If $n_1$  denotes number of points of non-differentiability of $f\left(x\right)=\cos (2x-1)\frac{\pi }{2}.$$|x^2-4x-5|+|x^5+x^3|+x.e^{\left|x\right|}$
  and  $n_2$ denotes number of points of discontinuity of  $g\left(x\right)=\mathrm{sgn}({\sin }^{2}x-\sin x-1),$$x\in (0,4\pi ).$  [Note : $\mathrm{sgn}x$  denotes the signum function of $x.$ ]   Find the value of $(n_1+n_2).$   
 

 
$\underset{differentiableonR}{\underbrace{f\left(x\right)=\underset{\begin{array}{l}differentiableat\\ x=-1.5and\\ value=0at\\ thesepoints\end{array}}{\underbrace{\cos (2x-1)\frac{\pi }{2}}}.\underset{\begin{array}{l}continuous\\ atx=-5,5\end{array}}{\underbrace{\left|(x+1)(x-5)\right|}}+\underset{\begin{array}{l}differentiable\\ atx=0\end{array}}{\underbrace{x^2\left|x\right|(x^2+1)}}+\underset{\begin{array}{l}differentiable\\ atx=0and\\ hasvalue=0\\ atx=0\end{array}}{\underbrace{x}}.\underset{\begin{array}{l}continous\\ atx=0\end{array}}{\underbrace{e^{\left|x\right|}}}}}$
$\Rightarrow n_1=0$ $g\left(x\right)=\mathrm{sgn}({\sin }^{2}x-\sin x-1)$is discontinuous where
$$\begin{gathered} \therefore ({\sin }^{2}x-\sin x-1)=0 \\ \Rightarrow \sin x=\frac{1\pm \sqrt{5}}{2}\Rightarrow \sin x=\frac{1-\sqrt{5}}{2} \end{gathered}$$
i.e., it changes sign 4 times.
$\Rightarrow n_2=4\Rightarrow n_1+n_2=4$