Let `f(x)=[9^x-3^x+1]` for all `x in (-oo, 1),` then the range of `f(x)` is, ([.] denotes the greatest integer function).

Options

A. `{0, 1, 2, 3, 4, 5, 6, 7}`
B. `{0, 1, 2 , 3, 4, 5, 6}`
C. `{1, 2, 3, 4, 5, 6, 7}`
D. `{1, 2, 3, 4, 5, 6}`

Explanation

To solve the problem, we need to find the range of the function \( f(x) = \lfloor 9^x - 3^x + 1 \rfloor \) for \( x \in (-\infty, 1) \).

### Step-by-Step Solution:

1. **Rewrite the Function**:
We can express \( 9^x \) as \( (3^2)^x = (3^x)^2 \). Thus, we can rewrite the function:
\[ f(x) = \lfloor (3^x)^2 - 3^x + 1 \rfloor \]

2. **Let \( y = 3^x \)**:
Since \( x \) approaches \(-\infty\), \( y \) approaches \( 0 \) (as \( 3^x \) is a decreasing function). When \( x \) approaches \( 1 \), \( y \) approaches \( 3 \). Therefore, \( y \) ranges from \( 0 \) to \( 3 \) (not including \( 3 \)):
\[ y \in (0, 3) \]

3. **Substitute \( y \) into the Function**:
Now, substituting \( y \) into our function gives:
\[ f(x) = \lfloor y^2 - y + 1 \rfloor \]

4. **Analyze the Expression \( y^2 - y + 1 \)**:
We need to find the minimum and maximum values of \( y^2 - y + 1 \) as \( y \) varies from \( 0 \) to \( 3 \).

- **At \( y = 0 \)**:
\[ f(0) = 0^2 - 0 + 1 = 1 \]
- **At \( y = 3 \)** (approaching but not including):
\[ f(3) = 3^2 - 3 + 1 = 9 - 3 + 1 = 7 \]

5. **Find the Range**:
The function \( g(y) = y^2 - y + 1 \) is a quadratic function that opens upwards. The vertex of this quadratic can be found using the formula \( y = -\frac{b}{2a} \):
\[ y = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]
Evaluating \( g\left(\frac{1}{2}\right) \):
\[ g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

Therefore, as \( y \) ranges from \( 0 \) to \( 3 \), \( g(y) \) ranges from \( g(0) = 1 \) to just below \( g(3) = 7 \).

6. **Apply the Greatest Integer Function**:
Since \( g(y) \) ranges from \( \frac{3}{4} \) to just below \( 7 \), the greatest integer function \( \lfloor g(y) \rfloor \) will take values from \( 0 \) to \( 6 \):
\[ f(x) \in [0, 6] \]

### Conclusion:
Thus, the range of \( f(x) \) is:
\[ \{0, 1, 2, 3, 4, 5, 6\} \]