If `f(x)={{:([x]+sqrt({x})",", xlt1),((1)/([x]+{x}^(2))",",xge1):}`, then
[where [.] and {.} represent the greatest integer and fractional part functions respectively]

Options

A. f(x) is continuous at x = 1 but not differentiable
B. f(x) is not continuous at x = 1
C. f(x) is differentiable at x = 1
D. `underset(xrarr1)(lim)f(x)` does not exist

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise as follows:

\[ f(x) = \begin{cases} [x] + \sqrt{x} & \text{if } x < 1 \\ \frac{1}{[x] + x^2} & \text{if } x \geq 1 \end{cases} \]

where \([x]\) is the greatest integer function and \(\{x\}\) is the fractional part function.

### Step 1: Evaluate \( f(1) \)

For \( x \geq 1 \):
\[ f(1) = \frac{1}{[1] + 1^2} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Step 2: Evaluate the limit as \( x \) approaches 1 from the left

For \( x < 1 \):
\[ f(x) = [x] + \sqrt{x} \]
Since \( [x] = 0 \) for \( x < 1 \), we have:
\[ f(x) = \sqrt{x} \]

Now, we calculate the limit as \( x \) approaches 1 from the left:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \sqrt{x} = \sqrt{1} = 1 \]

### Step 3: Evaluate the limit as \( x \) approaches 1 from the right

For \( x \geq 1 \):
\[ f(x) = \frac{1}{[x] + x^2} \]
Since \( [x] = 1 \) for \( x \) in the interval \([1, 2)\), we have:
\[ f(x) = \frac{1}{1 + x^2} \]

Now, we calculate the limit as \( x \) approaches 1 from the right:
\[ \lim_{x \to 1^+} f(x) = \frac{1}{1 + 1^2} = \frac{1}{2} \]

### Step 4: Check continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), the following must hold:
\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]

From our calculations:
- \( \lim_{x \to 1^-} f(x) = 1 \)
- \( f(1) = \frac{1}{2} \)
- \( \lim_{x \to 1^+} f(x) = \frac{1}{2} \)

Since \( \lim_{x \to 1^-} f(x) \neq f(1) \), \( f(x) \) is not continuous at \( x = 1 \).

### Step 5: Check differentiability at \( x = 1 \)

Since \( f(x) \) is not continuous at \( x = 1 \), it cannot be differentiable at that point.

### Conclusion

The function \( f(x) \) is continuous at \( x = 1 \) but not differentiable. Therefore, the correct option is:

**Answer: \( f(x) \) is continuous at \( x = 1 \) but not differentiable.**