If $m$ and $n$ are the least and greatest values of $|z|$ respectively and $|z-4+3 i| \leq 1$. Let $k$ be the least value of $\frac{x^4+x^2+4}{x}$ on the interval $(0, \infty)$. Then $k=$

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Aicharya · Accepted Answer

$\begin{aligned} |z-4+3 i| & \leq 1 \ |Z-(4-3 i)| & \leq 1 \end{aligned}$ It represent a circle of radius less than or equal to 1 unit Minimum value of $|z|=O P=O C-C P=5-1$ $m=4$ Maximum value of $|z|=O Q=O C+C Q=5+1$ $\begin{aligned} n & =6 \ \frac{x^4+x^2+4}{x} & =x^3+x+\frac{4}{x} \ \frac{x^4+x^2+4}{x} & =x^3+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x} \end{aligned}$ Since, $\mathrm{AM} \geq \mathrm{GM}$ $\begin{aligned} \frac{x^3+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}}{6} & \geq \sqrt[6]{x^3 \cdot x \cdot \frac{1}{x} \cdot \frac{1}{x} \cdot \frac{1}{x} \cdot \frac{1}{x}} \geq \sqrt[6]{1} \ x^3+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x} & \geq 6 \ \frac{x^4+x^2+4 x}{4} & \geq 6 \end{aligned}$ $\therefore$ Minimum Value of $\frac{x^4+x^2+x}{4}=6$ $\begin{array}{rr} k=6 & \text { [given] } \ k=n & {[\because 6=n]} \end{array}$ Hence, option (a) is correct.

If \(m\) and \(n\) are the least and greatest values of \(|z|\) respectively and \(|z-4+3 i| \leq 1\). Let…

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