જો શૂન્યતર વાસ્તવિક સંખ્યા b અને c છે કે જેથી | vedclass

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Question

We have 

$f\left( x \right) = {x^2} + 2bx + 2{c^2}$

and $g\left( x \right) =  - {x^2} - 2cx + {b^2},\left( {x \in R} \right)$

Vedclass · Accepted Answer

$\left( {\sqrt 2 \,,\,\infty } \right)$

Vedclass · Answer

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)$

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