A) \[1<x<3\]
B) \[1<x<3\]but \[x\ne 0\]
C) \[0<x,2\]
D) \[-1<x<3\]
Correct Answer: B
Solution :
(b): \[(x-1)(3-x){{(x-2)}^{2}}>0\] \[\Rightarrow (x-1)(3-x)>0(\therefore {{(x-2)}^{2}}\] is always >0, it being perfect square for \[x\ne 0\]) \[\Rightarrow (x-1)(x-3)<0\] and proceed to get the answer.You need to login to perform this action.
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