Question:
Suppose $f(x)$ is a polynomial of degree four, having critical points at $-1,0,1$. If $T=\{x \in \mathbf{R} \mid f(x)=f(0)\}$, then the sum of squares of all the elements of $T$ is :
Correct Option: 1
Solution:
$\because$ The critical points are $-1,0,1$
$\therefore f^{\prime}(x)=k \cdot x(x+1)(x-1)=k\left(x^{3}-x\right)$
$\Rightarrow f(x)=k\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right)+C$\
$\Rightarrow f(0)=C$
$\because f(x)=f(0)$
$\Rightarrow k \frac{\left(x^{4}-2 x^{2}\right)}{4}+C=C$
$\Rightarrow x^{2}\left(x^{2}-2\right)=0$
$\Rightarrow x=0, \sqrt{2},-\sqrt{2}$
$\Rightarrow T=\{0, \sqrt{2},-\sqrt{2}\}$