Which of the following is/are correct

A) $\text{Let }g\left(x\right)=f\left(f\right(x\left)\right)\Rightarrow g^{\text{'}}\left(x\right)=f^{\text{'}}\left(f\left(x\right)\right).f^{\text{'}}\left(x\right)$

$$\begin{gathered} f^{\text{'}}\left(x\right)=0\Rightarrow x=1,2\&f^{\text{'}}\left(f\left(x\right)\right)=0 \\ If f\left(x\right)=1\Rightarrow f^{\text{'}}\left(f\left(x\right)\right)=f^{\text{'}}\left(1\right)=0 \\ f\left(x\right)=2\Rightarrow f^{\text{'}}\left(f\left(x\right)\right)=f^{\text{'}}\left(2\right)=0 \end{gathered}$$

$\therefore$the number of values of  x  for which  $y=f\left(f\right(x\left)\right)$ attains local  extremum is 4

${\int }_0^{\frac{(\alpha +{\tan }^{-1})}{2}}\left[\tan x\right]dx=\frac{\pi }{2}-{\tan }^{-1}2$

B) $\begin{array}{l}\text{ Area }=\left(2\alpha \right)\sin (\frac{\pi }{2}+\alpha )=2\alpha \cos \alpha \\ \frac{dA}{d\alpha }=2(-\alpha \sin \alpha +\cos \alpha )=2\sin \alpha (-\alpha +\cot \alpha \\ \frac{dA}{d\alpha }=0\Rightarrow \alpha =\cot \alpha \{\because \alpha \in (0,\frac{\pi }{2})\}\end{array}$

C) Solving given equations, we get $a(x^3-6x^2+12x-8)=x^3-12x+16\Rightarrow a{(x-2)}^3={(x-2)}^2(x+4)$

$\therefore x=2$ is a repeated root and hence the two curves touches each other at $x=2\forall a\in R$

D) $f^{\text{'}}\left(x\right)=k(x-1)(x-3){(x-2)}^2\text{ and }k>0\Rightarrow f\left(x\right)=k\int (x^2-4x+3)(x^2-4x+4)dx$

$=k\int x^4+16x^2-8x^3+7(x^2-4x)+12dx=k(\frac{x^5}{5}-2x^4+\frac{23x^3}{3}-14x^2+12x)+c$

$\because f\left(0\right)=2\Rightarrow c=2\therefore f\left(1\right)=\frac{88}{15}\Rightarrow \frac{88}{15}=k(\frac{1}{5}-2+\frac{23}{3}-14+12)+c$

$\Rightarrow \frac{58}{15}=k\times \frac{58}{15}\Rightarrow k=1\therefore a_5=\frac{1}{5}$