If non-zero real numbers b and c are such that $$\min f(x) > \max g(x)$$, where $$f(x) = x^2 + 2bx + 2c^2$$ and $$g(x) = -x^2 - 2cx + b^2$$, $$(x \in R)$$; then $$\left|\frac{c}{b}\right|$$ lies in the interval:

We are given two quadratic functions: $$ f(x) = x^2 + 2bx + 2c^2 $$ and $$ g(x) = -x^2 - 2cx + b^2 $$, where $$ b $$ and $$ c $$ are non-zero real numbers. The condition is that the minimum value of $$ f(x) $$ is greater than the maximum value of $$ g(x) $$, written as $$ \min f(x) > \max g(x) $$. We need to find the interval for $$ \left| \frac{c}{b} \right| $$.

First, recall that for a quadratic function $$ ax^2 + bx + c $$, the vertex occurs at $$ x = -\frac{b}{2a} $$. If $$ a > 0 $$, the parabola opens upwards and the vertex is the minimum point. If $$ a < 0 $$, it opens downwards and the vertex is the maximum point.

For $$ f(x) = x^2 + 2bx + 2c^2 $$, the coefficient of $$ x^2 $$ is $$ a = 1 $$, which is positive. So, it has a minimum value at the vertex. The x-coordinate of the vertex is:

$$ x = -\frac{2b}{2 \cdot 1} = -\frac{2b}{2} = -b $$

Substitute $$ x = -b $$ into $$ f(x) $$ to find the minimum value:

$$ f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = -b^2 + 2c^2 $$

So, $$ \min f(x) = -b^2 + 2c^2 $$.

For $$ g(x) = -x^2 - 2cx + b^2 $$, the coefficient of $$ x^2 $$ is $$ a = -1 $$, which is negative. So, it has a maximum value at the vertex. The x-coordinate of the vertex is:

$$ x = -\frac{-2c}{2 \cdot (-1)} = -\frac{-2c}{-2} = -\frac{2c}{-2} = -c $$

Note: The formula is $$ x = -\frac{B}{2A} $$ where $$ B $$ is the coefficient of $$ x $$ and $$ A $$ is the coefficient of $$ x^2 $$. Here, $$ A = -1 $$ and $$ B = -2c $$, so:

$$ x = -\frac{(-2c)}{2 \cdot (-1)} = -\frac{-2c}{-2} = -\frac{2c}{-2} = -c $$

Substitute $$ x = -c $$ into $$ g(x) $$:

$$ g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2 $$

So, $$ \max g(x) = b^2 + c^2 $$.

The condition $$ \min f(x) > \max g(x) $$ gives:

$$ -b^2 + 2c^2 > b^2 + c^2 $$

Bring all terms to one side:

$$ -b^2 + 2c^2 - b^2 - c^2 > 0 $$

$$ -2b^2 + c^2 > 0 $$

Rearrange:

$$ c^2 - 2b^2 > 0 $$

$$ c^2 > 2b^2 $$

Since $$ b $$ and $$ c $$ are non-zero real numbers, $$ b^2 > 0 $$. Divide both sides by $$ b^2 $$:

$$ \frac{c^2}{b^2} > 2 $$

$$ \left( \frac{c}{b} \right)^2 > 2 $$

Taking square roots on both sides (and noting that the square root function is increasing and $$ \left| \frac{c}{b} \right| $$ is non-negative):

$$ \left| \frac{c}{b} \right| > \sqrt{2} $$

Thus, $$ \left| \frac{c}{b} \right| $$ lies in the interval $$ (\sqrt{2}, \infty) $$.

Now, comparing with the options:

A. $$ (\sqrt{2}, \infty) $$

B. $$ \left[ \frac{1}{2}, \frac{1}{\sqrt{2}} \right) $$

C. $$ \left( 0, \frac{1}{2} \right) $$

D. $$ \left[ \frac{1}{\sqrt{2}}, \sqrt{2} \right] $$

Hence, the correct answer is Option A.