A) \[\alpha \]
B) \[\beta \]
C) \[\alpha +\beta \]
D) \[\alpha -\beta \]
Correct Answer: A
Solution :
\[\because \,\,|Z-(4-3i)|\le 1\] \[|Z{{|}_{\max }}=OC+AC=5+1=6=\alpha \] \[|Z{{|}_{\min }}=OC-BC=5-1=4=\beta \] Also, get \[y=\frac{{{x}^{4}}+{{x}^{2}}+4}{x}={{x}^{3}}+x+\frac{4}{x}\] \[={{x}^{3}}+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\] For \[x\in (0,\,\infty )\,AM\ge GM\] \[\frac{{{x}^{3}}+x+\frac{4}{x}}{6}\ge \sqrt{1}\] \[\Rightarrow \,\,{{x}^{3}}+x+\frac{4}{x}\ge 6=k=\alpha \]You need to login to perform this action.
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