Question

Let $ f(x) = x^2 + 2bx + 2c^2 $ and $ g(x) = -x^2 - 2cx + b^2 $, $ x \in \mathbb{R} $. If $ b $ and $ c $ are non-zero real numbers such that $ \min f(x)>\max g(x) $, then $$ \left| \frac{c}{b} \right| $$ lies in the interval

• A. \( \left( \frac{1}{2}, \frac{1}{\sqrt{2}} \right) \)
• B. \( \left( \frac{1}{\sqrt{2}}, \sqrt{2} \right) \)
• C. \( \left( \sqrt{2}, \infty \right) \)
• D. \( (0, 1) \)

Hint:

For inequalities involving extrema of functions, apply vertex formulas and compare the resulting expressions carefully.

Answer 1

The Correct Option is C

Find minimum of \( f(x) \): \[ f(x) = x^2 + 2bx + 2c^2 \Rightarrow \text{min at } x = -b,\ f(-b) = b^2 + 2c^2 \] Find maximum of \( g(x) \): \[ g(x) = -x^2 - 2cx + b^2 \Rightarrow \text{max at } x = -c,\ g(-c) = -c^2 + b^2 \] Now the condition: \[ b^2 + 2c^2>-c^2 + b^2 \Rightarrow 3c^2>0 \Rightarrow \frac{c^2}{b^2}>\frac{1}{2} \Rightarrow \left| \frac{c}{b} \right|>\sqrt{2} \]