A) \[2\sqrt{3}\]
B) \[-2\sqrt{3}\]
C) does not exist
D) None of these
Correct Answer: C
Solution :
Given, \[P={{x}^{3}}-\frac{1}{{{x}^{3}}},\,\,Q=x-\frac{1}{x}\] \[\therefore \] \[\frac{P}{{{Q}^{2}}}=\frac{\left( x-\frac{1}{x} \right)\left( {{x}^{2}}+1+\frac{1}{{{x}^{2}}} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}}\] \[=\frac{{{\left( x-\frac{1}{x} \right)}^{2}}+3}{\left( x-\frac{1}{x} \right)}\] \[=\left( x-\frac{1}{x} \right)+\frac{3}{\left( x-\frac{1}{x} \right)}\] Clearly, the minimum does not exist.
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