Question

$\mathrm{If}f\left(x\right)=(\begin{array}{c}{x}^3+x^2,for0\le x\le 2\\ x+2,for2\le x\le 4\end{array}$
then the odd extension of f(x) would be -

• A. $f\left(x\right)=(\begin{array}{c}{x}^3+x^2,for-2\le x\le 0\\ x+2,for-4\le x\le -2\end{array}$
• B. $f\left(x\right)=(\begin{array}{c}-x^3+x^2,for-2\le x\le 0\\ -x+2,for-4\le x\le -2\end{array}$
• C.
  $f\left(x\right)=(\begin{array}{c}{x}^3-x^2,for-2\le x\le 0\\ x-2,for-4\le x\le -2\end{array}$
• D. $\text{None of these}$
  

Answer 1

The correct option is C

$f\left(x\right)=(\begin{array}{c}{x}^3-x^2,for-2\le x\le 0\\ x-2,for-4\le x\le -2\end{array}$

Given,

$x^3+x^2\text{for}0\le x\le 2$

$f\left(x\right)=x+2\text{for}2<x\le 4$

The function is defined only on positive side of X- axis. Now in order to extend the function we'll have to extend the domain on negative side as well.

To make it an odd function we'll replace x by –x and multiply the whole expression by $(-1)$. So we'll have
$(-1)(-x^3+x^2)\text{for}-2\le x\le 0$

$f\left(x\right)=(-1)(-x+2)\text{for}-4\le x<-2$

Or the complete function defined would be
$x^3+x^2\text{for}0\le x\le 2$

$f\left(x\right)=x+2\text{for}2<x\le 4$

$x^3-x^2\text{for}-2\le x\le 0$

$f\left(x\right)=x-2\text{for}-4\le x<-2$