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Range of y=\sqrt{x^{2}-4x+3} is

Option: 1

[0,\infty)

 

 

 


Option: 2

\left ( 0,\infty \right )


Option: 3

R


Option: 4

None of these


Answers (1)

best_answer

We can use simple manipulations to find its range

\begin{aligned} & x^{2}-4 x+3 \\ =& x^{2}-4 x+4-4+3 \\ =&(x-2)^{2}-1 . \end{aligned}

Now, \begin{gathered} (x-2)^{2} \geqslant 0 \\ \end{gathered} (for any x)

        \begin{gathered} (x-2)^{2}-1 \geqslant-1 \end{gathered}(for any x)

But only positive values \begin{gathered} (x-2)^{2}-1 \end{gathered} are allowed for \sqrt{x^{2}-4x+3}

\therefore \begin{gathered} (x-2)^{2}-1 \geqslant 0 \end{gathered} (for any x in domain)

\begin{aligned} &\Rightarrow \sqrt{(x-2)^{2}-1} \geqslant 0 \\ &\Rightarrow \sqrt{x^{2}-4 x+3} \geqslant 0 \\ &\Rightarrow y \geqslant 0 \end{aligned}

So, range is \left [\, 0,\infty \right )

Posted by

jitender.kumar

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