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 :
Solution
Critical points = $-$1, 0, 1.<br><br>$\therefore$ f'(x) = a(x $-$ 1)(x + 1)x<br><br>$\therefore$ f(x) = a$\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C$<br><br>$\because$ f(x) = f(0)<br><br>$- a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C$<br><br>$a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0$<br><br>$\therefore$ x = 0, $\sqrt 2$, $- \sqrt 2$<br><br>$\therefore$ T = $\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}$<br><br>Sum of square of elements of $T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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