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 different intervals and ensure that it is one-one and monotonically increasing across all \( x \in \mathbb{R} \).

### Given Function:
\[ f(x) = \begin{cases} -k - e^{-x} & \text{for } x \leq 0 \\ e^{x} + 1 & \text{for } 0 < x < 1 \\ e^{x^2} + \lambda & \text{for } x \geq 1 \end{cases} \]

### Step 1: Check Monotonicity for Each Interval

1. **For \( x \leq 0 \)**:
\[ f(x) = -e^{-x} + k \]
The derivative is:
\[ f'(x) = e^{-x} > 0 \quad \text{(since \( e^{-x} \) is always positive)} \]
Therefore, \( f(x) \) is increasing for \( x \leq 0 \).

2. **For \( 0 < x < 1 \)**:
\[ f(x) = e^{x} + 1 \]
The derivative is:
\[ f'(x) = e^{x} > 0 \quad \text{(since \( e^{x} \) is always positive)} \]
Therefore, \( f(x) \) is increasing for \( 0 < x < 1 \).

3. **For \( x \geq 1 \)**:
\[ f(x) = e^{x^2} + \lambda \]
The derivative is:
\[ f'(x) = 2xe^{x^2} > 0 \quad \text{(since \( x \geq 1 \) and \( e^{x^2} \) is always positive)} \]
Therefore, \( f(x) \) is increasing for \( x \geq 1 \).

### Step 2: Ensure Continuity at Transition Points

1. **At \( x = 0 \)**:
\[ f(0^-) = -1 + k \quad \text{and} \quad f(0^+) = 1 \]
For continuity:
\[ -1 + k \leq 1 \implies k \leq 3 \]

2. **At \( x = 1 \)**:
\[ f(1^-) = e + 1 \quad \text{and} \quad f(1^+) = e + \lambda \]
For continuity:
\[ e + 1 \leq e + \lambda \implies \lambda \geq 1 \]

### Step 3: Determine Maximum \( k \) and Minimum \( \lambda \)

- The maximum value of \( k \) is \( 3 \).
- The minimum value of \( \lambda \) is \( 1 \).

### Final Answer:
- Maximum value of \( k \) is \( 3 \).
- Minimum value of \( \lambda \) is \( 1 \).