Question

If $f\left(x\right)=max(y^2-2xy+1)$ for $0\le y\le 2,$ then the minimum value of $f\left(x\right)$ $\forall x\in R$ is

• A. $1$
• B. $2$
• C. $0$
• D. Can not be determined.

Answer 1

The correct option is A $1$

Let $g\left(y\right)=y^2-2xy+1\{0\le y\le 2\}$
$g^{\prime }\left(y\right)=2y-2x$
$g^{\prime }\left(y\right)=0\Rightarrow y=x.$
$g\left(0\right)=1,g\left(2\right)=5-4x$
$g\left(x\right)=1-x^2$
$f\left(x\right)=max\{1,5-4x,1-x^2\}$

$f\left(x\right)=\{\begin{array}{cc}1,& x\ge 1\\ 5-4x,& x<1\end{array}$

$\therefore$ Minimum value of $f\left(x\right)$ = 1.