1. For continuity of the piecewise function on $[0,\infty)$, we only need to check the junction points:

   • $x=1$
   • $x=\sqrt{2}$

   since each piece is continuous on its own interval.

2. The function is

   \begin{cases} \dfrac{2x^2}{a}, & 0\le x<1 \\ a, & 1\le x<\sqrt{2} \\ \dfrac{2b^2-4b}{x^3}, & \sqrt{2}\le x<\infty \end{cases}$$ with $a\ne 0$.

3. Continuity at $x=1$:

   Left-hand limit at $x=1$: $\lim_{x\to 1^-} f(x)=\frac{2(1)^2}{a}=\frac{2}{a}$

   Value at $x=1$ from the second piece: $f(1)=a$

   For continuity, $\frac{2}{a}=a$ $a^2=2$ $a=\pm \sqrt{2}$

4. Continuity at $x=\sqrt{2}$:

   Left-hand limit from the second piece: $\lim_{x\to (\sqrt{2})^-} f(x)=a$

   Value at $x=\sqrt{2}$ from the third piece: $f(\sqrt{2})=\frac{2b^2-4b}{(\sqrt{2})^3}$

   Now, $(\sqrt{2})^3=2\sqrt{2}$ so $f(\sqrt{2})=\frac{2b^2-4b}{2\sqrt{2}}=\frac{b^2-2b}{\sqrt{2}}$

   For continuity, $a=\frac{b^2-2b}{\sqrt{2}}$

5. Use the two possible values of $a$.

   Case 1: $a=\sqrt{2}$

   $\sqrt{2}=\frac{b^2-2b}{\sqrt{2}}$ Multiply by $\sqrt{2}$: $2=b^2-2b$ $b^2-2b-2=0$ $b=\frac{2\pm\sqrt{4+8}}{2}=\frac{2\pm 2\sqrt{3}}{2}=1\pm\sqrt{3}$

   Among the options, with $a=\sqrt{2}$ we have:

   • $\left(\sqrt{2},1-\sqrt{3}\right)$
   • $\left(\sqrt{2},-1+\sqrt{3}\right)$, and note $-1+\sqrt{3}=\sqrt{3}-1\ne 1+\sqrt{3}$

   So only $b=1-\sqrt{3}$ matches option A.

   Case 2: $a=-\sqrt{2}$

   $-\sqrt{2}=\frac{b^2-2b}{\sqrt{2}}$ Multiply by $\sqrt{2}$: $-2=b^2-2b$ $b^2-2b+2=0$

   Discriminant: $\Delta = (-2)^2-4(1)(2)=4-8=-4<0$ So there is no real value of $b$.

6. Therefore the real ordered pairs satisfying continuity are obtained from $a=\sqrt{2},\quad b=1\pm\sqrt{3}.$

   From the given options, only $\boxed{\left(\sqrt{2},1-\sqrt{3}\right)}$ appears.

7. Check options:

   • A: $\left(\sqrt{2},1-\sqrt{3}\right)$ ✅
   • B: $a=-\sqrt{2}$ gives no real $b$ ❌
   • C: $b=-1+\sqrt{3}=\sqrt{3}-1$, which does not satisfy $b=1\pm\sqrt{3}$ ❌
   • D: $a=-\sqrt{2}$ gives no real $b$ ❌

Hence the correct option is A.