Question

If $|z-4+3i|\le 1$ and $m$ and $n$ are the least and greatest values of $|z|$ 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

Answer: (B)

$n$

We have,

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

$\Rightarrow 1\ge \left|z\right|-|4-3i|,|4-3i|-\left|z\right|$

$\Rightarrow 1\ge \left|z\right|-5,5-\left|z\right|$

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

Let $y=\frac{4+x^2+x^4}{x}=x^3+x+\frac{4}{x}=x^3+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}$

$\because x\in (0,\infty )$, then $x^3,x,\frac{1}{x},\frac{1}{x},\frac{1}{x},\frac{1}{x}$ are all positive numbers whose product is 1.

Thus their sum y will be least when

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

So least value of $y=6,k=6$

So $k=n$