Question

$f\left(x\right)=\{\begin{array}{cc}2x^2-5,& x\le -8\\ 3x+p,& -8<x<-3\\ (k-7)(|3-x|+|3+x|),& -3\le x\le 3\\ 3x+9,& 3<x<8\\ 5-2x^2,& x\ge 8\end{array}$

If $f\left(x\right)$ is an odd function then the value of $k-p$ is

Answer 1

$f\left(x\right)=\{\begin{array}{cc}2x^2-5,& x\le -8\\ 3x+p,& -8<x<-3\\ (k-7)(|3-x|+|3+x|),& -3\le x\le 3\\ 3x+9,& 3<x<8\\ 5-2x^2,& x\ge 8\end{array}$

$f(-x)=\{\begin{array}{cc}2x^2-5,& -x\le -8\Rightarrow x\ge 8\\ p-3x,& -8<-x<-3\Rightarrow 3<x<8\\ (k-7)(|3+x|+|3-x|),& -3\le -x\le 3\\ 9-3x,& 3<-x<8\Rightarrow -8<x<-3\\ 5-2x^2,& -x\ge 8\Rightarrow x\le -8\end{array}$

$\Rightarrow f(-x)=\{\begin{array}{cc}-(2x^2-5),& x\le -8\\ -3x+9,& -8<x<-3\\ (k-7)(|3+x|+|3-x|),& -3\le -x\le 3\\ -3x+p,& 3<x<8\\ -(5-2x^2),& x\ge 8\end{array}$

$f\left(x\right)+f(-x)=\{\begin{array}{cc}0,& x\le -8\\ 9+p,& -8<x<-3\\ 2(k-7)(|3-x|+|3+x|),& -3\le x\le 3\\ p+9,& 3<x<8\\ 0,& x\ge 8\end{array}$

For a function to be odd $f\left(x\right)+f(-x)=0$
$\Rightarrow p+9=0\Rightarrow p=-9$
Also $k-7=0\Rightarrow k=7$
$\therefore k-p=16$