A) \[1<m<3\]
B) \[-3<m<-2\]
C) \[m\in (-3,-2)\cup (1,2)\]
D) None of these
Correct Answer: C
Solution :
(c): \[\frac{{{m}^{2}}+m-6}{{{m}^{2}}+m-2}<0\Rightarrow \frac{(m+3)(m-2)}{(m-1)(m+2)}<0\] \[\Rightarrow \frac{(m+3)(m-2)(m-1)(m+2)}{{{(m-1)}^{2}}{{(m+2)}^{2}}}<0\] (multiplying numerator and denominator by (m - 1) (m+2) \[\Rightarrow (m+3)(m-2)(m-1)(m+2)<0\] \[\Rightarrow (m-2)(m-1)(m+2)(m+3)<0\] \[\Rightarrow m\in (-3,-2)\cup (1,2)\]You need to login to perform this action.
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