Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if and only if the following three conditions are met:

  1. $f(c)$ is defined.
  2. $\mathop {\lim }\limits_{x \to c} f(x)$ exists.
  3. $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$. For a piecewise function, this implies that the left-hand limit, the right-hand limit, and the function value at the point must all be equal: $\mathop {\lim }\limits_{x \to c^-} f(x) = \mathop {\lim }\limits_{x \to c^+} f(x) = f(c)$.

• Standard Limit of Sine Function: A fundamental limit in calculus is $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

• Binomial Approximation: For small values of $x$, the binomial expansion $(1+x)^n \approx 1 + nx$ can be used.

Step-by-Step Solution

Step 1: Evaluate the left-hand limit of $f(x)$ at $x=0$. We are given that for $x < 0$, $f(x) = \frac{\sin((a+2)x) + \sin x}{x}$. To find the left-hand limit, we set $x \to 0^-$. $f(0^-) = \mathop {\lim }\limits_{x \to {0^- }} \frac{\sin \left( {a + 2} \right)x + \sin x}{x}$ We can split this into two separate limits: $f(0^-) = \mathop {\lim }\limits_{x \to {0^- }} \frac{\sin \left( {a + 2} \right)x}{x} + \mathop {\lim }\limits_{x \to {0^- }} \frac{\sin x}{x}$ To use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we multiply and divide the first term by $(a+2)$: $f(0^-) = \mathop {\lim }\limits_{x \to {0^- }} \frac{\sin \left( {a + 2} \right)x}{\left( {a + 2} \right)x} \times \left( {a + 2} \right) + \mathop {\lim }\limits_{x \to {0^- }} \frac{\sin x}{x}$ As $x \to 0^-$, $(a+2)x \to 0$ and $x \to 0$. Therefore, applying the standard limit: $f(0^-) = 1 \times (a+2) + 1$ $f(0^-) = a + 2 + 1 = a + 3$

Step 2: Evaluate the right-hand limit of $f(x)$ at $x=0$. We are given that for $x > 0$, $f(x) = \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}}$. To find the right-hand limit, we set $x \to 0^+$. $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{\left( {x + 3{x^2}} \right)^{{1 \over 3}} - {x^{{1 \over 3}}}}{{{x^{{4 \over 3}}}}}$ We can factor out $x^{1/3}$ from the term in the numerator: $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{\left( {x(1 + 3x)} \right)^{{1 \over 3}} - {x^{{1 \over 3}}}}{{{x^{{4 \over 3}}}}}$ $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{x^{{1 \over 3}}}(1 + 3x)^{{1 \over 3}} - {x^{{1 \over 3}}}}{{{x^{{4 \over 3}}}}}$ Now, factor out $x^{1/3}$ from the numerator: $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{x^{{1 \over 3}}}((1 + 3x)^{{1 \over 3}} - 1)}{{{x^{{4 \over 3}}}}}$ We can simplify the powers of $x$: $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{(1 + 3x)^{{1 \over 3}} - 1}{{{x^{{4 \over 3}}} / {x^{{1 \over 3}}}}}$ $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{(1 + 3x)^{{1 \over 3}} - 1}{{{x^{{4 \over 3} - {1 \over 3}}}}}$ $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{(1 + 3x)^{{1 \over 3}} - 1}{x}$ Now, we can use the binomial approximation $(1+y)^n \approx 1 + ny$ for small $y$. Here, $y=3x$ and $n=1/3$. As $x \to 0^+$, $3x \to 0$, so the approximation is valid. $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{(1 + \frac{1}{3}(3x)) - 1}{x}$ $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{(1 + x) - 1}{x}$ $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{x}$ $f(0^+) = 1$

Alternatively, we can use L'Hopital's rule or the definition of the derivative. Let $g(x) = (1+3x)^{1/3}$. Then $f(0^+) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{g(x) - g(0)}{x} = g'(0)$. $g'(x) = \frac{1}{3}(1+3x)^{-2/3} \cdot 3 = (1+3x)^{-2/3}$. So, $g'(0) = (1+0)^{-2/3} = 1$. Thus, $f(0^+) = 1$.

Step 3: Evaluate the function value at $x=0$. We are given that for $x=0$, $f(x) = b$. $f(0) = b$

Step 4: Apply the condition for continuity. Since the function $f(x)$ is continuous at $x=0$, the left-hand limit, the right-hand limit, and the function value at $x=0$ must be equal. $f(0^-) = f(0) = f(0^+)$ From Step 1, Step 2, and Step 3, we have: $a + 3 = b = 1$

Step 5: Solve for the values of $a$ and $b$. From the equality $a + 3 = 1$, we get: $a = 1 - 3$ $a = -2$ From the equality $b = 1$, we have: $b = 1$

Step 6: Calculate the value of $a + 2b$. Now we substitute the values of $a$ and $b$ that we found: $a + 2b = (-2) + 2(1)$ $a + 2b = -2 + 2$ $a + 2b = 0$

Common Mistakes & Tips

• Incorrect application of standard limits: Ensure that the argument of the sine function in the numerator matches the denominator when using $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. This often requires multiplying and dividing by a constant or an expression involving $x$.
• Algebraic errors in simplifying expressions: Pay close attention to exponent rules and factoring when dealing with terms like $x^{4/3}$ and $(x+3x^2)^{1/3}$. Simplifying the expression before applying limits can prevent errors.
• Misapplication of binomial approximation: The approximation $(1+x)^n \approx 1 + nx$ is valid only for small values of $x$. In this case, $3x$ approaches 0 as $x$ approaches 0, so the approximation is appropriate.

Summary

To determine the values of $a$ and $b$ for the given piecewise function to be continuous at $x=0$, we first calculated the left-hand limit, the right-hand limit, and the function value at $x=0$. By setting these three values equal to each other, we formed a system of equations that allowed us to solve for $a$ and $b$. The left-hand limit involved using the standard sine limit, while the right-hand limit required simplification and the binomial approximation. Once $a$ and $b$ were found, we computed the required expression $a + 2b$.

The final answer is \boxed{0}. This corresponds to option (A).