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

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(D) Given $f(x) = x^2 + 2bx + 2c^2$. Completing the square,we get $f(x) = (x + b)^2 + 2c^2 - b^2$.Since the coefficient of $x^2$ is positive,the minim

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$|c| > |b| \sqrt{2}$

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(D) Given $f(x) = x^2 + 2bx + 2c^2$. Completing the square,we get $f(x) = (x + b)^2 + 2c^2 - b^2$.Since the coefficient of $x^2$ is positive,the minimum value of $f(x)$ is $2c^2 - b^2$.Given $g(x) = -x^2 - 2cx + b^2$. Completing the square,we get $g(x) = -(x^2 + 2cx) + b^2 = -(x + c)^2 + c^2 + b^2$.Since the coefficient of $x^2$ is negative,the maximum value of $g(x)$ is $c^2 + b^2$.According to the problem,$\min f(x) > \max g(x)$.Substituting the values,we get $2c^2 - b^2 > c^2 + b^2$.Rearranging the terms,we get $c^2 > 2b^2$.Taking the square root on both sides,we get $|c| > |b| \sqrt{2}$.

If $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ such that $\min f(x) > \max g(x)$,then the relation between $b$ and $c$ is:

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