If `f(x) = {{:(x+{x}+x sin {x}",","for",x ne 0),(0",","for",x = 0):}`, where {x} denotes the fractional part function, then

Options

A. f is continuous and differentiable at x = 0
B. f is continuous but not differentiable at x = 0
C. f is continuous and differentiable at x = 2
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} x + \{x\} + x \sin(\{x\}) & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} \]

where \(\{x\}\) denotes the fractional part of \(x\).

### Step 1: Check Continuity at \(x = 0\)

To check the continuity of \(f(x)\) at \(x = 0\), we need to evaluate:

\[ \lim_{x \to 0} f(x) \]

For \(x \neq 0\):

\[ f(x) = x + \{x\} + x \sin(\{x\}) \]

The fractional part function \(\{x\} = x - \lfloor x \rfloor\). For \(x\) approaching \(0\), \(\{x\} = x\) when \(x\) is positive and \(\{x\} = 0\) when \(x\) is negative. Thus, we can analyze the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( x + \{x\} + x \sin(\{x\}) \right) \]

As \(x \to 0\):

- For \(x > 0\): \(\{x\} = x\), so \(f(x) = x + x + x \sin(x) = 2x + x \sin(x)\).
- For \(x < 0\): \(\{x\} = 0\), so \(f(x) = x + 0 + x \sin(0) = x\).

Now, we compute the limits:

1. As \(x \to 0^+\):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x + x \sin(x)) = 0 \]

2. As \(x \to 0^-\):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0 \]

Since both one-sided limits equal \(0\), we have:

\[ \lim_{x \to 0} f(x) = 0 \]

Thus, \(f(0) = 0\) and \(f(x)\) is continuous at \(x = 0\).

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

To check differentiability, we need to find the left-hand derivative (LHD) and right-hand derivative (RHD) at \(x = 0\).

**Left-Hand Derivative (LHD)**:

\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h)}{h} \]

For \(h < 0\):

\[ f(h) = h \quad \text{(since } \{h\} = 0\text{)} \]

Thus,

\[ \text{LHD} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]

**Right-Hand Derivative (RHD)**:

\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h)}{h} \]

For \(h > 0\):

\[ f(h) = 2h + h \sin(h) \]

Thus,

\[ \text{RHD} = \lim_{h \to 0^+} \frac{2h + h \sin(h)}{h} = \lim_{h \to 0^+} (2 + \sin(h)) = 2 \]

### Conclusion

Since the left-hand derivative (1) and the right-hand derivative (2) are not equal, \(f(x)\) is not differentiable at \(x = 0\).

### Final Answer

The function \(f(x)\) is continuous at \(x = 0\) but not differentiable at \(x = 0\). Therefore, the correct option is:

**Option 2: f is continuous but not differentiable at \(x = 0\)**.