JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 4)
If ƒ(x) is a non-zero polynomial of degree four,
having local extreme points at x = –1, 0, 1; then
the set
S = {x $$ \in $$ R : ƒ(x) = ƒ(0)}
Contains exactly :
S = {x $$ \in $$ R : ƒ(x) = ƒ(0)}
Contains exactly :
four rational numbers.
four irrational numbers.
two irrational and one rational number.
two irrational and two rational numbes.
Explanation
Local extreme points of f(x) is at x = –1, 0, 1.
$$ \therefore $$ f'(x) = 0 has three solutions x = –1, 0, 1.
$$ \therefore $$ f'(x) = k(x + 1)x(x - 1)
$$\int {f'(x)dx} = \int {k(x + 1)x(x - 1)dx} $$
f(x) = $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$
Also given that
ƒ(x) = ƒ(0)
$$ \therefore $$ $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$ = C
$$ \Rightarrow $$ $${{{x^2}} \over 2}\left( {{{{x^2}} \over 2} - 1} \right)$$ = 0
$$ \Rightarrow $$ x = 0, $$ - \sqrt 2 $$, $$ \sqrt 2 $$
$$ \therefore $$ x has two irrational and one rational value.
$$ \therefore $$ f'(x) = 0 has three solutions x = –1, 0, 1.
$$ \therefore $$ f'(x) = k(x + 1)x(x - 1)
$$\int {f'(x)dx} = \int {k(x + 1)x(x - 1)dx} $$
f(x) = $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$
Also given that
ƒ(x) = ƒ(0)
$$ \therefore $$ $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$ = C
$$ \Rightarrow $$ $${{{x^2}} \over 2}\left( {{{{x^2}} \over 2} - 1} \right)$$ = 0
$$ \Rightarrow $$ x = 0, $$ - \sqrt 2 $$, $$ \sqrt 2 $$
$$ \therefore $$ x has two irrational and one rational value.
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