Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given $f\left(x\right)=\left\{\begin{array}{ll}x+a;& x\le 0\\ \left|x-4\right|;& x>0\end{array}\right.$and $g\left(x\right)=\left\{\begin{array}{ll}x+1;& x<0\\ {\left(x-4\right)}^2+b;& x\ge 0\end{array}\right.$

Given $f\left(x\right) \& g\left(x\right)$ are continuous on $R$

So, for continuity

$f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right)$

$\Rightarrow a=\left|0-4\right|$

$\Rightarrow a=4$

And $g\left(0^{-}\right)=g\left(0\right)=g\left(0^{+}\right)$

$\Rightarrow 1={\left(0-4\right)}^2+b$

$\Rightarrow b=1-16=-15$

Then, $g\left(f\left(2\right)\right)+f\left(g\left(-2\right)\right)$

$=g\left(2\right)+f\left(-1\right)$

$={\left(2-4\right)}^2-15+\left(-1+4\right)$

$=4-15+3=-8$