Statement-1: If a continuous funtion on [0,1] satisfy ` 0 le f(x) le 1`, then there exist `c in [0,1]` such that f(c )=c
Statement-2: `lim_(x to c) f(x)=f(c)`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the two statements provided and establish their validity.

### Step 1: Understanding the Statements

**Statement 1:** If a continuous function \( f \) on the interval \([0, 1]\) satisfies \( 0 \leq f(x) \leq 1 \), then there exists \( c \in [0, 1] \) such that \( f(c) = c \).

**Statement 2:** \( \lim_{x \to c} f(x) = f(c) \) is always true for continuous functions.

### Step 2: Analyzing Statement 2

Statement 2 is a fundamental property of continuous functions. By definition, a function \( f \) is continuous at a point \( c \) if:

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

Thus, Statement 2 is true.

### Step 3: Analyzing Statement 1

To prove Statement 1, we will define a new function:

\[ g(x) = x - f(x) \]

### Step 4: Determine the Values of \( g(x) \)

Now, we analyze \( g(x) \) over the interval \([0, 1]\):

- At \( x = 0 \):
\[ g(0) = 0 - f(0) = -f(0) \quad \text{(since } 0 \leq f(0) \leq 1 \Rightarrow g(0) \leq 0\text{)} \]

- At \( x = 1 \):
\[ g(1) = 1 - f(1) \quad \text{(since } 0 \leq f(1) \leq 1 \Rightarrow g(1) \geq 0\text{)} \]

### Step 5: Applying the Intermediate Value Theorem

Since \( g(0) \leq 0 \) and \( g(1) \geq 0 \), by the Intermediate Value Theorem, since \( g(x) \) is continuous (as \( f(x) \) is continuous), there exists some \( c \in [0, 1] \) such that:

\[ g(c) = 0 \]

This implies:

\[ c - f(c) = 0 \quad \Rightarrow \quad f(c) = c \]

### Conclusion

Thus, we have shown that there exists \( c \in [0, 1] \) such that \( f(c) = c \). Therefore, Statement 1 is true.

### Final Result

Both statements are true:

- Statement 1: True
- Statement 2: True