Question:

Step 1: Determine $f(x)$. Since $f:(0,\infty)\to[-2,\infty)$ is surjective and $f(x)=ax^2-bx+c$, the minimum value is $-2$. Also $\lim_{x\to 0}f(x)=3$, so $c=3$ (since $f(0^+)=c=3$... but $0$ is not in domain). Actually $\lim_{x\to 0^+}f(x) = c = 3$. Step 2: The minimum of $f(x) = ax^2 - bx + 3$ on $(0,\infty)$ is $-2$. Minimum occurs at $x = \frac{b}{2a}$ and equals $c - \frac{b^2}{4a} = 3 - \frac{b^2}{4a} = -2$, so $\frac{b^2}{4a} = 5$. Step 3: For $g:[1,\infty)\to[-2,\infty)$ to be invertible (one-to-one and onto), $g$ must be monotonically increasing on $[1,\infty)$. This means the vertex of the parabola is at $x=1$: $\frac{b}{2a}=1 \Rightarrow b=2a$. Substituting: $\frac{(2a)^2}{4a} = a = 5$. So $a=5, b=10, c=3$. Thus $f(x) = 5x^2 - 10x + 3$ and $g(x) = 5x^2 - 10x + 3$ for $x\in[1,\infty)$. Step 4: Check option (a): $g'(x) = 10x - 10$, so $g'(1) = 0$. Thus $40\cdot g'(1) = 0$. ✓ Step 5: Find $g^{-1}(x)$. From $y = 5x^2 - 10x + 3 = 5(x-1)^2 - 2$: $y + 2 = 5(x-1)^2 \Rightarrow x = 1 + \sqrt{\frac{y+2}{5}}$ (taking $+$ since $x\geq 1$). So $g^{-1}(x) = 1 + \sqrt{\frac{x+2}{5}}$, domain $x\in[-2,\infty)$. Step 6: Check option (b): Domain of $g(g(x))$ requires $g(x)\in[1,\infty)$, i.e., $5x^2-10x+3\geq 1 \Rightarrow 5x^2-10x+2\geq 0$. Roots: $x = \frac{10\pm\sqrt{100-40}}{10} = 1\pm\frac{\sqrt{60}}{10} = 1\pm\frac{\sqrt{15}}{5}$. For $x\geq 1$: $x\geq 1+\frac{\sqrt{15}}{5} = 1+\sqrt{\frac{15}{25}} = 1+\sqrt{\frac{3}{5}}$. So domain is $\left[1+\sqrt{\frac{3}{5}},\infty\right)$, meaning $p=3, q=5$, so $q-p=2$. ✓ Step 7: Check option (c): $g(x) = g^{-1}(x)$ means $5x^2-10x+3 = 1+\sqrt{\frac{x+2}{5}}$. Since $g$ is increasing on $[1,\infty)$ and $g^{-1}$ is also increasing, solutions occur where $g(x)=x$ (on the line $y=x$). $5x^2-10x+3=x \Rightarrow 5x^2-11x+3=0$, discriminant $=121-60=61>0$, giving 2 roots, but check if both are $\geq 1$: roots $=\frac{11\pm\sqrt{61}}{10}$. $\frac{11-\sqrt{61}}{10}\approx\frac{11-7.81}{10}\approx 0.32 < 1$. So only 1 solution in $[1,\infty)$. Option (c) is incorrect. Step 8: Check option (d): $\frac{d}{dx}(90g^{-1}(x)) = 90\cdot(g^{-1})'(x) = \frac{90}{g'(g^{-1}(x))}$. At $x=43$: $g^{-1}(43) = 1+\sqrt{\frac{45}{5}} = 1+3=4$. $g'(4) = 10(4)-10=30$. So $\frac{90}{30}=3$. ✓