If f(x) is a non-zero polynomial of degree fo | vedclass

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(C) Since $f(x)$ is a polynomial of degree $4$ and has local extreme points at $x = -1, 0, 1$,its derivative $f'(x)$ must be a polynomial of degree $3

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two irrational and one rational number

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(C) Since $f(x)$ is a polynomial of degree $4$ and has local extreme points at $x = -1, 0, 1$,its derivative $f'(x)$ must be a polynomial of degree $3$ with roots at $-1, 0, 1$.Thus,$f'(x) = \lambda(x + 1)(x)(x - 1) = \lambda(x^3 - x)$,where $\lambda 
eq 0$.Integrating $f'(x)$,we get $f(x) = \lambda(\frac{x^4}{4} - \frac{x^2}{2}) + \mu$,where $\mu$ is the constant of integration.We are given the set $S = \{x \in R; f(x) = f(0)\}$.Substituting $f(0) = \mu$ into the equation $f(x) = f(0)$,we get $\lambda(\frac{x^4}{4} - \frac{x^2}{2}) + \mu = \mu$.This simplifies to $\lambda(\frac{x^4}{4} - \frac{x^2}{2}) = 0$.Since $\lambda 
eq 0$,we have $\frac{x^4}{4} - \frac{x^2}{2} = 0$,which implies $x^2(\frac{x^2}{4} - \frac{1}{2}) = 0$.This gives $x^2 = 0$ or $x^2 = 2$.Thus,the roots are $x = 0, 0, \sqrt{2}, -\sqrt{2}$.The distinct elements in the set $S$ are $0, \sqrt{2}, -\sqrt{2}$.Here,$0$ is a rational number,and $\sqrt{2}, -\sqrt{2}$ are irrational numbers.Therefore,the set $S$ contains two irrational numbers and one rational number.

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

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