If `f(x)={{:(-e^(-x)+k,",",xle0),(e^(x)+1,",",0ltxlt1),(ex^(2)+lambda,",",xge1):}` is one-one and monotonically increasing `AA x in R`, then

Options

A. maximum value of k is 1
B. maximum value of k is 3
C. minimum value of `lambda` is 0
D. minimum value of `lambda` is 1

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined in three parts and ensure that it is one-one and monotonically increasing for all \( x \in \mathbb{R} \).

### Step 1: Analyze the function in each interval

The function \( f(x) \) is defined as:
1. \( f(x) = -e^{-x} + k \) for \( x \leq 0 \)
2. \( f(x) = e^x + 1 \) for \( 0 < x < 1 \)
3. \( f(x) = e x^2 + \lambda \) for \( x \geq 1 \)

### Step 2: Check monotonicity in each interval

**For \( x \leq 0 \):**
- The derivative \( f'(x) = e^{-x} \) (since the derivative of \( -e^{-x} \) is \( e^{-x} \)).
- Since \( e^{-x} > 0 \) for all \( x \), \( f(x) \) is increasing in this interval.

**For \( 0 < x < 1 \):**
- The derivative \( f'(x) = e^x \).
- Since \( e^x > 0 \) for all \( x \), \( f(x) \) is increasing in this interval.

**For \( x \geq 1 \):**
- The derivative \( f'(x) = 2ex \).
- Since \( 2ex > 0 \) for all \( x \geq 1 \), \( f(x) \) is increasing in this interval.

### Step 3: Ensure continuity at the boundaries

To ensure that \( f(x) \) is one-one, we need to check the values at the boundaries \( x = 0 \) and \( x = 1 \).

**At \( x = 0 \):**
- \( f(0) = -e^0 + k = -1 + k \)

**At \( x = 1 \):**
- \( f(1) = e^1 + 1 = e + 1 \)

For \( f(x) \) to be continuous at \( x = 0 \):
\[ -1 + k < e + 1 \]
This implies:
\[ k < e + 2 \]

**At \( x = 1 \):**
- \( f(1) = e + 1 \) and \( f(1) = e(1)^2 + \lambda = e + \lambda \)

For continuity at \( x = 1 \):
\[ e + 1 = e + \lambda \]
This implies:
\[ \lambda = 1 \]

### Step 4: Ensure that \( f(x) \) is one-one

Since \( f(x) \) is increasing in all intervals and continuous, it is one-one.

### Conclusion

The conditions for \( f(x) \) to be one-one and monotonically increasing for all \( x \in \mathbb{R} \) are:
- \( k < e + 2 \)
- \( \lambda = 1 \)

### Final Result
Thus, the values of \( k \) and \( \lambda \) are:
- \( \lambda \leq 1 \)
- \( k < e + 2 \)