A) \[\alpha \]
B) \[\beta \]
C) \[\alpha \beta \]
D) None of these
Correct Answer: B
Solution :
[b]We have, |
\[\left| z-4+3i \right|\le 1\] |
\[\left| z-(4+3i) \right|\le 1\] |
\[-\,1+\left| 4-3i \right|\le 1+\left| 4-3i \right|\] |
\[-1+5\le \left| z \right|\le 1+5\] |
\[4\le \left| z \right|\le 6\] |
\[\because \]\[\alpha =4,\]\[\beta =6\] |
Let \[y=\frac{{{x}^{2}}+{{x}^{2}}+4}{x}\] |
\[\Rightarrow \]\[y={{x}^{3}}+x+\frac{4}{x}\] |
\[\Rightarrow \]\[y={{x}^{3}}+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\] |
Since, \[y\in (0,\infty ),\]therefore\[{{x}^{3}},\]\[x,\frac{1}{x}\] are positive. |
Sum will be least of |
\[\Rightarrow \] \[x=1\] |
\[\therefore \]Least value of y is 6 |
\[K=6\]\[\Rightarrow \]\[K=\beta \] |
You need to login to perform this action.
You will be redirected in
3 sec