If $$f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$$ is continuous at $$x = 0$$, then the ordered pair (p, q) is equal to:

We need to find the ordered pair $$(p, q)$$ such that the given piecewise function is continuous at $$x = 0$$.

For continuity at $$x = 0$$, we need:

$$\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$$

Step 1: Right-hand limit ($$x \to 0^+$$)

$$\lim_{x \to 0^+} \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}$$

Factor out $$\sqrt{x}$$ from the numerator:

$$= \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x}$$

Rationalize by multiplying numerator and denominator by $$(\sqrt{1+x} + 1)$$:

$$= \lim_{x \to 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$

Step 2: Value at $$x = 0$$

$$f(0) = q$$

From the continuity condition: $$q = \frac{1}{2}$$

Step 3: Left-hand limit ($$x \to 0^-$$)

$$\lim_{x \to 0^-} \frac{\sin(p+1)x + \sin x}{x}$$

Using the standard limit $$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$:

$$= \lim_{x \to 0^-} \left[\frac{\sin(p+1)x}{x} + \frac{\sin x}{x}\right] = (p+1) \cdot 1 + 1 = p + 2$$

Step 4: Apply continuity condition

$$p + 2 = q = \frac{1}{2}$$

$$p = \frac{1}{2} - 2 = -\frac{3}{2}$$

Therefore $$(p, q) = \left(-\frac{3}{2}, \frac{1}{2}\right)$$.

The correct answer is Option A.