Let f(x) = x^2 + 2bx + 2c^2 and g(x) = -x^2 - | vedclass

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(C) The minimum value of $f(x) = x^2 + 2bx + 2c^2$ is found at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.The

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

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(C) The minimum value of $f(x) = x^2 + 2bx + 2c^2$ is found at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.The maximum value of $g(x) = -x^2 - 2cx + b^2$ is found at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.Given $\min f(x) > \max g(x)$,we have $2c^2 - b^2 > c^2 + b^2$.This simplifies to $c^2 > 2b^2$,or $\frac{c^2}{b^2} > 2$.Taking the square root of both sides,we get $\left|\frac{c}{b}\right| > \sqrt{2}$.Thus,$\left|\frac{c}{b}\right| \in (\sqrt{2}, \infty)$.

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

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