Suppose f(x) is a polynomial of degree four, having critical points at -1,0,1.

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 :

  1. 6

  2. 8

  3. 4

  4. 2


Correct Option: , 3

Solution:

$f^{\prime}(x)=x(x+1)(x-1)=x^{3}-x$

$\int d f(x)=\int x^{3}-x d x$

$\mathrm{f}(\mathrm{x})=\mathrm{f}(0)$

$f(x)=f(0)$

$\frac{x^{4}}{4}-\frac{x^{2}}{2}=0$

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

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

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=0+2+2=4$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now