we are given
[tex] \frac{(n^4-10n^2+24)}{(n^4-9n^2+18)} [/tex]
Firstly, we will factor numerators and denominators
[tex] \frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)} [/tex]
we can see that
n^2 -6 is factor on both numerator and denominator
so, it will get cancelled
and n^2 -6 can not be equal to 0
so, one of restriction is
[tex] n^2-6\neq 0 [/tex]
[tex] n\neq -+ \sqrt{6} [/tex]
we can simplify it
[tex] \frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)} [/tex]
we know that denominator can not be zero
[tex] n^2-3\neq 0 [/tex]
[tex] n\neq -+ \sqrt{3} [/tex]
so, option-B.......Answer
