If non-zero real numbers b and c are such tha | vedclass

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(D) We have $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$.Completing the square for $f(x)$:$f(x) = (x + b)^2 + 2c^2 - b^2$.Thu

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

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(D) We have $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$.Completing the square for $f(x)$:$f(x) = (x + b)^2 + 2c^2 - b^2$.Thus,the minimum value of $f(x)$ is $f_{\min} = 2c^2 - b^2$.Completing the square for $g(x)$:$g(x) = -(x^2 + 2cx) + b^2 = -(x + c)^2 + c^2 + b^2$.Thus,the maximum value of $g(x)$ is $g_{\max} = c^2 + b^2$.Given the condition $\min f(x) > \max g(x)$:$2c^2 - b^2 > c^2 + b^2$.Subtracting $c^2$ and adding $b^2$ to both sides:$c^2 > 2b^2$.Since $b$ and $c$ are non-zero,we divide by $b^2$:$\frac{c^2}{b^2} > 2$.Taking the square root on both sides:$\left| \frac{c}{b} \right| > \sqrt{2}$.Therefore,$\left| \frac{c}{b} \right| \in (\sqrt{2}, \infty)$.

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 $g(x) = -x^2 - 2cx + b^2$ for $x \in R$; then $\left| \frac{c}{b} \right|$ lies in the interval

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