Suppose f x is a polynomial of degree four having critical points at - 1 ,   0 ,   1 . If T = x…

$f^{\text{'}}\left(x\right)=k.x\left(x+1\right)\left(x-1\right)$

$=k\left(x^3-x\right)$

Integrating both sides with respect to $x$, we get

$f\left(x\right)=k \left(\frac{x^4}{4}-\frac{x^2}{2}\right)+c$

$\Rightarrow f\left(0\right)=c$

$\because f\left(x\right)=f\left(0\right)$

$\Rightarrow k\frac{\left(x^4-2x^2\right)}{4}+c=c$

$\Rightarrow x^2\left(x^2-2\right)=0$

$\Rightarrow x=0, \sqrt{2}, -\sqrt{2}$

$\Rightarrow T=\left\{0, \sqrt{2}, -\sqrt{2}\right\}$

Thus, sum of squares of all the elements of $T$ is ${\left(0\right)}^2+{\left(\sqrt{2}\right)}^2+{\left(-\sqrt{2}\right)}^2=4$.