If `f(x)={{:(,e^(x),x lt 2),(,ax+b,x ge 2):}` is differentiable for all `x in R`, them

Options

A. `a=e^(2),b=-e^(2)`
B. `a=-e^(2),b=e^(2)`
C. `a=b=e^(2)`
D. none of these

Explanation

To solve the problem, we need to ensure that the piecewise function \( f(x) \) is both continuous and differentiable at the point where the pieces meet, which is at \( x = 2 \).

### Step 1: Ensure Continuity at \( x = 2 \)

For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit as \( x \) approaches 2 must equal the right-hand limit and the value of the function at that point.

1. **Left-hand limit** as \( x \to 2^- \):
\[ \lim_{x \to 2^-} f(x) = e^2 \]

2. **Right-hand limit** as \( x \to 2^+ \):
\[ \lim_{x \to 2^+} f(x) = a(2) + b = 2a + b \]

3. **Value of the function at \( x = 2 \)**:
\[ f(2) = 2a + b \]

Setting the left-hand limit equal to the right-hand limit gives us:
\[ e^2 = 2a + b \quad \text{(1)} \]

### Step 2: Ensure Differentiability at \( x = 2 \)

For \( f(x) \) to be differentiable at \( x = 2 \), the left-hand derivative must equal the right-hand derivative.

1. **Left-hand derivative** at \( x = 2 \):
\[ f'(x) = e^x \quad \Rightarrow \quad f'(2) = e^2 \]

2. **Right-hand derivative** at \( x = 2 \):
\[ f'(x) = a \quad \Rightarrow \quad f'(2) = a \]

Setting the left-hand derivative equal to the right-hand derivative gives us:
\[ e^2 = a \quad \text{(2)} \]

### Step 3: Solve the System of Equations

Now we have two equations:
1. \( e^2 = 2a + b \) (from continuity)
2. \( e^2 = a \) (from differentiability)

Substituting equation (2) into equation (1):
\[ e^2 = 2(e^2) + b \]
\[ e^2 = 2e^2 + b \]
\[ b = e^2 - 2e^2 = -e^2 \]

### Final Values

Thus, we have:
\[ a = e^2 \quad \text{and} \quad b = -e^2 \]

### Summary of the Solution

The values of \( a \) and \( b \) are:
\[ \boxed{a = e^2, \quad b = -e^2} \]