A) \[2\]
B) \[3\sqrt{2}\]
C) \[2\sqrt{2}\]
D) \[1\]
Correct Answer: A
Solution :
| Given, \[\left| {{Z}^{3}}+\frac{1}{{{Z}^{3}}} \right|\le 2\] |
| \[\left| {{Z}^{3}}+\frac{1}{{{Z}^{3}}} \right|\le \left| {{Z}^{3}} \right|+\frac{1}{\left| {{Z}^{3}} \right|}\le 2\] |
| By \[AM\ge GM\] |
| \[\therefore \] \[\left| {{Z}^{3}} \right|+\frac{1}{\left| {{Z}^{3}} \right|}=2\,\,\left| Z \right|=1\] |
| \[\therefore \] \[\left| Z+\frac{1}{Z} \right|\le \left| Z \right|+\frac{1}{\left| Z \right|}=2\] |
| \[\therefore \] Maximum value of \[\left| Z+\frac{1}{Z} \right|=2\] |
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