If `f(x)=3x^(2)+k|sin x|` is differentiable at `x=0` then the value of k is-

Options

A. `1`
B. `-1`
C. `0`
D. `-3`

Answer

To determine the value of \( k \) such that the function \( f(x) = 3x^2 + k|\sin x| \) is differentiable at \( x = 0 \), we will follow these steps:

### Step 1: Evaluate \( f(0) \)
First, we need to find \( f(0) \):
\[ f(0) = 3(0)^2 + k|\sin(0)| = 0 + k \cdot 0 = 0 \]

### Step 2: Define \( f(x) \) for \( x > 0 \) and \( x < 0 \)
Next, we need to express \( f(x) \) for \( x > 0 \) and \( x < 0 \):
- For \( x > 0 \):
\[ f(x) = 3x^2 + k\sin x \]
- For \( x < 0 \):
\[ f(x) = 3x^2 - k\sin x \]

### Step 3: Find the left-hand derivative at \( x = 0 \)
The left-hand derivative at \( x = 0 \) is given by:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(3h^2 - k\sin h) - 0}{h} \]
This simplifies to:
\[ f'(0^-) = \lim_{h \to 0^-} \left( 3h - \frac{k\sin h}{h} \right) \]
Using the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \):
\[ f'(0^-) = 0 - k = -k \]

### Step 4: Find the right-hand derivative at \( x = 0 \)
The right-hand derivative at \( x = 0 \) is given by:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(3h^2 + k\sin h) - 0}{h} \]
This simplifies to:
\[ f'(0^+) = \lim_{h \to 0^+} \left( 3h + \frac{k\sin h}{h} \right) \]
Again, using the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \):
\[ f'(0^+) = 0 + k = k \]

### Step 5: Set the left-hand and right-hand derivatives equal
For \( f(x) \) to be differentiable at \( x = 0 \), the left-hand and right-hand derivatives must be equal:
\[ -k = k \]
This implies:
\[ 2k = 0 \implies k = 0 \]

### Conclusion
The value of \( k \) for which \( f(x) \) is differentiable at \( x = 0 \) is:
\[ \boxed{0} \]