Question

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

• A. \(\left(0,\frac12\right)\)
• B. \(\left[\frac12,\frac{1}{\sqrt2}\right)\)
• C. \(\left[\frac{1}{\sqrt2},\sqrt2\right]\)
• D. \(\left(\sqrt2,\infty\right)\)

Hint:

Complete the square for both quadratics to get the minimum of \(f\) and the maximum of \(g\).

Answer 1

The Correct Option is D

\(f(x)=(x+b)^2+2c^2-b^2\), so \(\min f=2c^2-b^2\).
\(g(x)=-(x+c)^2+b^2+c^2\), so \(\max g=b^2+c^2\).
\(\min f>\max g\Rightarrow 2c^2-b^2>b^2+c^2\Rightarrow c^2>2b^2\Rightarrow\left|\frac cb\right|>\sqrt2\).
Answer: (4) \(\left(\sqrt2,\infty\right)\)