If f ( x ) , defined below, is continuous at x = 4 , then f x = x 2 + a x + b ,   0 ≤ x < 2 3…

$f\left(x\right)=\left\{\begin{array}{l}{x}^2+ax+b, 0\le x<2\\ 3x+2, 2\le x\le 4\\ 2ax+5b, 4<x\le 8\end{array}\right.$

If function is continuous at $x=2$ and $x=4$, then

$L.H.L=f\left(x\right)=R.H.L$

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=f\left(2\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)$

$4+2a+b=8$

$\Rightarrow 2a+b=4.........\left(i\right)$

$\underset{x\to 4^{-}}{\lim }f\left(x\right)=f\left(4\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)$

$14=8a+5b.........\left(ii\right)$

Solve $\left(i\right) \& \left(ii\right)$ by elimination,we get

$a=11,b=-18$