Let `f(x)=a+b|x|+c|x|^(2)`, where a,b,c are real constants. The, f'(0) exists if

Options

A. b=0
B. c=0
C. a=0
D. b=c

Explanation

To determine the conditions under which the derivative \( f'(0) \) exists for the function \( f(x) = a + b|x| + c|x|^2 \), we will analyze the function step by step.

### Step 1: Rewrite the Function
The function can be rewritten by recognizing that \( |x|^2 = x^2 \) for all \( x \). Therefore, we have:
\[ f(x) = a + b|x| + cx^2 \]

### Step 2: Identify the Non-Differentiable Point
The absolute value function \( |x| \) is not differentiable at \( x = 0 \). Thus, we need to check the differentiability of \( f(x) \) at this point.

### Step 3: Calculate the Left-Hand Derivative
To find \( f'(0) \), we calculate the left-hand derivative:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} \]
For \( h < 0 \), \( |h| = -h \), so:
\[ f(h) = a + b(-h) + c(h^2) = a - bh + ch^2 \]
Thus,
\[ f(0) = a \]
Now, substituting into the limit:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{(a - bh + ch^2) - a}{h} = \lim_{h \to 0^-} \frac{-bh + ch^2}{h} = \lim_{h \to 0^-} (-b + ch) = -b \]

### Step 4: Calculate the Right-Hand Derivative
Next, we calculate the right-hand derivative:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} \]
For \( h > 0 \), \( |h| = h \), so:
\[ f(h) = a + bh + ch^2 \]
Thus,
\[ f'(0^+) = \lim_{h \to 0^+} \frac{(a + bh + ch^2) - a}{h} = \lim_{h \to 0^+} \frac{bh + ch^2}{h} = \lim_{h \to 0^+} (b + ch) = b \]

### Step 5: Set the Left-Hand and Right-Hand Derivatives Equal
For \( f'(0) \) to exist, the left-hand and right-hand derivatives must be equal:
\[ -b = b \]
This implies:
\[ 2b = 0 \quad \Rightarrow \quad b = 0 \]

### Conclusion
Thus, the derivative \( f'(0) \) exists if and only if \( b = 0 \).

### Final Answer
The condition for \( f'(0) \) to exist is:
\[ \boxed{b = 0} \]