The function `f (x)=[sqrt(1-sqrt(1-x ^(2)))],`(where [.] denotes greatest integer function):
i)has domain [−1,1]
ii)is discontinous at two points in its domain
iii) is discontinous at x=0
iv)is discontinous at x = 1

Options

A. has domain `[-1,1]`
B. is discontinous at two points in its domain
C. is discontinous at `x=0`
D. is discontinous at `x=1`

Answer

To solve the problem, we need to analyze the function \( f(x) = \lfloor \sqrt{1 - \sqrt{1 - x^2}} \rfloor \) where \( \lfloor . \rfloor \) denotes the greatest integer function. We will go through the steps to determine the domain, points of discontinuity, and check the specific points mentioned in the question.

### Step 1: Determine the Domain of \( f(x) \)

The function involves a square root, which requires the expression inside to be non-negative. Thus, we need to ensure:

\[ 1 - x^2 \geq 0 \]

This implies:

\[ x^2 \leq 1 \quad \Rightarrow \quad -1 \leq x \leq 1 \]

So, the domain of \( f(x) \) is:

\[ \text{Domain} = [-1, 1] \]

### Step 2: Check for Discontinuity at \( x = 0 \)

To check for continuity at \( x = 0 \), we need to evaluate the left-hand limit (LHL), right-hand limit (RHL), and the function value at \( x = 0 \).

1. **Right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lfloor \sqrt{1 - \sqrt{1 - (0^+)^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{1}} \rfloor = \lfloor \sqrt{0} \rfloor = 0 \]

2. **Left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lfloor \sqrt{1 - \sqrt{1 - (0^-)^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{1}} \rfloor = \lfloor \sqrt{0} \rfloor = 0 \]

3. **Function value at \( x = 0 \)**:
\[ f(0) = \lfloor \sqrt{1 - \sqrt{1 - 0^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{1}} \rfloor = \lfloor \sqrt{0} \rfloor = 0 \]

Since LHL = RHL = \( f(0) = 0 \), the function is continuous at \( x = 0 \).

### Step 3: Check for Discontinuity at \( x = 1 \)

1. **Right-hand limit as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lfloor \sqrt{1 - \sqrt{1 - (1^+)^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{1 - 1}} \rfloor = \lfloor \sqrt{1 - 0} \rfloor = \lfloor 1 \rfloor = 1 \]

2. **Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lfloor \sqrt{1 - \sqrt{1 - (1^-)^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{1 - 1}} \rfloor = \lfloor \sqrt{1 - 0} \rfloor = \lfloor 1 \rfloor = 1 \]

3. **Function value at \( x = 1 \)**:
\[ f(1) = \lfloor \sqrt{1 - \sqrt{1 - 1^2}} \rfloor = \lfloor \sqrt{1 - \sqrt{0}} \rfloor = \lfloor \sqrt{1} \rfloor = \lfloor 1 \rfloor = 1 \]

Since LHL = RHL = \( f(1) = 1 \), the function is continuous at \( x = 1 \).

### Conclusion

- The domain of \( f(x) \) is \( [-1, 1] \).
- The function is continuous at both \( x = 0 \) and \( x = 1 \).
- Therefore, the function is not discontinuous at the points specified in the question.

### Final Answer

i) The domain of \( f(x) \) is \( [-1, 1] \).
ii) The function is continuous at both \( x = 0 \) and \( x = 1 \), hence it is not discontinuous at these points.