If `|z-4+3i| leq 1 and m and n` be the least and greatest values of `|z| and K` be the least value of `(x^4+x^2+4)/x` on the interval `(0,oo)`, then `K=`

Options

A. m
B. n
C. `m+n`
D. mn

Answer

To solve the given problem, we need to break it down into two parts:

1. Finding the least and greatest values of \( |z| \) given the inequality \( |z - 4 + 3i| \leq 1 \).
2. Finding the least value of the function \( \frac{x^4 + x^2 + 4}{x} \) on the interval \( (0, \infty) \).

### Step 1: Finding the least and greatest values of \( |z| \)

Given the inequality:
\[ |z - (4 - 3i)| \leq 1 \]
This represents a circle in the complex plane with center at \( (4, -3) \) and radius \( 1 \).

#### Finding the center and radius:
- Center \( C = (4, -3) \)
- Radius \( r = 1 \)

#### Finding the least value of \( |z| \):
The least value of \( |z| \) occurs at the point on the circle closest to the origin. We can find this by calculating the distance from the center to the origin and then subtracting the radius.

1. Distance from the center to the origin:
\[ d = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

2. Least value \( m \):
\[ m = d - r = 5 - 1 = 4 \]

#### Finding the greatest value of \( |z| \):
The greatest value of \( |z| \) occurs at the point on the circle farthest from the origin. This can be found by adding the radius to the distance from the center to the origin.

1. Greatest value \( n \):
\[ n = d + r = 5 + 1 = 6 \]

### Step 2: Finding the least value of \( K = \frac{x^4 + x^2 + 4}{x} \)

We can rewrite the function:
\[ f(x) = x^3 + x + \frac{4}{x} \]

#### Finding the derivative:
To find the critical points, we differentiate \( f(x) \):
\[ f'(x) = 3x^2 + 1 - \frac{4}{x^2} \]

Setting the derivative to zero:
\[ 3x^2 + 1 - \frac{4}{x^2} = 0 \]
Multiplying through by \( x^2 \) to eliminate the fraction:
\[ 3x^4 + x^2 - 4 = 0 \]

Let \( t = x^2 \):
\[ 3t^2 + t - 4 = 0 \]

#### Solving the quadratic:
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
\[ t = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6} \]
Calculating the roots:
1. \( t = 1 \) (valid since \( t = x^2 \geq 0 \))
2. \( t = -\frac{4}{3} \) (not valid)

Thus, \( x^2 = 1 \) gives \( x = 1 \) (since \( x > 0 \)).

#### Finding the second derivative:
To confirm if this is a minimum:
\[ f''(x) = 6x + \frac{8}{x^3} \]
Evaluating at \( x = 1 \):
\[ f''(1) = 6 \cdot 1 + \frac{8}{1^3} = 6 + 8 = 14 > 0 \]
This indicates a local minimum.

#### Finding the minimum value:
Now, substituting \( x = 1 \) back into the function:
\[ f(1) = 1^3 + 1 + \frac{4}{1} = 1 + 1 + 4 = 6 \]

### Final Result:
Thus, the least value \( K \) is:
\[ K = 6 \]