Question

If $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$ such that $\min f\left(x\right)>\max g\left(x\right)$, then the relation between $b$ and $c$ is

• A. No real value of $b$ and $c$
• B. $0<c<b\sqrt{2}$
• C. $\left|c\right|<\left|b\right|\sqrt{2}$
• D. $\left|c\right|>\left|b\right|\sqrt{2}$

Answer 1

The correct option is D

$\left|c\right|>\left|b\right|\sqrt{2}$

Explanation for the correct option.

Step 1. Find the maximum value of $f\left(x\right)$.

The function $f\left(x\right)=x^2+2bx+2c^2$ maximum value is given as $-\frac{D}{4a}=-\frac{b^2-4ac}{2a}$. So

$\begin{array}{rcl}\max f\left(x\right)& =& -\frac{{\left(2b\right)}^2-4\times 1\times 2c^2}{4\times 1}\left[\mathbf{Here}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathbf{2}\mathit{b}\mathbf{,}\mathbf{}\mathit{c}\mathbf{=}\mathbf{2}{\mathit{c}}^{\mathbf{2}}\right]\\ & =& -\frac{4b^2-8c^2}{4}\\ & =& -b^2+2c^2\end{array}$

Step 2. Find the minimum value of $g\left(x\right)$.

The function $g\left(x\right)=-x^2-2cx+b^2$ minimum value is given as $-\frac{D}{4a}=-\frac{b^2-4ac}{2a}$. So

$\begin{array}{rcl}\min g\left(x\right)& =& -\frac{{\left(-2c\right)}^2-4\times \left(-1\right)\times b^2}{4\times \left(-1\right)}\left[\mathbf{Here}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{c}\mathbf{,}\mathbf{}\mathit{c}\mathbf{=}{\mathit{b}}^{\mathbf{2}}\right]\\ & =& -\frac{4c^2+4b^2}{-4}\\ & =& c^2+b^2\end{array}$

Step 3. Find the relation between $b$ and $c$ .

It is given that $\min f\left(x\right)>\max g\left(x\right)$, so putting the values in the inequality:

$$\begin{gathered} -b^2+2c^2>c^2+b^2 \\ \Rightarrow 2c^2-c^2>b^2+b^2 \\ \Rightarrow c^2>2b^2 \\ \Rightarrow \left|c\right|>\sqrt{2}\left|b\right| \end{gathered}$$

Thus, the relation between $b$ and $c$ is $\left|c\right|>\left|b\right|\sqrt{2}$.

Hence, the correct option is D.