Consider the function `f(x)=cos^(-1)([2^(x)])+sin^(-1)([2^(x)]-1)`, then
(where `[.]` represents the greatest integer part function)

Options

A. Domain of f(x) is `x in (-oo,0]`
B. Range of `f(x)` is singleton
C. `f(x)` is an even function
D. `f(x)` is an odd function

Answer

To solve the problem, we need to analyze the function given:

\[ f(x) = \cos^{-1}(\lfloor 2^x \rfloor) + \sin^{-1}(\lfloor 2^x \rfloor - 1) \]

where \(\lfloor . \rfloor\) denotes the greatest integer function.

### Step 1: Determine the range of \( \lfloor 2^x \rfloor \)

The expression \( 2^x \) can take values from \( 0 \) to \( 2 \) as \( x \) varies. Thus, the greatest integer function \( \lfloor 2^x \rfloor \) can take the values:
- \( 0 \) when \( 0 \leq 2^x < 1 \) (i.e., \( x < 0 \))
- \( 1 \) when \( 1 \leq 2^x < 2 \) (i.e., \( 0 \leq x < 1 \))
- \( 2 \) when \( 2^x \geq 2 \) (i.e., \( x \geq 1 \))

### Step 2: Evaluate \( f(x) \) for different cases of \( \lfloor 2^x \rfloor \)

#### Case 1: \( \lfloor 2^x \rfloor = 0 \)

When \( \lfloor 2^x \rfloor = 0 \):
\[ f(x) = \cos^{-1}(0) + \sin^{-1}(0 - 1) \]
\[ = \frac{\pi}{2} + \sin^{-1}(-1) \]
\[ = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

#### Case 2: \( \lfloor 2^x \rfloor = 1 \)

When \( \lfloor 2^x \rfloor = 1 \):
\[ f(x) = \cos^{-1}(1) + \sin^{-1}(1 - 1) \]
\[ = 0 + \sin^{-1}(0) = 0 \]

#### Case 3: \( \lfloor 2^x \rfloor = 2 \)

When \( \lfloor 2^x \rfloor = 2 \):
\[ f(x) = \cos^{-1}(2) + \sin^{-1}(2 - 1) \]
However, \( \cos^{-1}(2) \) is undefined in the real number system, so this case does not contribute to the function.

### Conclusion

From the above evaluations, we find that for \( x < 0 \) and \( 0 \leq x < 1 \), \( f(x) = 0 \). The case where \( \lfloor 2^x \rfloor = 2 \) does not yield a valid output.

Thus, we conclude that the function \( f(x) \) is constant and equal to \( 0 \) for all \( x < 1 \).

### Final Answer

The range of \( f(x) \) is \( \{0\} \), which indicates that it is a single-valued function.