Question

The range of the function $f\left(x\right)=\sqrt{(x-1)(3-x)}$ is

• A. $[0,1]$
• B. $(-1,1)$
• C. $(-3,3)$
• D. $(-3,1)$

Answer 1

The correct option is A $[0,1]$

Given :

$f\left(x\right)=\sqrt{(x-1)(3-x)}$

$=\sqrt{-x^2+4x-3}$

$=\sqrt{-x^2+4x-4+1}$

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

maximum value at $f\left(x\right)$ will be '1'

when $(x-2)=0$

so, $f_{max}=1$

minimum when $(x-2{)}^2=1$

$x-2=\pm 1$

$x=3$ or $x=1$

so, minimum = $\sqrt{1-1}=0$

so Range = $[0,1]\to$ option 'A'