we are given
![((n^4-10n^2+24))/((n^4-9n^2+18))](https://img.qammunity.org/2019/formulas/mathematics/college/fug0yu7cxw6ogeuri2pd58epezllnseb15.png)
Firstly, we will factor numerators and denominators
![((n^4-10n^2+24))/((n^4-9n^2+18))=((n^2-6)(n^2-4))/((n^2-3)(n^2-6))](https://img.qammunity.org/2019/formulas/mathematics/college/bf93bgyl2h5s8xq6p1t9evpup5lj8gzj9h.png)
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
![n^2-6\\eq 0](https://img.qammunity.org/2019/formulas/mathematics/college/vtqzi94ay5ta196icdv0c26hprx4e3hve9.png)
![n\\eq -+ √(6)](https://img.qammunity.org/2019/formulas/mathematics/college/qaqa2oftuiin324jqf967v707vxwd2mhfj.png)
we can simplify it
![((n^4-10n^2+24))/((n^4-9n^2+18))=((n^2-4))/((n^2-3))](https://img.qammunity.org/2019/formulas/mathematics/college/uy5vxnrqwxjxvkpnggk12eqruf0msnmpa3.png)
we know that denominator can not be zero
![n^2-3\\eq 0](https://img.qammunity.org/2019/formulas/mathematics/college/6opx0blspiua060ym8bvo540i8jsuok9ge.png)
![n\\eq -+ √(3)](https://img.qammunity.org/2019/formulas/mathematics/college/yba1m8n9nhlwza5cvveka8gkvvj3ers1s1.png)
so, option-B.......Answer