Let $$a, b$$ be two real numbers such that $$ab < 0$$. If the complex number $$\frac{1+ai}{b+i}$$ is of unit modulus and $$a + ib$$ lies on the circle $$|z - 1| = |2z|$$, then a possible value of $$\frac{1+[a]}{4b}$$, where $$[t]$$ is greatest integer function, is:

We need to find a possible value of $$\frac{1+[a]}{4b}$$ given the conditions on $$a$$ and $$b$$.

1. $$ab < 0$$ ($$a$$ and $$b$$ have opposite signs)

2. $$\left|\frac{1+ai}{b+i}\right| = 1$$ (unit modulus)

3. $$a + ib$$ lies on the circle $$|z - 1| = |2z|$$

Use the unit modulus condition.

$$\left|\frac{1+ai}{b+i}\right| = 1 \implies |1+ai| = |b+i|$$

$$\sqrt{1 + a^2} = \sqrt{b^2 + 1}$$

$$1 + a^2 = b^2 + 1$$

$$a^2 = b^2$$

$$|a| = |b|$$

Since $$ab < 0$$, we have $$a = -b$$ (they have opposite signs).

Use the circle condition.

The point $$z = a + ib$$ lies on $$|z - 1| = |2z|$$:

$$|a + ib - 1| = |2(a + ib)|$$

$$|(a-1) + ib| = |2a + 2ib|$$

$$(a-1)^2 + b^2 = 4a^2 + 4b^2$$

Since $$a = -b$$, substitute $$b = -a$$:

$$(a-1)^2 + a^2 = 4a^2 + 4a^2$$

$$a^2 - 2a + 1 + a^2 = 8a^2$$

$$2a^2 - 2a + 1 = 8a^2$$

$$6a^2 + 2a - 1 = 0$$

Solve the quadratic.

$$a = \frac{-2 \pm \sqrt{4 + 24}}{12} = \frac{-2 \pm \sqrt{28}}{12} = \frac{-2 \pm 2\sqrt{7}}{12} = \frac{-1 \pm \sqrt{7}}{6}$$

So $$a = \frac{-1 + \sqrt{7}}{6} \approx \frac{-1 + 2.646}{6} \approx \frac{1.646}{6} \approx 0.274$$

or $$a = \frac{-1 - \sqrt{7}}{6} \approx \frac{-3.646}{6} \approx -0.608$$

Since $$ab < 0$$ and $$b = -a$$:

If $$a > 0$$, then $$b < 0$$, and $$ab = -a^2 < 0$$ (valid)

If $$a < 0$$, then $$b > 0$$, and $$ab = -a^2 < 0$$ (valid)

Both values are valid.

Calculate $$\frac{1+[a]}{4b}$$.

Case 1: $$a \approx 0.274$$, so $$[a] = 0$$ and $$b = -a \approx -0.274$$

$$\frac{1 + 0}{4 \times (-0.274)} = \frac{1}{-1.096} \approx -0.912$$

Case 2: $$a \approx -0.608$$, so $$[a] = -1$$ and $$b = -a \approx 0.608$$

$$\frac{1 + (-1)}{4 \times 0.608} = \frac{0}{2.432} = 0$$

In Case 2, the value is exactly $$0$$.

The correct answer is Option A: $$0$$.