The function `f(x) ={{:(e^(2x)-1,","x lt 0),(ax+(bx^2)/2-,","xlt0):}` continuous and differentiable for

Options

A. `a=1,b=2`
B. `a=2,b=1`
C. `a=2`, any b
D. any `a,b=4`

Explanation

To find the values of \( a \) and \( b \) for the function

\[ f(x) = \begin{cases} e^{2x} - 1 & \text{if } x < 0 \\ ax + \frac{bx^2}{2} & \text{if } x \geq 0 \end{cases} \]

such that \( f(x) \) is continuous and differentiable, we will follow these steps:

### Step 1: Ensure Continuity at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \]

Calculating \( \lim_{x \to 0^-} f(x) \):

\[ \lim_{x \to 0^-} f(x) = e^{2 \cdot 0} - 1 = 1 - 1 = 0 \]

Calculating \( \lim_{x \to 0^+} f(x) \):

\[ \lim_{x \to 0^+} f(x) = a \cdot 0 + \frac{b \cdot 0^2}{2} = 0 \]

Setting these equal for continuity:

\[ 0 = 0 \]

This condition is satisfied for any \( a \) and \( b \).

### Step 2: Ensure Differentiability at \( x = 0 \)

For \( f(x) \) to be differentiable at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x) \]

Calculating \( f'(x) \) for \( x < 0 \):

\[ f'(x) = \frac{d}{dx}(e^{2x} - 1) = 2e^{2x} \]

Thus,

\[ \lim_{x \to 0^-} f'(x) = 2e^{2 \cdot 0} = 2 \cdot 1 = 2 \]

Calculating \( f'(x) \) for \( x \geq 0 \):

\[ f'(x) = \frac{d}{dx}\left(ax + \frac{bx^2}{2}\right) = a + bx \]

Thus,

\[ \lim_{x \to 0^+} f'(x) = a + b \cdot 0 = a \]

Setting these equal for differentiability:

\[ 2 = a \]

### Step 3: Conclusion about \( b \)

Since the continuity condition did not impose any restrictions on \( b \), \( b \) can take any real value. Therefore, we conclude:

\[ a = 2 \quad \text{and} \quad b \in \mathbb{R} \]

### Final Answer:

\[ a = 2, \quad b \text{ can be any real number} \]

---