The function 'g' defined by `g(x)= sin(sin^(-1)sqrt({x}))+cos(sin^(-1)sqrt({x}))-1` (where {x} denotes the functional part function) is (1) an even function (2) a periodic function (3) an odd function (4) neither even nor odd

Options

A. an even function
B. periodic function
C. odd function
D. Neither even nor odd

Explanation

To determine the nature of the function \( g(x) = \sin(\sin^{-1}(\sqrt{x})) + \cos(\sin^{-1}(\sqrt{x})) - 1 \), we will analyze it step by step.

### Step 1: Simplify the Function
We start with the function:
\[ g(x) = \sin(\sin^{-1}(\sqrt{x})) + \cos(\sin^{-1}(\sqrt{x})) - 1 \]
Using the property of the inverse sine function, we know that:
\[ \sin(\sin^{-1}(y)) = y \quad \text{for } y \in [0, 1] \]
Thus, we can simplify:
\[ g(x) = \sqrt{x} + \cos(\sin^{-1}(\sqrt{x})) - 1 \]

### Step 2: Simplify the Cosine Term
Next, we simplify \( \cos(\sin^{-1}(\sqrt{x})) \). We use the identity:
\[ \cos(\sin^{-1}(y)) = \sqrt{1 - y^2} \]
So, we have:
\[ \cos(\sin^{-1}(\sqrt{x})) = \sqrt{1 - (\sqrt{x})^2} = \sqrt{1 - x} \]
Now substituting this back into our function:
\[ g(x) = \sqrt{x} + \sqrt{1 - x} - 1 \]

### Step 3: Check for Evenness
To check if \( g(x) \) is an even function, we need to verify if \( g(-x) = g(x) \).
Calculating \( g(-x) \):
\[ g(-x) = \sqrt{-x} + \sqrt{1 - (-x)} - 1 \]
However, \( \sqrt{-x} \) is not defined for real \( x \) when \( x > 0 \). Thus, we cannot directly compare \( g(-x) \) with \( g(x) \).

### Step 4: Check for Oddness
Next, we check if \( g(-x) = -g(x) \):
From the previous step, since \( g(-x) \) is not defined for \( x > 0 \), we cannot conclude that \( g(x) \) is odd.

### Step 5: Determine Periodicity
To check if the function is periodic, we need to find if there exists a period \( T \) such that \( g(x + T) = g(x) \) for all \( x \). Since the function involves square roots and is defined for \( x \in [0, 1] \) (due to the square root and inverse sine), it does not exhibit periodic behavior.

### Conclusion
Based on the analysis:
- The function is not even because \( g(-x) \) is not defined for all \( x \).
- The function is not odd for the same reason.
- The function is not periodic as it does not repeat values over an interval.

Thus, the correct answer is:
**(4) Neither even nor odd.**