JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 1)
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 :
T = {x $$ \in $$ R | f(x) = f(0)}, then the sum of squares of all the elements of T is :
6
2
8
4
Explanation
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 $$T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$$
$$ \therefore $$ 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 $$T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$$
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