Question

Suppose $f\left(x\right)$ is a polynomial of degree four having critical points at $-1,0,1.$ If $T=\{x\in R\left|f\right(x)=f(0)\}$, then the sum of squares of all the elements of $T$ is:

• A. $6$
• B. $2$
• C. $8$
• D. $4$

Answer 1

The correct option is D $4$

$f^{\prime }\left(x\right)=k\cdot x(x+1)(x-1)$
$f^{\prime }\left(x\right)=k(x^3-x)$
Integrating both sides
$f\left(x\right)=k[\frac{x^4}{4}-\frac{x^2}{2}]+C$
$f\left(0\right)=C$
$f\left(x\right)=f\left(0\right)$
$\Rightarrow k(\frac{x^4}{4}-\frac{x^2}{2})+C=C$
$\Rightarrow k\frac{x^2}{4}(x^2-2)=0$
$\Rightarrow x=0,\pm \sqrt{2}$
$\therefore T=\{0,\sqrt{2},-\sqrt{2}\}$
Hence, sum of squares of all the elements of $T$ is $0+2+2=4$