JEE Mains · Maths · STD 12 - 1. relation and function
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
- 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 \(\left( {\sqrt 2 \,,\,\infty } \right)\)
Answer & Solution
Correct Answer
(D) \(\left( {\sqrt 2 \,,\,\infty } \right)\)
Step-by-step Solution
Detailed explanation
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…
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