Suppose f (x ) is a polynomial of degree four, having critical points at -1, 0, 1 If T = x ∈ RI f ( x )= f (0) then the sum of squares of all the elements of T is:

Question

Suppose f (x ) is a polynomial of degree four, having critical points at -1, 0, 1 If T = x ∈ RI f ( x )= f (0) then the sum of squares of all the elements of T is: (A) 6 (B) 8 (C) 4 (D) 2

Tardigrade Admin · Accepted Answer

f prime(x)=x(x+1)(x-1)=x3-x ∫ d f(x)=∫ x3-x d x f(x)=(x4/4)-(x2/2)+C f(x)=f(0) (x4/4)-(x2/2)=0 x2(x2-2)=0 x=0,0, √2,-√2 x12+x22+x32=0+2+2=4

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

6

8

4

2

$f^{'} (x) = x (x + 1) (x - 1) = x^{3} - x$
$\int df (x) = \int x^{3} - x d x$
$f (x) = \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + C$
$f (x) = f (0)$
$\frac{x ^{4}}{4} - \frac{x ^{2}}{2} = 0$
$x^{2} (x^{2} - 2) = 0$
$x = 0, 0, 2, - 2$
$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 0 + 2 + 2 = 4$

Q. Suppose f(x) is a polynomial of degree four, having critical points at −1,0,1 If T={x∈RIf(x)=f(0)}, then the sum of squares of all the elements of T is:

6

8

4

2

f′(x)=x(x+1)(x−1)=x3−x ∫df(x)=∫x3−xdx f(x)=4x4​−2x2​+C f(x)=f(0) 4x4​−2x2​=0 x2(x2−2)=0 x=0,0,2​,−2​ x12​+x22​+x32​=0+2+2=4

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

$f^{'} (x) = x (x + 1) (x - 1) = x^{3} - x$
$\int df (x) = \int x^{3} - x d x$
$f (x) = \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + C$
$f (x) = f (0)$
$\frac{x ^{4}}{4} - \frac{x ^{2}}{2} = 0$
$x^{2} (x^{2} - 2) = 0$
$x = 0, 0, 2, - 2$
$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 0 + 2 + 2 = 4$