Find the values of p and q for which the function f(x) is continuous for all x in R, where:

f(x) =
begin{cases}
\sin((p+1)x) + \frac{\sin x}{x}, & x < 0 \
q, & x = 0 \
\sqrt{x} + x^2 - \sqrt{ax}x^{3/2}, & x > 0
end{cases}

For f(x) to be continuous at x=0, the left-hand limit, right-hand limit, and the value of the function at x=0 must be equal. Let's analyze the limits:

Left-hand limit (x→0⁻):

lim_(x→0⁻) [sin((p+1)x) + sin(x)/x] = sin(0) + lim_(x→0⁻) [sin(x)/x] = 0 + 1 = 1

Right-hand limit (x→0⁺):

lim_(x→0⁺) [√x + x² - √(ax)x³/²] = 0 + 0 - 0 = 0

Function value at x=0:

f(0) = q

For continuity at x=0, we must have:

1 = 0 = q
This is not possible. There must be a mistake in the problem statement or our understanding of the function definition. Let's assume there's a typo in the provided function. Perhaps the 'a' in the third part is supposed to be '1', resulting in the term √x * x³/² simplifying to x². In this case, the right-hand limit will also be 1 and thus will be continuous at x=0.

Let's re-examine the right-hand limit with the assumption a=1:

lim_(x→0⁺) [√x + x² - √x * x³/²] = lim_(x→0⁺) [√x + x² - x²] = lim_(x→0⁺) √x = 0

This still leads to an inconsistency (1 ≠ 0). Let's check if there's a different interpretation. Let's consider the case where the term is \sqrt{a x} x^{3/2}. If a=1, then the term becomes x². Then the right hand limit would be:

lim_(x→0⁺) (√x + x² - x²) = lim_(x→0⁺) √x = 0

This is inconsistent with the left-hand limit, so there must be an error in the question. If a=0 then the right hand limit is 0. Therefore there is no such p and q values that would make the function continuous at x=0. If we change the third function to √x + x^2 - x^2 we then have:

lim_(x→0⁺) √x = 0
This is still not consistent with the left hand limit which was 1. The question is likely flawed.