If non-zero real numbers p and q exist such t | vedclass

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(A) For $f(x) = x^2 + 2px + 2q^2$,the minimum value is at $x = -p$,which is $f(-p) = (-p)^2 + 2p(-p) + 2q^2 = p^2 - 2p^2 + 2q^2 = 2q^2 - p^2$.For $g(x

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$[0, \sqrt{2})$

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(A) For $f(x) = x^2 + 2px + 2q^2$,the minimum value is at $x = -p$,which is $f(-p) = (-p)^2 + 2p(-p) + 2q^2 = p^2 - 2p^2 + 2q^2 = 2q^2 - p^2$.For $g(x) = -x^2 - 2qx + p^2$,the maximum value is at $x = -q$,which is $g(-q) = -(-q)^2 - 2q(-q) + p^2 = -q^2 + 2q^2 + p^2 = q^2 + p^2$.The condition $\min f(x) > \max g(x)$ implies $2q^2 - p^2 > q^2 + p^2$.Rearranging the terms,we get $q^2 > 2p^2$,which implies $\frac{q^2}{p^2} > 2$.Taking the reciprocal,$\frac{p^2}{q^2} Multiplying by $2$,we get $|\frac{2p}{q}| Since $p, q$ are non-zero,$|\frac{2p}{q}| > 0$.Thus,the set of values is $[0, \sqrt{2})$.

If non-zero real numbers $p$ and $q$ exist such that $\min f(x) > \max g(x)$,where $f(x) = x^2 + 2px + 2q^2$ and $g(x) = -x^2 - 2qx + p^2$ for $x \in \mathbb{R}$,find the set of values containing $|\frac{2p}{q}|$.

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