Question:

Step 1: For $g: R \to (-\infty, -1]$ to be surjective, the range of $g(x)$ must be exactly $(-\infty, -1]$. This means $g(x)$ must have a maximum value of $-1$ and must be an onto function onto $(-\infty, -1]$. Step 2: For $g(x)$ to map onto $(-\infty, -1]$, the odd-degree terms must vanish (so the function is bounded above), meaning the coefficients of $x^5$ and $x^3$ must be zero, and the coefficient of $x^2$ must be negative (so the function goes to $-\infty$). Step 3: Set coefficient of $x^5$ equal to zero: $$pq + 2p - q - 2 = 0$$ $$p(q+2) - (q+2) = 0$$ $$(p-1)(q+2) = 0$$ So $p = 1$ or $q = -2$. Step 4: Set coefficient of $x^3$ equal to zero: $$p^3 - 2p + 1 = 0$$ $$(p-1)(p^2 + p - 1) = 0$$ So $p = 1$ or $p = \dfrac{-1 \pm \sqrt{5}}{2}$. Step 5: Since $p$ must be rational, from Step 4, $p = 1$. Step 6: With $p = 1$, check coefficient of $x^2$: $$p^2 - 2p - 3 = 1 - 2 - 3 = -4 < 0$$ ✓ Step 7: With $p = 1$, the function becomes: $$g(x) = -4x^2 + (1 + 2q)x - 5$$ For surjectivity onto $(-\infty, -1]$, the maximum value must equal $-1$. Step 8: Maximum of $g(x) = -4x^2 + (1+2q)x - 5$ occurs at $x = \dfrac{1+2q}{8}$: $$g_{\max} = -5 + \frac{(1+2q)^2}{16} = -1$$ $$\frac{(1+2q)^2}{16} = 4$$ $$(1+2q)^2 = 64$$ $$1 + 2q = \pm 8$$ $$q = \frac{7}{2} \text{ or } q = -\frac{9}{2}$$ Step 9: Compute $p + q$ for each case: - If $q = \dfrac{7}{2}$: $p + q = 1 + \dfrac{7}{2} = \dfrac{9}{2}$ — but wait, check option: this gives $\dfrac{9}{2}$. - If $q = -\dfrac{9}{2}$: $p + q = 1 - \dfrac{9}{2} = -\dfrac{7}{2}$. Step 10: Also check $q = -2$ from Step 3 with rational $p$ from Step 4. Since $p$ must be rational and $p=1$ is the only rational solution, the valid pairs give $p+q = \dfrac{9}{2}$ or $p+q = -\dfrac{7}{2}$, corresponding to options (a) and (c), i.e., options 1 and 3.