Let f(x) be a function defined by `f(x)={{:(,e^(x),x lt 2),(,a+bx,x ge 2):}` It f(x) is differentiably for all `x in R`, then

Options

A. `a+b=0`
B. `a+2b=e^(2)`
C. `b=e^(2)`
D. None of these

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is differentiable for all \( x \in \mathbb{R} \). The function is defined as follows:

\[ f(x) = \begin{cases} e^x & \text{if } x < 2 \\ a + bx & \text{if } x \geq 2 \end{cases} \]

### Step 1: Check differentiability at \( x = 2 \)

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

### Step 2: Find the left-hand limit and right-hand limit at \( x = 2 \)

**Left-hand limit (as \( x \) approaches 2 from the left):**
\[ f(2^-) = e^2 \]

**Right-hand limit (as \( x \) approaches 2 from the right):**
\[ f(2^+) = a + 2b \]

### Step 3: Set the left-hand limit equal to the right-hand limit for continuity

For continuity at \( x = 2 \):
\[ e^2 = a + 2b \quad \text{(1)} \]

### Step 4: Find the derivatives at \( x = 2 \)

**Left-hand derivative (as \( x \) approaches 2 from the left):**
\[ f'(2^-) = \frac{d}{dx}(e^x) \bigg|_{x=2} = e^2 \]

**Right-hand derivative (as \( x \) approaches 2 from the right):**
\[ f'(2^+) = \frac{d}{dx}(a + bx) \bigg|_{x=2} = b \]

### Step 5: Set the left-hand derivative equal to the right-hand derivative for differentiability

For differentiability at \( x = 2 \):
\[ e^2 = b \quad \text{(2)} \]

### Step 6: Substitute \( b \) into equation (1)

From equation (2), we have \( b = e^2 \). Substitute this into equation (1):
\[ e^2 = a + 2(e^2) \]

### Step 7: Solve for \( a \)

Rearranging gives:
\[ e^2 = a + 2e^2 \implies a = e^2 - 2e^2 = -e^2 \]

### Conclusion

The values of \( a \) and \( b \) that make \( f(x) \) differentiable for all \( x \in \mathbb{R} \) are:
\[ a = -e^2, \quad b = e^2 \]