Consider the function `f(x) = {{:(x{x}+1",","if",0 le x lt 1),(2-{x}",","if",1 le x le 2):}`, where {x} denotes the fractional part function. Which one of the following statements is not correct ?

Options

A. `underset(x rarr 1)(lim) f(x)` exists
B. `f(0) ne f(2)`
C. f(x) is continuous in [0, 2]
D. Rolle's theorem is not applicable to f(x) in [0, 2]

Answer

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

\[ f(x) = \begin{cases} \{x\} + 1 & \text{if } 0 \leq x < 1 \\ 2 - \{x\} & \text{if } 1 \leq x \leq 2 \end{cases} \]

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

### Step 1: Check the limit as \( x \) approaches 1 from the left (\( f(1^-) \))

To find \( f(1^-) \):

\[ f(1^-) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} (\{1 - h\} + 1) \]

Since \( 1 - h \) is slightly less than 1, the fractional part \( \{1 - h\} = 1 - h \).

Thus,

\[ f(1^-) = \lim_{h \to 0} (1 - h + 1) = \lim_{h \to 0} (2 - h) = 2 \]

### Step 2: Check the limit as \( x \) approaches 1 from the right (\( f(1^+) \))

To find \( f(1^+) \):

\[ f(1^+) = \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} (2 - \{1 + h\}) \]

Since \( 1 + h \) is slightly more than 1, the fractional part \( \{1 + h\} = h \).

Thus,

\[ f(1^+) = \lim_{h \to 0} (2 - h) = 2 \]

### Step 3: Verify if the limit exists at \( x = 1 \)

Since both the left-hand limit and the right-hand limit at \( x = 1 \) are equal:

\[ f(1^-) = f(1^+) = 2 \]

Thus, the limit exists at \( x = 1 \).

### Step 4: Check the function value at \( x = 1 \)

Now, we find \( f(1) \):

\[ f(1) = 2 - \{1\} = 2 - 0 = 2 \]

Since \( f(1^-) = f(1^+) = f(1) = 2 \), the function is continuous at \( x = 1 \).

### Step 5: Check continuity at \( x = 2 \)

Now, we need to check the continuity at \( x = 2 \).

#### Left-hand limit at \( x = 2 \):

\[ f(2^-) = \lim_{h \to 0} f(2 - h) = \lim_{h \to 0} (2 - \{2 - h\}) = \lim_{h \to 0} (2 - (1 - h)) = \lim_{h \to 0} (1 + h) = 1 \]

#### Function value at \( x = 2 \):

\[ f(2) = 2 - \{2\} = 2 - 0 = 2 \]

Since \( f(2^-) = 1 \) and \( f(2) = 2 \), the function is not continuous at \( x = 2 \).

### Conclusion

Now we can summarize the findings:

1. The limit \( \lim_{x \to 1} f(x) \) exists and equals \( 2 \).
2. \( f(0) = 1 \) and \( f(2) = 2 \), so \( f(0) \neq f(2) \).
3. The function is continuous at \( x = 1 \) but not at \( x = 2 \).

Thus, the statement that is **not correct** is that \( f(x) \) is continuous on the interval \( [0, 2] \).