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}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval
If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$ and $g\left( x \right) = {-x^2} - 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval
- [JEE MAIN 2014]
- 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
$\left( {\sqrt 2 \,,\,\infty } \right)$
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- [JEE MAIN 2022]
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