For a real number x,if the minimum value of f | vedclass

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(A) The function $f(x) = x^2 + 2bx + 2c^2$ is an upward-opening parabola. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^

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$c^2 > 2b^2$

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(A) The function $f(x) = x^2 + 2bx + 2c^2$ is an upward-opening parabola. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$. The function $g(x) = -x^2 - 2cx + b^2$ is a downward-opening parabola. Its maximum value occurs at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$. According to the problem,the minimum value of $f(x)$ is greater than the maximum value of $g(x)$: $2c^2 - b^2 > c^2 + b^2$ $2c^2 - c^2 > b^2 + b^2$ $c^2 > 2b^2$.

For a real number $x$,if the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$,then:

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