A) \[(3,+\infty )\]
B) \[(-1,\,\,1)\cup (3,\,+\infty )\]
C) \[[-1,\,\,1]\cup (3,+\infty )\]
D) None of these
Correct Answer: B
Solution :
\[\frac{{{x}^{2}}-3x+4}{x+1}>1\Rightarrow \frac{{{x}^{2}}-3x+4}{x+1}-1>0\] \[\Rightarrow \,\,\frac{{{x}^{2}}-4x+3}{x+1}>0\Rightarrow \,\,\frac{(x+1)(x-1)(x-3)}{{{\left( x+1 \right)}^{2}}}>0\] \[\Rightarrow \,\,\,(x+1)(x-1)(x-3)>0\,\,and\,\,x\ne -1\] Using method of interval, we get, \[\operatorname{x}\in \left( -1,\,\,1 \right)\cup \left( 3,\,\,\infty \right))\]
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