Official solution
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
f(x) = f(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.