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.

 

Answers (1)
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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