$f\left(x\right)=\left\{\begin{array}{c}ax+b if x\le 1\\ a{x}^2+c if 1<x\le 2\\ \frac{d{x}^2+1}{x} if x\ge 2\end{array}\right.$ differentiable on R then ad-bc = 

$f\left(1^{-}\right)=f\left(1^{.}\right)$

$\begin{array}{c}Lt\\ x\to 1^{-}\end{array}ax+b=\begin{array}{c}Lt\\ x\to 1^{+}\end{array}a{x}^2+c$

$a+b=a+c\Rightarrow b=c$

$\begin{array}{c}Lt\\ x\to 2^{-}\end{array}f\left(x\right)=\begin{array}{c}Lt\\ x\to 2^{+}\end{array}f\left(x\right)$

$\begin{array}{c}Lt\\ x\to 2^{-}\end{array}a{x}^2+c=\begin{array}{c}Lt\\ x\to 2^{+}\end{array}\frac{d{x}^2+1}{x}$

$4a+c=\frac{4d+1}{2}$

$\boxed{8a+2c=4d+1}$

$f^1\left(x\right)=\left\{\begin{array}{cc}a& x\le 1\\ 2ax& 1<x\le 2\\ d-\frac{1}{x^2}& n\ge 2\end{array}\right.$

$f^1\left(1^{-}\right)=f^1\left(1^{+}\right)$

$\begin{array}{c}Lt\\ x\to 1^{-}\end{array}a=\begin{array}{c}Lt\\ x\to 1^{+}\end{array}2ax$

$a=2a\Rightarrow a=0$

$f^1\left(2^{-}\right)=f^1\left(2^{+}\right)$

$\begin{array}{c}Lt\\ x\to 2^{-}\end{array}2ax=\begin{array}{c}Lt\\ x\to 2^{+}\end{array}d-\frac{1}{x^2}$

$4a=d-\frac{1}{4}$     $O=d-\frac{1}{4}$

$d=\frac{1}{4}$

$\therefore b=c,a=0,d=\frac{1}{4}$

$\Rightarrow 8a+2c=4d+1$

$0+2c=1+1=2$

$\boxed{c=1}$

$\therefore a=0,b=1=c,d=\frac{1}{4}$

$ad-bc=0\left(\frac{1}{4}\right)-1\left(1\right)=-1$