Question

Number of solutions of the equation $|x-1|+|x-2|+|x-3|=7$ is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is B $2$

$|x-1|+|x-2|+|x-3|=7$
critical points
$x-1=0\Rightarrow x=1$
$x-2=0\Rightarrow x=2$
$x-3=0\Rightarrow x=3$

On the basis of above points, we have Four regions
$x<1,x\in [1,2],x\in (2,3],x>3$

$\text{case-1},x<1$
For $x<1,$
$|x-1|=-(x-1),|x-2|=-(x-2),|x-3|=-(x-3)$
Now equation will be :
$-(x-1)-(x-2)-(x-3)=7$
$-3x+6=7$
$x=-\frac{1}{3}$ Also $-\frac{1}{3}<1$
So,$x=-\frac{1}{3}\cdots \left(1\right)$

$\text{case-2},x\in [1,2]$
For $x\in [1,2],$
$|x-1|=(x-1),|x-2|=-(x-2),|x-3|=-(x-3)$
Now equation will be :
$(x-1)-(x-2)-(x-3)=7$
$-x+4=7$
$x=-3$ but $-3\notin [1,2]$
So, $x=\varphi \cdots \left(2\right)$

$\text{case-3},x\in (2,3]$
For $x\in (2,3],$
$|x-1|=(x-1),|x-2|=(x-2),|x-3|=-(x-3)$
Now equation will be :
$(x-1)+(x-2)-(x-3)=7$
$x=7$
$x=7$ but $7\notin (2,3]$
So, $x=\varphi \cdots \left(3\right)$

$\text{case-4},x>3$
For $x>3,$
$|x-1|=(x-1),|x-2|=(x-2),|x-3|=(x-3)$
Now equation will be :
$(x-1)+(x-2)+(x-3)=7$
$3x-6=7$
$x=\frac{13}{3}$ Also $\frac{13}{3}>3$
So, $x=\frac{13}{3}\cdots \left(4\right)$

By $\left(1\right)\cup \left(2\right)\cup \left(3\right)\cup \left(4\right)$
$x=\{-\frac{1}{3},\frac{13}{3}\}$