Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2-3,& x\le -5\\ x+\lambda ,& -5<x<-1\\ (\mu -7)(|1-x|+|1+x|),& -1\le x\le 1\\ x+6,& 1<x<5\\ 3-x^2,& x\ge 5\end{array}$
If $f\left(x\right)$ is an odd function, then the value of $\lambda +\mu$ is

Answer 1

$f\left(x\right)=\{\begin{array}{cc}{x}^2-3,& x\le -5\\ x+\lambda ,& -5<x<-1\\ (\mu -7)(|1-x|+|1+x|),& -1\le x\le 1\\ x+6,& 1<x<5\\ 3-x^2,& x\ge 5\end{array}$

$f(-x)=\{\begin{array}{cc}{x}^2-3,& -x\le -5\Rightarrow x\ge 5\\ -x+\lambda ,& -5<-x<-1\Rightarrow 1<x<5\\ (\mu -7)(|1+x|+|1-x|),& -1\le x\le 1\\ -x+6,& -5<x<-1\\ 3-x^2,& x\le -5\end{array}$

$f\left(x\right)+f(-x)=\{\begin{array}{cc}0,& x\le -5\\ \lambda +6,& -5<x<-1\\ 2(\mu -7)(|1-x|+|1+x|),& -1\le x\le 1\\ \lambda +6,& 1<x<5\\ 0,& x\ge 5\end{array}$

For $f\left(x\right)$ to be an odd function,
$f\left(x\right)+f(–x)=0\forall x\in R$
$\therefore \lambda =-6$ and $\mu =7$
$\Rightarrow \lambda +\mu =1$