The range of the real valued function $f(x)=|x-2|+|x-3|$ is

Given function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+|\mathrm{x}-3|$ We will thick on three point $\mathrm{x} \leq 2,2 < \mathrm{x} < 3$ and $\mathrm{x} \geq 3$ For $\mathrm{x} \leq 2$ $$ f(x)=-x+2-x+3=-2 x+5 $$ For $2 < \mathrm{x} < 3$ $$ f(x)=x-2-x+3=1 $$ For, $x \geq 3$, $$ \begin{aligned} & f(x)=x-2+x-3 \\ & =2 x-5 \\ & \begin{cases}-2 x+5, & x \leq 2 \\ 1, & 2 < x < 3 \\ 2 x-5, & x \geq 3\end{cases} \end{aligned} $$ Now, take $\mathrm{x}=1,2$ for $\mathrm{x} \leq 2$ then $\mathrm{f}(\mathrm{x})=3$ and $\mathrm{f}(\mathrm{x})=1$ respectively. So, It will give always greater than or equal to 1 . Similarly, for $\mathrm{x} \geq 3$ function always gives values greater than or equal to 1 . So, Range is $[1, \infty)$ Therfore, option (b) is correct.