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

T = {x $$ \in $$ R |
f(x) = f(0)}, then the sum of squares of all the
elements of T is :

Accepted Answer

Critical points = $-$1, 0, 1.

$$ \therefore $$ f'(x) = a(x $-$ 1)(x + 1)x

$$ \therefore $$ f(x) = a$$\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C$$

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

$$ - a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C$$

$$a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0$$

$$ \therefore $$ x = 0, $$\sqrt 2 $$, $$ - \sqrt 2 $$

$$ \therefore $$ T = $$\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}$$

Sum of square of elements of $$T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$$

Solutions · 1