The function g, defined by g(x) `= sinalpha + cosalpha - 1, alpha = sin^(-1) sqrt({x})` , {.} denotes fractional part function is

Options

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

Answer

To solve the problem, we need to analyze the function \( g(x) = \sin(\alpha) + \cos(\alpha) - 1 \) where \( \alpha = \sin^{-1}(\sqrt{\{x\}}) \) and \(\{x\}\) denotes the fractional part of \(x\).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( g(x) \) is defined in terms of \( \alpha \), which is derived from the fractional part of \( x \). The fractional part function, denoted as \(\{x\}\), is defined as \( x - \lfloor x \rfloor \), where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \( x \).

2. **Substituting \( \alpha \)**:
We substitute \( \alpha \) into the function:
\[ g(x) = \sin(\sin^{-1}(\sqrt{\{x\}})) + \cos(\sin^{-1}(\sqrt{\{x\}})) - 1 \]
Using the identity \( \sin(\sin^{-1}(y)) = y \), we have:
\[ g(x) = \sqrt{\{x\}} + \cos(\sin^{-1}(\sqrt{\{x\}})) - 1 \]

3. **Finding \( \cos(\sin^{-1}(y)) \)**:
We can find \( \cos(\sin^{-1}(y)) \) using the Pythagorean identity:
\[ \cos(\sin^{-1}(y)) = \sqrt{1 - y^2} \]
Therefore, substituting \( y = \sqrt{\{x\}} \):
\[ \cos(\sin^{-1}(\sqrt{\{x\}})) = \sqrt{1 - \{x\}} \]

4. **Final Expression for \( g(x) \)**:
Now substituting back into the equation for \( g(x) \):
\[ g(x) = \sqrt{\{x\}} + \sqrt{1 - \{x\}} - 1 \]

5. **Checking for Evenness**:
To check if \( g(x) \) is even, we need to evaluate \( g(-x) \):
\[ g(-x) = \sqrt{\{-x\}} + \sqrt{1 - \{-x\}} - 1 \]
Since \(\{-x\} = 1 - \{x\}\) when \(x\) is not an integer, we can substitute:
\[ g(-x) = \sqrt{1 - \{x\}} + \sqrt{\{x\}} - 1 = g(x) \]
Thus, \( g(-x) = g(x) \), indicating that \( g(x) \) is an even function.

6. **Checking for Periodicity**:
To check if \( g(x) \) is periodic, we evaluate \( g(x + t) \):
\[ g(x + t) = \sqrt{\{x + t\}} + \sqrt{1 - \{x + t\}} - 1 \]
If \( t \) is an integer, then \(\{x + t\} = \{x\}\). Therefore:
\[ g(x + t) = \sqrt{\{x\}} + \sqrt{1 - \{x\}} - 1 = g(x) \]
Hence, \( g(x) \) is periodic with period \( t \) being any integer.

### Conclusion:
The function \( g(x) \) is an even function and is periodic with integer periods.