A) no relation
B) \[0<c<\frac{b}{2}\]
C) \[|c|<\sqrt{2}|b|\]
D) \[|c|\,\,>\sqrt{2}|b|\]
Correct Answer: D
Solution :
We have, \[f(x)={{(x+b)}^{2}}+2{{c}^{2}}-{{b}^{2}}\] \[\Rightarrow \] \[\min f(x)=2{{c}^{2}}-{{b}^{2}}\] and \[g(x)={{b}^{2}}+{{c}^{2}}-{{(x+c)}^{2}}\] \[\Rightarrow \] \[\max g(x)={{b}^{2}}+{{c}^{2}}\] Thus, \[\max f(x)>\max g(x)\] \[\Rightarrow \] \[2{{c}^{2}}-{{b}^{2}}>{{b}^{2}}+{{c}^{2}}\] \[\Rightarrow \] \[{{c}^{2}}>2{{b}^{2}}\] \[\Rightarrow \] \[|c|\,\,>\sqrt{2}|b|\]
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