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

$[0,1]$

Find the range of the given function

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

$$\begin{gathered} \Rightarrow f\left(x\right)=\sqrt{-x^2+4x-3} \\ \Rightarrow f\left(x\right)=\sqrt{-x^2+4x-4+1} \\ \Rightarrow f\left(x\right)=\sqrt{1-{(x-2)}^2} \end{gathered}$$

Now, $1-{(x-2)}^2\ge 0$

$\Rightarrow 1\le x\le 3.....\left(1\right)$

Let $f\left(x\right)$ be $y$

$$\begin{gathered} \Rightarrow y=\sqrt{1-{(x-2)}^2} \\ \Rightarrow y^2=1-{(x-2)}^2 \\ \Rightarrow x-2=\sqrt{1-y^2} \\ \Rightarrow x=\sqrt{1-y^2}+2 \end{gathered}$$

from $\left(1\right)$

$$\begin{gathered} \Rightarrow -1\le \sqrt{1-y^2}\le 1 \\ \Rightarrow \left(1-y^2\right)\le 1 \\ \Rightarrow 1-1\le y^2 \\ \Rightarrow y^2\ge 0 \\ \Rightarrow y\ge 0 \end{gathered}$$

$\therefore$The maximum value of $f\left(x\right)$ is when $(x-2)=0$

$\Rightarrow f_{max}=\sqrt{1-0^2}=1$

$\therefore$ Range of $f\left(x\right)$ is $\left[0,1\right]$.

Hence, the correct option is $A$, $[0,1]$.