Question:
Suppose $f(x)$ is a polynomial of degree four, having critical points at $-1,0,1$. If $\mathrm{T}=\{\mathrm{x} \in \mathrm{R} \mid \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\}$, then the sum of squares of all the elements of $\mathrm{T}$ is :
Correct Option: , 3
Solution:
$f^{\prime}(x)=x(x+1)(x-1)=x^{3}-x$
$\int d f(x)=\int x^{3}-x d x$
$\mathrm{f}(\mathrm{x})=\mathrm{f}(0)$
$f(x)=f(0)$
$\frac{x^{4}}{4}-\frac{x^{2}}{2}=0$
$x^{2}\left(x^{2}-2\right)=0$
$x=0,0, \sqrt{2},-\sqrt{2}$
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=0+2+2=4$