \):
\[ 9 > x^2 \implies -3 < x < 3 \]
This means \( x \) must be in the interval \( (-3, 3) \).

#### Combining Conditions:
From \( -3 < x < 3 \) and \( x > 0 \), we get:
\[ 0 < x < 3 \]
Additionally, since \( x \neq 1 \), we exclude 1 from the interval:
\[ x \in (0, 1) \cup (1, 3) \]

### Step 2: Evaluate the Function
Now we will evaluate the function \( f(x) = \log_{x}(9 - x^2) \) over the intervals \( (0, 1) \) and \( (1, 3) \).

#### Interval \( (0, 1) \):
As \( x \) approaches 0 from the right, \( 9 - x^2 \) approaches 9, and since the base \( x \) is less than 1, the logarithm will yield positive values:
\[ \lim_{x \to 0^+} f(x) = \log_{0}(9) \to +\infty \]
At \( x = 1 \), we cannot evaluate \( f(1) \) since the base cannot be 1.

#### Interval \( (1, 3) \):
As \( x \) approaches 3, \( 9 - x^2 \) approaches 0:
\[ \lim_{x \to 3^-} f(x) = \log_{3}(0) \to -\infty \]
At \( x = 1 \), as stated, \( f(1) \) is undefined.

### Step 3: Finding the Range
Now we need to find the values of \( f(x) \) in both intervals:
- In the interval \( (0, 1) \), \( f(x) \) ranges from \( +\infty \) to a finite value as \( x \) approaches 1.
- In the interval \( (1, 3) \), \( f(x) \) ranges from a finite value (which we will calculate next) to \( -\infty \).

#### Evaluating at the Endpoints:
1. **At \( x = 2 \)**:
\[ f(2) = \log_{2}(9 - 2^2) = \log_{2}(5) \]

2. **As \( x \) approaches 1 from the right**:
\[ \lim_{x \to 1^+} f(x) = \log_{1}(8) \text{ (undefined, but approaches a finite value)} \]

### Final Range Calculation:
Combining the ranges from both intervals, we can conclude:
- From \( (0, 1) \): \( f(x) \) approaches \( +\infty \).
- From \( (1, 3) \): \( f(x) \) approaches \( -\infty \) and has a maximum value of \( \log_{2}(5) \).

Thus, the range of \( f(x) \) is:
\[ (-\infty, \log_{2}(5)] \text{ (since we take the greatest integer function)} \]

### Conclusion:
The range of \( f(x) = \log_{x}(9 - x^2) \) is \( (-\infty, \lfloor \log_{2}(5) \rfloor] \).