$$\begin{gathered} Let f\left(x\right)=\{\begin{array}{l}\max .\{t^3-t^2+t+1.0\le t\le x\},0\le x\le 1\\ \min .\{3-t,1<t\le x\}1<x\le 2\end{array} and \\ g\left(x\right)=\{\begin{array}{l}\max .\{\frac{3}{8}{t}^4+\frac{1}{2}{t}^3-\frac{3}{2}{t}^2+1,0\le t\le x\},0\le x<1\\ \min .\{\frac{3}{8}t+\frac{1}{32}{\sin }^2\pi t+\frac{5}{8},1\le t\le x\}1\le x\le 2\end{array} \end{gathered}$$

Which of the following options are does not holds good with respect to the given functions  $f\left(x\right)$  and  $g\left(x\right)$

consider f(x) in [0,1]
$f^1\left(t\right)=3t^2-2t+1>0\forall t\in (0,1)\Rightarrow fis\uparrow in(0,1)$
Maximum occurs at  $t=x.$
And hence  $f\left(x\right)is\{\begin{array}{l}{x}^3-x^2+x+10\le x\le 1\\ 3-x1<x\le 2\end{array}$
Again consider   $g\left(x\right)=in[0,1)$
$$\begin{gathered} g\left(t\right)=\frac{3}{8}{t}^4+\frac{1}{2}{t}^3-\frac{3}{2}{t}^2+1 \\ {g}^1\left(t\right)=\frac{3}{2}{t}^3+\frac{3}{-}{t}^2-3t=\frac{3}{2}t(t^2+t-2)=\frac{3}{2}t(t-1)(t+2) \end{gathered}$$
g(t) decreases in [0,1)
maximum occurs when t=0 and g(0)=1
again consider g(x) function in [1,2]
$$\begin{gathered} g\left(t\right)=\frac{3}{8}t+\frac{1}{32}su{n}^2\pi t+\frac{5}{8} \\ {g}^1\left(t\right)=\frac{3}{8}+\frac{\pi }{32}\sin \left(2\pi t\right)>0\forall t\in R \end{gathered}$$
G is an increasing function in [1,2]
$\Rightarrow$ minimum occurs when t=1   And g(1)=1
hence  $g\left(x\right)=\{\begin{array}{l}10\le x<1\\ 11<x\le 2\end{array}=\forall x\in [0,2]$
f(x) is continuous but not differentiable at x=1 (B)
$$\begin{gathered} \underset{x\to x^{-}}{\lim }fog\left(x\right)=f\left(1\right)and\underset{x\to x^{+}}{\lim }gof\left(x\right)=1alsof\left(1\right)>1\left(A\right) \\ Z\left(x\right)=\frac{d}{dx}f{\left(x\right)}^{g\left(x\right)}=\frac{d}{dx}f{\left(x\right)}^1\forall x\in [0,1)\cup (1,2] \\ {f}^1\left(x\right)\forall x\in [0,1)\cup (1,2] \\ \&Y\left(x\right)=\frac{d}{dx}g{\left(x\right)}^{f\left(x\right)}=\frac{d}{dx}{1}^{f\left(x\right)}=0\forall x\in [0,1)\cup (1,2] \end{gathered}$$
Hence the function Y(x)and Z(x) can vanish simultaneously at   which is not possible for any real X.