A) \[\alpha \]
B) \[\beta \]
C) \[\alpha +\beta \]
D) None of these
Correct Answer: B
Solution :
Given, |
\[\Rightarrow \left| \,z\, \right|\le 6\]and \[\left| \,z\, \right|\ge 4\]\[\Rightarrow 4\le \left| \,z\, \right|\le 6\]\[\Rightarrow \alpha =4,\beta =6\] |
Let \[y=\frac{{{x}^{4}}+{{x}^{2}}+4}{x}={{x}^{3}}+x+\frac{4}{x}\]\[={{x}^{3}}+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\] |
Since \[x\in (0,\infty ),\] therefore \[{{x}^{3}},x,\frac{1}{x}\] are positive. |
Sum will be least when \[{{x}^{3}}=x=\frac{1}{x}\]\[\Rightarrow x=1\] |
\[\therefore k=6\] |
Hence, \[k=\beta \] |
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