Q62 / 200 · Manipal (MET) 2012 · Mathematics

If $f(x)=x^2+2 b x+2 c^2$ and $g(x)=-x^2-2 c x+b^2$ are such that $\min f(x) \gt \max g(x)$, then relation between $b$ and $c$ is

  • A
    no relation
  • B
    $0 \lt c \lt \frac{b}{2}$
  • C
    $|c| \lt \sqrt{2}|b|$
  • Correct answer
    D
    $|c| \gt \sqrt{2}|b|$

Solution

$\begin{aligned} \text {We have, } f(x)=(x+b)^2+2 c^2-b^2 \\ \Rightarrow \quad \min f(x)=2 c^2-b^2 \\ \text {and } \quad g(x)=b^2+c^2-(x+c)^2 \\ \Rightarrow \quad \max g(x)=b^2+c^2 \\ \text { Thus, } \quad \min f(x) \gt \max g(x) \\ \Rightarrow 2 c^2-b^2 \gt b^2+c^2 \\ \Rightarrow c^2 \gt 2 b^2 \quad \Rightarrow|c| \gt \sqrt{2}|b|\end{aligned}$

Hence, the correct answer is option: 'D'.

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