A) \[\left( 0,\frac{1}{2} \right)\]
B) \[\left[ \frac{1}{2},\frac{1}{\sqrt{2}} \right)\]
C) \[\left[ \frac{1}{\sqrt{2},}\sqrt{2} \right]\]
D) \[\left( \sqrt{2},\infty \right)\]
Correct Answer: D
Solution :
We have \[f(x)={{x}^{2}}+2bx+2{{c}^{2}}\] and \[g(x)=-{{x}^{2}}-2cx+{{b}^{2}},(x\in R)\] \[\Rightarrow \]\[f(x)={{(x+b)}^{2}}+2{{c}^{2}}-{{b}^{2}}\] and \[g(x)=-{{(x+c)}^{2}}+{{b}^{2}}+{{c}^{2}}\] Now, \[{{f}_{\min }}=2{{c}^{2}}-{{b}^{2}}\]and \[{{g}_{\max }}={{b}^{2}}+{{c}^{2}}\] Given : \[\min f(x)>\max g(x)\] \[\Rightarrow \]\[2{{c}^{2}}-{{b}^{2}}>{{b}^{2}}+{{c}^{2}}\] \[\Rightarrow \]\[{{c}^{2}}>2{{b}^{2}}\] \[\Rightarrow \]\[|c|>|b|\sqrt{2}\] \[\Rightarrow \]\[\frac{|c|}{|b|}>\sqrt{2}\Rightarrow \left| \frac{c}{b} \right|>\sqrt{2}\] \[\Rightarrow \]\[\left| \frac{c}{b} \right|\in (\sqrt{2},\infty ).\]
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