Let $f\left(x\right)=\mathrm{sgn}(x^2-4x+4+k^2),x\in R.f\left(x\right)$ is discontinuous at exactly one point then the value of $\left({\tan }^{-1}\right(k)+{\cos }^{-1}(k)+\cos e{c}^{-1}(2k-1\left)\right)$ is equal to $\frac{m\pi }{2}$ where $m$ is a whole number.  Then the value of $m$ is_____ $\left(\mathrm{sgn}\right(x)$ is represents signum function)

$f\left(x\right)=\mathrm{sgn}(x^2-4x+4+k^2)$

$\mathrm{sgn}(\left[{(x-2)}^2\right]+k^2)$

If $k\ne 0,{(x-2)}^2+k^2>0\Rightarrow \mathrm{sgn}[{(x-2)}^2+k^2]=1\Rightarrow f$ is constant function $\forall x\in R$

$\Rightarrow f$ is continuous $\forall x\in R$

If $k=0f\left(x\right)=\mathrm{sgn}\left({(x-2)}^2\right)=\{\begin{array}{l}1ifx\ne 2\\ 0x=2\end{array}$

Is discontinuous at $x=2$

$\therefore Ta{n}^{-1}\left(k\right)+{\cos }^{-1}\left(k\right)+\cos e{c}^{-1}(2k-1)={\tan }^{-1}\left(0\right)+{\cos }^{-1}\left(0\right)+\cos e{c}^{-1}(-1)$

$=0+\frac{\pi }{2}-\frac{\pi }{2}=0=\frac{m\pi }{2}\Rightarrow m=0$