Question

If $|z-4+3i|\le 1$ and $m$ and $n$ are the least and greatest values of $\left|z\right|$ and $k$ is the least value of $\frac{x^4+x^2+4}{x}$ on the interval $(0,\infty ),$ then $k$ is equal to

• A. $m$
• B. $n$
• C. $m+n$
• D. None of these

Answer 1

The correct option is B $n$

We have, $1\ge |z-(4-3i)|$

$\ge \{\begin{array}{c}\left|z\right|-|4-3i|\\ |4-3i|-\left|z\right|\end{array}\Rightarrow 1\ge \{\begin{array}{c}\left|z\right|-5\\ 5-\left|z\right|\end{array}$

$\Rightarrow \left|z\right|\le 6$ and $\left|z\right|\ge 4$

$\Rightarrow 4\le \left|z\right|\le 6\Rightarrow m=4,n=6$

Let $y=\frac{x^4+x^3+4}{x}=x^3+x+\frac{4}{x}$

$=x^3+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}$

Since $x\in (0,\infty ),$

$\therefore x^3,x,\frac{1}{x},\frac{1}{x},\frac{1}{x},\frac{1}{x}$ are all positive number whose product is $1.$

$\therefore$ their sum $y$ will be least when

$x^3=x=\frac{1}{x}\Rightarrow x=1$

$\therefore$ least value of $y=6\therefore k=6$

Hence, $k=n$