Key Concepts and Formulas

• Continuity at a point: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$. 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: $f(c^-) = f(c^+) = f(c)$.
• Standard Limits:
  • $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$
  • $\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \frac{1}{2}$

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x=0$. For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function value at that point must be equal. Therefore, we must have: $f(0^-) = f(0) = f(0^+)$

Step 2: Calculate the Left-Hand Limit, $f(0^-)$ The left-hand limit is the limit of the function as $x$ approaches 0 from values less than 0. For $x < 0$, $f(x) = \frac{\sin((p+1)x) + \sin x}{x}$. $f(0^-) = \lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin x}{x}$ We can split this into two separate limits: $f(0^-) = \lim_{x \to 0^-} \frac{\sin((p+1)x)}{x} + \lim_{x \to 0^-} \frac{\sin x}{x}$ Using the standard limit $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$, we get: $f(0^-) = (p+1) \lim_{x \to 0^-} \frac{\sin((p+1)x)}{(p+1)x} + \lim_{x \to 0^-} \frac{\sin x}{x}$ As $x \to 0^-$, $(p+1)x \to 0$. So, $\lim_{x \to 0^-} \frac{\sin((p+1)x)}{(p+1)x} = 1$ and $\lim_{x \to 0^-} \frac{\sin x}{x} = 1$. $f(0^-) = (p+1) \cdot 1 + 1 = p+1+1 = p+2$

Step 3: Determine the Function Value at $x=0$, $f(0)$ For $x=0$, the function is defined as $f(x) = q$. $f(0) = q$

Step 4: Calculate the Right-Hand Limit, $f(0^+)$ The right-hand limit is the limit of the function as $x$ approaches 0 from values greater than 0. For $x > 0$, $f(x) = \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}$. $f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}$ First, we can factor out $\sqrt{x}$ from the numerator: $f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x^{3/2}}$ Now, simplify the expression by canceling $\sqrt{x}$ with $x^{3/2} = x \cdot x^{1/2}$: $f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \sqrt{x}}$ $f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x}$ This limit is in the indeterminate form $\frac{0}{0}$. To evaluate it, we can multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{1+x} + 1$: $f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}$ $f(0^+) = \lim_{x \to 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)}$ $f(0^+) = \lim_{x \to 0^+} \frac{x}{x(\sqrt{1+x} + 1)}$ Cancel out $x$: $f(0^+) = \lim_{x \to 0^+} \frac{1}{\sqrt{1+x} + 1}$ Now, substitute $x=0$: $f(0^+) = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1+1} = \frac{1}{2}$

Step 5: Equate the Limits and Function Value to Find p and q From Step 1, we have $f(0^-) = f(0) = f(0^+)$. Substituting the values calculated in Steps 2, 3, and 4: $p+2 = q = \frac{1}{2}$ Equating $q$ and $\frac{1}{2}$: $q = \frac{1}{2}$ Equating $p+2$ and $\frac{1}{2}$: $p+2 = \frac{1}{2}$ $p = \frac{1}{2} - 2$ $p = \frac{1}{2} - \frac{4}{2}$ $p = -\frac{3}{2}$ Thus, the ordered pair $(p,q)$ is $\left(-\frac{3}{2}, \frac{1}{2}\right)$.

Common Mistakes & Tips

• Incorrectly applying standard limits: Ensure that the argument of the sine function matches the denominator when using $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$. In Step 2, we correctly multiplied and divided by $(p+1)$.
• Algebraic errors in rationalizing: When dealing with square roots, be careful with the conjugate multiplication and expansion. Errors in this step can lead to an incorrect right-hand limit.
• Forgetting the function value at the point: Continuity requires $f(0^-) = f(0) = f(0^+)$. While $f(0^-)$ and $f(0^+)$ are often used to find $p$ and $q$, it's crucial to remember that $f(0)$ must also be equal to them. In this case, $f(0)=q$ directly provides one of the values.

Summary To ensure continuity at $x=0$, we equated the left-hand limit, the function value at $x=0$, and the right-hand limit. We calculated the left-hand limit using the standard sine limit formula, found the function value directly from the definition, and evaluated the right-hand limit by rationalizing the numerator. Equating these expressions allowed us to solve for the unknown values of $p$ and $q$.

The final answer is \boxed{\left( { - {3 \over 2}, {1 \over 2}} \right)}. This corresponds to option (C).