If f (x) = `ae^(|x|)+b|x|^(2); a","b epsilon R` and f(x) is differentiable at x=0.Then ,a and b are

Options

A. a= 0,`b epsilon R`
B. a=1,b=2
C. b=0,`a epsilon R`
D. a=4 , b=5

Answer

To determine the values of \( a \) and \( b \) such that the function \( f(x) = ae^{|x|} + b|x|^2 \) is differentiable at \( x = 0 \), we will follow these steps:

### Step 1: Define the function for positive and negative \( x \)

For \( x \geq 0 \):
\[ f(x) = ae^x + bx^2 \]

For \( x < 0 \):
\[ f(x) = ae^{-x} + b(-x)^2 = ae^{-x} + bx^2 \]

### Step 2: Calculate \( f(0) \)

To find \( f(0) \):
\[ f(0) = ae^{0} + b(0)^2 = a \]

### Step 3: Calculate the right-hand derivative at \( x = 0 \)

Using the definition of the derivative:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} \]
Substituting \( f(h) \):
\[ f'(0^+) = \lim_{h \to 0^+} \frac{(ae^h + bh^2) - a}{h} \]
This simplifies to:
\[ = \lim_{h \to 0^+} \frac{a(e^h - 1) + bh^2}{h} \]
Using the fact that \( \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \):
\[ = \lim_{h \to 0^+} \left( a + bh \right) = a \]

### Step 4: Calculate the left-hand derivative at \( x = 0 \)

Using the definition of the derivative:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} \]
Substituting \( f(h) \):
\[ f'(0^-) = \lim_{h \to 0^-} \frac{(ae^{-h} + bh^2) - a}{h} \]
This simplifies to:
\[ = \lim_{h \to 0^-} \frac{a(e^{-h} - 1) + bh^2}{h} \]
Using the fact that \( \lim_{h \to 0} \frac{e^{-h} - 1}{h} = -1 \):
\[ = \lim_{h \to 0^-} \left( -a + bh \right) = -a \]

### Step 5: Set the derivatives equal for differentiability

For \( f(x) \) to be differentiable at \( x = 0 \), the left-hand and right-hand derivatives must be equal:
\[ f'(0^+) = f'(0^-) \implies a = -a \]
This implies:
\[ 2a = 0 \implies a = 0 \]

### Step 6: Determine the value of \( b \)

Since there are no conditions on \( b \), it can be any real number. Thus, \( b \in \mathbb{R} \).

### Final Result

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