Let $f\left(x\right)=\left\{\begin{array}{c}|x-3|,\text{ if }x<-1\\ 3x+4,\text{ if }x\ge -1\end{array},g\left(x\right)=x^2-bx-2\right.$ 

$x \in R$ and $b$ is a real constant. If $gof$ is continuous at $x =-1$, then $b$ is equal to 

Clearly ,

$gof\left(x\right)=\left\{\begin{array}{l}|x-3{|}^2-b|x-3|-2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x<-1\\ (3x+4{)}^2-b(3x+4)-2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\ge -1\end{array}\right.$

It is given that$gof\left(x\right)$  is continuous at $x = -1$. 

$\begin{array}{r}\therefore \underset{x\to -1^{-}}{lim} g\circ f\left(x\right)=\underset{x\to -1^{+}}{lim} gof\left(x\right)\\ \Rightarrow \underset{x\to -1}{lim} \left\{|x-3{|}^2-b|x-3|-2\right\}\end{array}$

$=\underset{x\to -1}{lim} \left\{(3x+4{)}^2-b(3x+4)-2\right\}$ 

$\Rightarrow 16-4b-2=1-b-2\Rightarrow 3b=15\Rightarrow b=5$