Suppose f(x) is a polynomial of degree four
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 :

  1. (1) 4

  2. (2) 6

  3. (3) 2

  4. (4) 8


Correct Option: 1

Solution:

$\because$ The critical points are $-1,0,1$

$\therefore f^{\prime}(x)=k \cdot x(x+1)(x-1)=k\left(x^{3}-x\right)$

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

$\Rightarrow f(0)=C$

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

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

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

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

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

 

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