let A be the greatest value of the function `f(x)=log _(x) [x],` (where `[.]` denotes gratest integer function) and B be the least value of the function `g (x) = |sin x | +|cos x| ,` then :

Options

A. `A gtB`
B. `A lt B`
C. `A =B`
D. `2A+ B=4`

Answer

To solve the problem, we need to find the values of \( A \) and \( B \) as defined in the question.

### Step 1: Determine \( A \) for the function \( f(x) = \log_{\lfloor x \rfloor}(x) \)

1. **Understanding the function**: The function \( f(x) = \log_{\lfloor x \rfloor}(x) \) means we take the logarithm of \( x \) with the base being the greatest integer less than or equal to \( x \), denoted by \( \lfloor x \rfloor \).
2. **Range of \( \lfloor x \rfloor \)**: For \( x \) in the interval \( [n, n+1) \) where \( n \) is a positive integer, \( \lfloor x \rfloor = n \).
3. **Evaluating \( f(x) \)**: In this interval, we have:
\[ f(x) = \log_{n}(x) \]
This can be rewritten using the change of base formula:
\[ f(x) = \frac{\log(x)}{\log(n)} \]
4. **Finding the maximum**: As \( x \) approaches \( n+1 \), \( f(x) \) approaches:
\[ f(n+1) = \log_{n}(n+1) = \frac{\log(n+1)}{\log(n)} \]
We need to find the maximum value of \( \frac{\log(n+1)}{\log(n)} \) for positive integers \( n \).
5. **Behavior of the function**: As \( n \) increases, \( \frac{\log(n+1)}{\log(n)} \) approaches 1. Therefore, the maximum value of \( f(x) \) approaches 1.

Thus, \( A = 1 \).

### Step 2: Determine \( B \) for the function \( g(x) = |\sin x| + |\cos x| \)

1. **Understanding the function**: The function \( g(x) = |\sin x| + |\cos x| \) represents the sum of the absolute values of sine and cosine functions.
2. **Finding the minimum value**: The minimum value occurs when both \( |\sin x| \) and \( |\cos x| \) are at their lowest. The minimum value of either function is 0.
3. **Analyzing the function**: The values of \( |\sin x| \) and \( |\cos x| \) can be analyzed using the identity:
\[ |\sin x|^2 + |\cos x|^2 = 1 \]
The minimum occurs when both are equal, which happens at \( x = \frac{\pi}{4} + k\frac{\pi}{2} \) for integers \( k \).
4. **Calculating the minimum**: At \( x = \frac{\pi}{4} \):
\[ g\left(\frac{\pi}{4}\right) = |\sin(\frac{\pi}{4})| + |\cos(\frac{\pi}{4})| = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \]
The minimum value of \( g(x) \) is \( 1 \) when \( x = 0, \frac{\pi}{2}, \pi, \ldots \).

Thus, \( B = 1 \).

### Conclusion

Now we have:
- \( A = 1 \)
- \( B = 1 \)

Therefore, \( A = B \).