Application Of Derivatives Ques 73
73. If $f(x)=x^{2}+2 b x+2 c^{2}$ and $g(x)=-x^{2}-2 c x+b^{2}$, such that $\min f(x)>\max g(x)$, then the relation between $b$ and $c$, is
$(2003,2 \mathrm{M})$
(a) No real value of $b$ and $c$
(b) $0<c<b \sqrt{2}$
(c) $|c|<|b| \sqrt{2}$
(d) $|c|>|b| \sqrt{2}$
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Answer:
Correct Answer: 73.(d)
Solution:
Formula:
Maxima and Minima of functions of one variable :
- Given $f(x)=x^{2}+2 b x+2 c^{2}$ and $g(x)=-x^{2}-2 c x+b^{2}$
Then, $f(x)$ is minimum and $g(x)$ is maximum at $(x=\frac{-b}{4 a}$ and $f(x)=\frac{-D}{4 a})0$, respectively.
$\therefore \quad \min f(x)=\frac{-\left(4 b^{2}-8 c^{2}\right)}{4}=\left(2 c^{2}-b^{2}\right)$
and $\max g(x)=-\frac{\left(4 c^{2}+4 b^{2}\right)}{4(-1)}=\left(b^{2}+c^{2}\right)$
Now, $\min f(x)>\max g(x)$
$ \begin{array}{lrl} \Rightarrow & 2 c^{2}-b^{2}>b^{2}+c^{2} \\ \Rightarrow & c^{2}>2 b^{2} \\ \Rightarrow & |c|>\sqrt{2}|b| \end{array} $
BETA
