# If $f(x)$ is non-zero polynomial of degree four,having local extreme points at $x=-1,0,1;$ then the set    $S=\left \{ x\; \epsilon\; R:f(x)=f(0) \right \}$contains exactly : Option 1) four irrational numbers. Option 2) four rational numbers. Option 3) two irrational and two rational numbers. Option 4) two irrational and one rational number.

S solutionqc

$f(x)$ has exterme points at $\left \{ -1,0,1 \right \}$

So, ${f}'\left ( o \right )=0$ at $-1,0,1$

we can write

${f}'\left ( x \right )=ax\left ( x^{2}-1 \right )$

$f\left ( x \right )=\int ax\left ( x^{2}-1 \right )dx=\frac{ax^{4}}{4}-\frac{ax^{2}}{2}+c$

given that

$f\left ( x \right )=f\left ( 0 \right )$

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

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

$\Rightarrow x^{2}\left ( x^{2}-2 \right )=0$

$x=0,\pm \sqrt{2}$

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

Option 1)

four irrational numbers.

Option 2)

four rational numbers.

Option 3)

two irrational and two rational numbers.

Option 4)

two irrational and one rational number.

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