If f (x)= ` ae^(|x|)+b |x|^(2) ,a ,b in ` R and f(x) is differentiable at x =0 ,then

Options

A. a=1,b=3
B. a=0 ,b` in ` R
C. a=1 ,b=2
D. a=2,b=3

Answer

To solve the problem, we need to analyze the function \( f(x) = ae^{|x|} + b|x|^2 \) and determine the conditions under which it is differentiable at \( x = 0 \).

### Step-by-Step Solution:

1. **Define the Function for Different Intervals**:
- For \( x \geq 0 \):
\[ f(x) = ae^x + b x^2 \]
- For \( x < 0 \):
\[ f(x) = ae^{-x} + b (-x)^2 = ae^{-x} + b x^2 \]

2. **Check Continuity at \( x = 0 \)**:
- We need to check if \( f(0) \) is the same from both sides:
\[ f(0) = ae^0 + b(0)^2 = a \]
- Left-hand limit as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = a e^0 + b(0)^2 = a \]
- Right-hand limit as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = a e^0 + b(0)^2 = a \]
- Since both limits equal \( f(0) \), the function is continuous at \( x = 0 \).

3. **Find the Derivative**:
- For \( x \geq 0 \):
\[ f'(x) = a e^x + 2bx \]
- For \( x < 0 \):
\[ f'(x) = -a e^{-x} + 2bx \]

4. **Evaluate the Derivative at \( x = 0 \)**:
- Left-hand derivative as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f'(x) = -a e^0 + 2b(0) = -a \]
- Right-hand derivative as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f'(x) = a e^0 + 2b(0) = a \]

5. **Set the Left-hand and Right-hand Derivatives Equal**:
- For \( f(x) \) to be differentiable at \( x = 0 \):
\[ -a = a \]
- This implies:
\[ 2a = 0 \Rightarrow a = 0 \]

6. **Conclusion about \( b \)**:
- Since \( b \) does not appear in the differentiability condition, \( b \) can be any real number.
- Thus, \( b \in \mathbb{R} \).

### Final Answer:
- The values of \( a \) and \( b \) are:
\[ a = 0, \quad b \in \mathbb{R} \]