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If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$ and $g\left( x \right) = {-x^2} - 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval
$\left( {0\,,\,\frac{1}{2}} \right)$
$\left[ {\frac{1}{2}\,,\,\frac{1}{{\sqrt 2 }}} \right)$
$\left[ {\frac{1}{{\sqrt 2 }}\,,\,\sqrt 2 } \right]$
$\left( {\sqrt 2 \,,\,\infty } \right)$
Solution
We have
$f\left( x \right) = {x^2} + 2bx + 2{c^2}$
and $g\left( x \right) = – {x^2} – 2cx + {b^2},\left( {x \in R} \right)$
$ \Rightarrow f\left( x \right) = {\left( {x + b} \right)^2} + 2{c^2} – {b^2}$
and $g\left( x \right) = – {\left( {x + c} \right)^2} + {b^2} + {c^2}$
Now, ${f_{\min }} = 2{c^2} – {b^2}$ and ${g_{\max }} = {b^2} + {c^2}$
Given $:\min f\left( x \right) > \max g\left( x \right)$
$ \Rightarrow 2{c^2} – {b^2} > {b^2} + {c^2}$
$ \Rightarrow {c^2} > 2{b^2}$
$ \Rightarrow \left| c \right| > \left| b \right|\sqrt 2 $
$ \Rightarrow \frac{{\left| c \right|}}{{\left| b \right|}} > \sqrt 2 \Rightarrow \left| {\frac{c}{b}} \right| > \sqrt 2 $
$ \Rightarrow \left| {\frac{c}{b}} \right| \in \left( {\sqrt 2 ,\infty } \right)$
