Also $f\left(x\right)$ must be continuous at $x=1$
$\therefore f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$
$\frac{1}{2}=a+{\tan }^{-1}\left(1-b\right)\dots \left(\mathrm{i}\right)$
$f^{\prime }\left(x\right)=\left\{\begin{array}{ccc}\frac{1}{1+{\left(x-b\right)}^2}& ;& x\ge 1\\ \frac{1}{2}& ;& x<1\end{array}\right.$
Now for $f\left(x\right)$ to be differentiable at $x=1$
$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$
$\frac{1}{1+{\left(1-b\right)}^2}=\frac{1}{2}\Rightarrow 1+{\left(1-b\right)}^2=2$

$\Rightarrow b=2$ or $0\dots \left(\mathrm{i}\mathrm{i}\right)$
From equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{i}\mathrm{i}\right)$, we get
$\frac{1}{2}=a\pm \frac{\pi }{4}\Rightarrow a=\frac{1}{2}-\frac{\pi }{4}$ or $\frac{1}{2}+\frac{\pi }{4}$
$\left(a,b\right)\equiv \left(\frac{1}{2}-\frac{\pi }{4},0\right)$ or $\left(\frac{1}{2}+\frac{\pi }{4},2\right)$
Hence, $4a-b=2-\pi ,\pi$