F(x) = `{( (sin(a+2)x + sin x)/x , x lt 0), (b , x = 0), (((x + 3x^2 )^(1/3) - x^(1/3))/x^(4/3)), x gt 0):}`
Function is continuous at x = 0, find `a + 2b`.

Options

A. `-1`
B. `0`
C. `1`
D. `-2`

Answer

To solve the problem, we need to ensure that the function \( F(x) \) is continuous at \( x = 0 \). This means that the left-hand limit as \( x \) approaches 0 must equal the right-hand limit as \( x \) approaches 0, and both must equal \( F(0) \).

### Step 1: Find the left-hand limit as \( x \) approaches 0
For \( x < 0 \):
\[ F(x) = \frac{\sin(a + 2x) + \sin x}{x} \]
We need to find:
\[ \lim_{x \to 0^-} F(x) = \lim_{x \to 0^-} \frac{\sin(a + 2x) + \sin x}{x} \]
Using the Taylor series expansion for \( \sin \):
\[ \sin(a + 2x) \approx \sin a + 2x \cos a \quad \text{and} \quad \sin x \approx x \]
Thus,
\[ \sin(a + 2x) + \sin x \approx \sin a + 2x \cos a + x = \sin a + x(2\cos a + 1) \]
Now substituting this back into the limit:
\[ \lim_{x \to 0^-} \frac{\sin a + x(2\cos a + 1)}{x} = \lim_{x \to 0^-} \left(\frac{\sin a}{x} + 2\cos a + 1\right) \]
As \( x \to 0^- \), \( \frac{\sin a}{x} \) diverges unless \( \sin a = 0 \). Therefore, we require:
\[ \sin a = 0 \implies a = n\pi \quad (n \in \mathbb{Z}) \]

### Step 2: Find the right-hand limit as \( x \) approaches 0
For \( x > 0 \):
\[ F(x) = \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}} \]
We need to find:
\[ \lim_{x \to 0^+} F(x) = \lim_{x \to 0^+} \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}} \]
Factoring out \( x^{1/3} \):
\[ = \lim_{x \to 0^+} \frac{x^{1/3} \left((1 + 3x)^{1/3} - 1\right)}{x^{4/3}} = \lim_{x \to 0^+} \frac{(1 + 3x)^{1/3} - 1}{x} \]
Using the binomial expansion:
\[ (1 + 3x)^{1/3} \approx 1 + \frac{1}{3}(3x) = 1 + x \]
Thus,
\[ \lim_{x \to 0^+} \frac{x}{x} = 1 \]

### Step 3: Set the limits equal to each other
Since the function is continuous at \( x = 0 \), we have:
\[ \lim_{x \to 0^-} F(x) = \lim_{x \to 0^+} F(x) = F(0) \]
This gives us:
\[ 2\cos a + 1 = b \]
And since \( \sin a = 0 \), we have \( a = n\pi \). For \( n = 0 \):
\[ a = 0 \implies b = 2 \cdot 1 + 1 = 3 \]

### Step 4: Calculate \( a + 2b \)
Now substituting the values:
\[ a + 2b = 0 + 2 \cdot 3 = 6 \]

### Final Answer:
The value of \( a + 2b \) is \( 6 \).