Let `f:[1,3] to R` be a function satisfying `(x)/([x]) le f (x) le sqrt(6-x),"for all "x ne 2 and f(2)=1`, Where R is the set of all real number and [x] denotes the largest integer less than or equal to x.
Statement-1: `lim_(x to 2) f(x)` exists.
Statement-2: F is continuous at x=2.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1
B. Statement 1 is false, Statement 2 is true
C. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1
D. Statement 1 is true, Statement 2 is false

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined on the interval \([1, 3]\) with the given conditions. Let's break it down step by step.

### Step 1: Understand the function and its bounds
The function \( f(x) \) is defined such that:
\[ \frac{x}{[x]} \leq f(x) \leq \sqrt{6 - x} \quad \text{for all } x \neq 2 \]
where \([x]\) denotes the greatest integer less than or equal to \(x\).

### Step 2: Determine the values of \([x]\)
For \( x \in [1, 3] \):
- When \( 1 \leq x < 2 \), \([x] = 1\)
- When \( 2 < x \leq 3 \), \([x] = 2\)

### Step 3: Analyze the bounds
1. For \( 1 \leq x < 2 \):
\[ \frac{x}{1} \leq f(x) \leq \sqrt{6 - x} \]
This simplifies to:
\[ x \leq f(x) \leq \sqrt{6 - x} \]

2. For \( 2 < x \leq 3 \):
\[ \frac{x}{2} \leq f(x) \leq \sqrt{6 - x} \]
This simplifies to:
\[ \frac{x}{2} \leq f(x) \leq \sqrt{6 - x} \]

### Step 4: Evaluate the limits as \( x \) approaches 2
We need to find:
\[ \lim_{x \to 2} f(x) \]

#### From the left (as \( x \to 2^- \)):
\[ x \to 2 \Rightarrow f(x) \text{ is bounded by } 2 \text{ and } \sqrt{6 - 2} = \sqrt{4} = 2 \]
Thus, as \( x \to 2^- \):
\[ f(x) \to 2 \]

#### From the right (as \( x \to 2^+ \)):
\[ x \to 2 \Rightarrow f(x) \text{ is bounded by } \frac{2}{2} = 1 \text{ and } \sqrt{6 - 2} = \sqrt{4} = 2 \]
Thus, as \( x \to 2^+ \):
\[ f(x) \to 2 \]

### Step 5: Conclusion for Statement 1
Since both the left-hand limit and right-hand limit as \( x \to 2 \) approach 2, we conclude:
\[ \lim_{x \to 2} f(x) = 2 \]
Thus, **Statement 1 is true**: \( \lim_{x \to 2} f(x) \) exists.

### Step 6: Check continuity at \( x = 2 \)
For \( f \) to be continuous at \( x = 2 \), we need:
\[ \lim_{x \to 2} f(x) = f(2) \]
We found \( \lim_{x \to 2} f(x) = 2 \) and given \( f(2) = 1 \).

Since:
\[ \lim_{x \to 2} f(x) \neq f(2) \]
Thus, **Statement 2 is false**: \( f \) is not continuous at \( x = 2 \).

### Final Answer
- Statement 1: True
- Statement 2: False