A) \[(-\infty ,-5)\cup (5,\infty )\]
B) \[(-3,3)\]
C) \[(-\infty ,-5)\cup (-3,3)\cup (5,\infty )\]
D) None of the above
Correct Answer: C
Solution :
We have, \[\frac{1}{|x|-3}<\frac{1}{2}\] Clearly, \[\frac{1}{|x|-3}\]is not defined for \[|x|=3,\]i.e. Now, \[\frac{1}{|x|-3}<\frac{1}{2}\] \[\Rightarrow \] \[\frac{1}{|x|-3}-\frac{1}{2}<0\] \[\Rightarrow \] \[\frac{2-|x|+3}{|x|-3}<0\] \[\Rightarrow \] \[\frac{|x|-5}{|x|-3}>0\] \[\Rightarrow \] \[|x|<3\] or \[|x|>5\] \[\Rightarrow \] \[x\in (-3,3)\]or \[x\in (-\infty ,-5)\cup (5,\infty )\] \[\Rightarrow \] \[x\in (-\infty ,-5)\cup (-3,3)\cup (5,\infty )\]
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