The domain of definiton of definition of `f(x)= log _(x) cosx`, is

Options

A. `(-pi//2, pi//2) - {1}`
B. ` [-pi//2, pi//2]-{1}`
C. `(-pi//2, pi//2)`
D. none of these

Explanation

To find the domain of the function \( f(x) = \log_x(\cos x) \), we need to ensure that both the argument of the logarithm and the base of the logarithm satisfy certain conditions.

### Step-by-Step Solution:

1. **Identify the conditions for the logarithm**:
The logarithmic function \( \log_b(a) \) is defined under the following conditions:
- \( a > 0 \) (the argument must be positive)
- \( b > 0 \) (the base must be positive)
- \( b \neq 1 \) (the base cannot be equal to 1)

In our case, \( a = \cos x \) and \( b = x \).

2. **Set up the inequalities**:
From the conditions above, we derive the following inequalities:
- \( \cos x > 0 \)
- \( x > 0 \)
- \( x \neq 1 \)

3. **Analyze \( \cos x > 0 \)**:
The cosine function is positive in the intervals:
- \( x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \)
This means that \( \cos x > 0 \) for \( x \) in this interval.

4. **Combine the conditions**:
Now we combine the conditions:
- From \( x > 0 \), we restrict our interval to \( (0, \frac{\pi}{2}) \).
- We also need to exclude \( x = 1 \) from our domain.

5. **Final domain**:
Thus, the domain of \( f(x) \) is:
\[ x \in (0, \frac{\pi}{2}) \quad \text{and} \quad x \neq 1 \]
In interval notation, this can be expressed as:
\[ (0, 1) \cup (1, \frac{\pi}{2}) \]