If f(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:f(x)=f(0)}` contains exactly (a) four rational numbers (b) two irrational and two rational numbers (c) four irrational numbers (d) two irrational and one rational number

Options

A. four rationl numbers
B. two irrational and two rational numbers
C. four irrational numbers
D. two irrational and one rational number

Answer

D

Explanation

Theb non-zero four degree polynomial f(x) has extremum points at `x=-1,0,1` so we can assume f`(x)=a(x+1)(x-0)=ax(x^(2)-1)`
where , a is non-zero constant
`f'(x)=ax^(3)-ax`
`rArr f(x)=(a)/(4)x^(4)-(a)/(2)x^(2)+C` [ intergrating both sides]
where, C si constant of integration.
Now, since, f(x)=f(0)
`rAr (a)/(4)x^(4)-(a)/(2)x^(2)+C=CrArr (x^(4))/(4)+(x^(2))/(2)`
`rArr x^(2)(x^(2)-2)=0 rArr x=-sqrt(2)0,sqrt(2)`
Thus, f(x)=f(0) has one rational and two irrational roots.