Question

The minimum value of
$f\left(x\right)=\sqrt{(x^2-2x+1)}+\sqrt{(x^2-4x+4)}$ is

• A. $0$
• B. $-\infty$
• C. $3$
• D. $1$

Answer 1

Answer: (D)

$1$

R.E.F image

$f\left(x\right)=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}$

$=\sqrt{(x-1{)}^2}+\sqrt{(x-2{)}^2}$

$=|x-1|+|x-2|$

$f\left(x\right)=(x-1)+(x-2);x\ge 2$

$=2x-3$

$x-1-x+2;1⩽x⩽2$

$=1$

$-(x-1)-(x-2);x⩽1$

$=-2x+3$

$\therefore$ Maximum value of f(x) is 1

option (D) is correct.