The domian of definition of `f (x) = log _((x ^(2) -x+1)) (2x ^(2)-7x+9)` is :

Options

A. R
B. `R-{0}`
C. `R-{0,1}`
D. `R-{1}`

Answer

To find the domain of the function \( f(x) = \log_{(x^2 - x + 1)}(2x^2 - 7x + 9) \), we need to ensure that both the base of the logarithm and the argument of the logarithm are valid.

### Step-by-Step Solution:

1. **Identify the conditions for the logarithm:**
- The base \( x^2 - x + 1 \) must be greater than 0 and not equal to 1.
- The argument \( 2x^2 - 7x + 9 \) must be greater than 0.

2. **Analyze the base \( x^2 - x + 1 \):**
- The expression \( x^2 - x + 1 \) is a quadratic function.
- To determine if it is always positive, we can find its discriminant:
\[ D = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 \]
- Since the discriminant is less than 0, the quadratic does not have real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( x^2 - x + 1 > 0 \) for all \( x \).

3. **Determine when the base is not equal to 1:**
- Set the base equal to 1:
\[ x^2 - x + 1 = 1 \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
- This gives us \( x = 0 \) or \( x = 1 \). Thus, the base is not equal to 1 when \( x \neq 0 \) and \( x \neq 1 \).

4. **Analyze the argument \( 2x^2 - 7x + 9 \):**
- We need to find when this expression is greater than 0. First, we can find its discriminant:
\[ D = (-7)^2 - 4(2)(9) = 49 - 72 = -23 \]
- Since the discriminant is negative, the quadratic \( 2x^2 - 7x + 9 \) does not cross the x-axis and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( 2x^2 - 7x + 9 > 0 \) for all \( x \).

5. **Combine the conditions:**
- From the analysis, we have:
- \( x^2 - x + 1 > 0 \) for all \( x \).
- \( 2x^2 - 7x + 9 > 0 \) for all \( x \).
- The base \( x^2 - x + 1 \) cannot equal 1, which gives us the restrictions \( x \neq 0 \) and \( x \neq 1 \).

6. **Final Domain:**
- The domain of the function \( f(x) \) is all real numbers except for \( x = 0 \) and \( x = 1 \):
\[ \text{Domain of } f(x) = \mathbb{R} \setminus \{0, 1\} \]