Question

Let $f\left(x\right)=|x-1|+|x-2|.$ Then which of the following is/are true?

• A. If $x\in R,$ then $f\left(x\right)\in [0,\infty )$
• B. If $-1\le x\le 3,$ then $f\left(x\right)\in [1,5]$
• C. $f\left(x\right)=1$ for all $x\in [\frac{11}{10},\frac{19}{10}]$
• D. $f(-2)+f\left(\frac{3}{2}\right)=f\left(5\right)$

Answer 1

Answer: (B), (C)

The correct options are
B If $-1\le x\le 3,$ then $f\left(x\right)\in [1,5]$
C $f\left(x\right)=1$ for all $x\in [\frac{11}{10},\frac{19}{10}]$

$f\left(x\right)=|x-1|+|x-2|$
$\Rightarrow f\left(x\right)=\{\begin{array}{c}3-2x,\text{if}x<1\\ 1,\text{if}1\le x\le 2\\ 2x-3,\text{if}x>2\end{array}$


Domain : $x\in R$
Range : $f\left(x\right)\in [1,\infty )$

If $-1\le x\le 3,$ then
$f(-1)=5,f\left(3\right)=3$
$\Rightarrow f\left(x\right)$ is decreasing from $5$ to $1$ and then increasing to $3.$
$\therefore f\left(x\right)\in [1,5]$

$f\left(x\right)=1\forall x\in [1,2]$

$f(-2)+f\left(\frac{3}{2}\right)=7+1=8$
and $f\left(5\right)=7$
$\therefore f(-2)+f\left(\frac{3}{2}\right)\ne f\left(5\right)$