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 $\left|z-1\right|=\left|2z\right|$, then a possible value of $\frac{1+\left[a\right]}{4b}$, where $\left[t\right]$ is greatest integer function, is :

Given,

$a,b$ be two real numbers such that $ab<0$, $\left|\frac{1+ai}{b+i}\right|=1$, $a+ib$ lies on the circle $\left|z-1\right|=\left|2z\right|$ and $ab<0$

Now using $\left|\frac{1+ai}{b+i}\right|=1$ we get,

$\Rightarrow \left|1+ai\right|=\left|b+i\right|$

Now squaring both side we get,

$\Rightarrow {\left|1+ai\right|}^2={\left|b+i\right|}^2$

$\Rightarrow \left(1+ai\right)\left(1-ai\right)=\left(b+i\right)\left(b-i\right) \left\{{\left|z\right|}^2=z\overline{z}\right\}$

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

$\Rightarrow a=\pm b\Rightarrow b=-a$ as $ab<0$

Now given,

$\left(a,b\right)$ lies on $\left|\mathrm{z}-1\right|=\left|2\mathrm{z}\right|$

So, $\left|a+ib-1\right|=2\left|a+ib\right|$

$\Rightarrow {\left|a-1+ib\right|}^2=4{\left|a+ib\right|}^2$

$\Rightarrow {\left(a-1\right)}^2+b^2=4\left(a^2+b^2\right)$

$\Rightarrow {\left(a-1\right)}^2+a^2=4\left(2a^2\right) \left\{\text{as} b=-a\right\}$

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

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

Taking positive sign we get,

$a=\frac{\sqrt{7}-1}{6} \& b=\frac{1-\sqrt{7}}{6}$ $\left\{\text{as} b=-a\right\}$

Now the value of $\left[a\right]=0$ as $0<\frac{\sqrt{7}-1}{6}<1$

$\therefore \frac{1+\left[a\right]}{4b}=\frac{6}{4\left(1-\sqrt{7}\right)}=-\left(\frac{1+\sqrt{7}}{4}\right)$

Now taking negative sign we get, $a=\frac{-1-\sqrt{7}}{6} \& b=\frac{\sqrt{7}+1}{6}$

Now  $\left[a\right]=-1 \left\{\text{as} -1<\frac{-1-\sqrt{7}}{6}<0\right\}$

So, the value of $\frac{1+\left[a\right]}{4b}= \frac{1+\left(-1\right)}{4b}=\frac{0}{4b}=0$

Note- Official answer key was drop by NTA, here we have modified the option.