1. Use the unit modulus condition

Given $\left|\frac{1+ai}{b+i}\right|=1$ so $|1+ai|=|b+i|.$

Now, $|1+ai|=\sqrt{1+a^2},\qquad |b+i|=\sqrt{b^2+1}.$ Hence, $1+a^2=b^2+1 \implies a^2=b^2.$ So, $a=\pm b.$

But it is also given that $ab<0.$ If $a=b$, then $ab=a^2\ge 0$, impossible. Therefore, $a=-b.$

2. Use the circle condition

Given that $a+ib$ lies on $|z-1|=|2z|.$ Let $z=a+ib$. Then $|z-1|=|a-1+ib|,\qquad |2z|=2|z|=2\sqrt{a^2+b^2}.$ So, $\sqrt{(a-1)^2+b^2}=2\sqrt{a^2+b^2}.$ Squaring, $(a-1)^2+b^2=4(a^2+b^2).$ Expand: $a^2-2a+1+b^2=4a^2+4b^2$ $3a^2+3b^2+2a-1=0.$

Using $a=-b$, we get $b^2=a^2$, so $3a^2+3a^2+2a-1=0$ $6a^2+2a-1=0.$ Solve: $6a^2+2a-1=0$ $a=\frac{-2\pm\sqrt{4+24}}{12}=\frac{-2\pm\sqrt{28}}{12}=\frac{-1\pm\sqrt7}{6}.$ Thus, $a=\frac{-1+\sqrt7}{6}\quad \text{or} \quad a=\frac{-1-\sqrt7}{6}.$ Since $b=-a$, $b=\frac{1-\sqrt7}{6}\quad \text{or} \quad b=\frac{1+\sqrt7}{6}.$

Check $ab<0$: since $b=-a$, indeed $ab=-a^2<0$ as long as $a\ne 0$, which is true here.

3. Compute $\dfrac{1+[a]}{4b}$

We test both possible values.

Case 1:

$a=\frac{-1+\sqrt7}{6}.$ Since $\sqrt7\approx 2.646$, $a\approx \frac{1.646}{6}\approx 0.274,$ so $[a]=0.$ Also, $b=-a=\frac{1-\sqrt7}{6}.$ Then $\frac{1+[a]}{4b}=\frac{1}{4b}=\frac{1}{4\cdot \frac{1-\sqrt7}{6}}=\frac{6}{4(1-\sqrt7)}=\frac{3}{2(1-\sqrt7)}.$ This is not among the options.

Case 2:

$a=\frac{-1-\sqrt7}{6}\approx -0.6077.$ So, $[a]=-1.$ Also, $b=-a=\frac{1+\sqrt7}{6}.$ Then $\frac{1+[a]}{4b}=\frac{1-1}{4b}=0.$

Thus a possible value is $\boxed{0}.$

4. Evaluate options

• A: $\dfrac{1+\sqrt7}{4}$ — not obtained
• B: $\dfrac12$ — not obtained
• C: $0$ — obtained
• D: $-1$ — not obtained

Therefore, the correct option is $\boxed{\text{C}}.$