Question:
If $\mathrm{f}(\mathrm{x})$ is a non-zero polynomial of degree four, having local extreme points at $x=-1,0,1$; then the set $S=\{x \in R: f(x)=f(0)\}$ contains exactly:
Correct Option: 4,
Solution:
Since, function $\mathrm{f}(\mathrm{x})$ have local extreem points at $\mathrm{x}=$
$-1,0,1$. Then
$f(x)=K(x+1) x(x-1)$
$=\mathrm{K}\left(\mathrm{x}^{3}-\mathrm{x}\right)$
$\Rightarrow \mathrm{f}(\mathrm{x})=K\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right)+C$ (using integration)
$\Rightarrow \mathrm{f}(0)=\mathrm{C}$
$\because \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \Rightarrow K\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right)=0$
$\Rightarrow \frac{x^{2}}{2}\left(\frac{x^{2}}{2}-1\right)=0$
$\Rightarrow x=0,0, \sqrt{2},-\sqrt{2}$
$\therefore S=\{0,-\sqrt{2}, \sqrt{2}\}$