$\text{ Let }f\left(x\right)=\left\{\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim} \frac{ax(x-1){\left(\cot \frac{\pi x}{4}\right)}^n+\left(p{x}^2+2\right)}{{\left(\cot \frac{\pi x}{4}\right)}^n+1},\text{ if }x\in (0,1)\cup (1,2)\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\text{ if }x=1\end{array}\text{ . If }f\right.\text{ is differentiable }$$\text{ at }x=1\text{ , then the value of }|a+p|\text{ is }$

$\begin{array}{l}\text{ as }n\to \mathrm{\infty }\\ \text{ If }x>1\text{ then }{x}^n\to \mathrm{\infty }\\ \text{ If }0<x<1\text{ then }{x}^n\to 0\\ \text{ Now, }f\left(1^{+}\right)=\frac{P{x}^2+2}{1}=0\\ \Rightarrow P=-2\\ \text{ And }f\left(1^{-}\right)=ax(x-1)=0\\ L\cdot H\cdot D=\underset{h\to 0}{lim} \frac{f(1-h)-f\left(1\right)}{-h}\\ =\underset{h\to 0}{lim} \frac{a(1-h)(-h)}{-h}=a\end{array}$

$\begin{array}{l}R\cdot H\cdot D=\underset{h\to 0}{\lim }\frac{f(1+h)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{P(1+h{)}^2+2}{h}\\ =\underset{h\to 0}{\lim }\frac{-2\left(1+2h+h^2\right)+2}{h}=-4\\ \therefore a=-4\\ \therefore |a+p|=|-4-2|=6\end{array}$