A) \[(-\infty ,\,1)\]
B) (2, 3)
C) \[(-\infty ,\,3)\]
D) \[(-\infty ,\,1)\cup (2,3)\]
Correct Answer: D
Solution :
[d] \[5x+2<3x+8\] \[\Rightarrow x<3\] (1) And \[\left| \frac{{{x}^{2}}+1}{x} \right|>2\] \[\Rightarrow \frac{x-2}{x-1}>0\] \[\Rightarrow x<1\]or \[x>2\] (2) From (1) and (2), we get \[x\in (-\infty ,1)\cup (2,3)\].
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