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If $f(x)=|x-1|+|x-2|+|x-3|, 2 < x < 3$, then $f$ is
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Verified Answer
The correct answer is:
a bijection
We have,
$\begin{aligned}
& f(x)=|x-1|+|x-2|+|x-3| \\
& \Rightarrow f(x)=\left\{\begin{array}{cc}
6-3 x, & x < 1 \\
4-x, & 1 < x < 2 \\
x, & 2 < x < 3 \\
3 x-6, & x>3
\end{array}\right.
\end{aligned}$
For $2 < x < 3$, then $f(x)=x$
So, $f(x)$ is one-one and onto function.
It implies that $f(x)$ is a bijection.
$\begin{aligned}
& f(x)=|x-1|+|x-2|+|x-3| \\
& \Rightarrow f(x)=\left\{\begin{array}{cc}
6-3 x, & x < 1 \\
4-x, & 1 < x < 2 \\
x, & 2 < x < 3 \\
3 x-6, & x>3
\end{array}\right.
\end{aligned}$
For $2 < x < 3$, then $f(x)=x$
So, $f(x)$ is one-one and onto function.
It implies that $f(x)$ is a bijection.
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