Question

If $ x $ satisfies $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ , then
A. $ x\in \varnothing $
B. $ x\in \left[ 1,2 \right] $
C. $ x\in \left[ 3,\infty \right) $
D. $ x\in \left( -\infty ,\infty \right) $

Answer 1

Hint: We express the whole domain into three parts for the equation $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ . We break it according to the equations. We then find the solutions for the equation and check if the intervals satisfy or not. Modulus function $ f\left( x \right)=\left| x \right| $ works as the distance of the number from 0. The number can be both positive and negative but the distance of that number will always be positive. Distance can never be negative.

Complete step-by-step answer:
We try to break the whole domain into three parts for the equation $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ .
The divisions are $ x\in \left( -\infty ,1 \right)\cup \left[ 1,2 \right)\cup \left[ 2,\infty \right) $ .
Therefore, for $ x\in \left( -\infty ,1 \right) $ , the values are
\[\begin{align}
  & \left| {{x}^{2}}-3x+2 \right|={{x}^{2}}-3x+2 \\
 & \left| x-1 \right|=-\left( x-1 \right)=1-x \\
\end{align}\]
The equation $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ becomes
 $ {{x}^{2}}-3x+2+1-x=x-3 $ .
Simplifying we get
\[\begin{align}
  & {{x}^{2}}-3x+2+1-x=x-3 \\
 & \Rightarrow {{x}^{2}}-5x+6=0 \\
 & \Rightarrow {{x}^{2}}-2x-3x+6=0 \\
 & \Rightarrow x\left( x-2 \right)-3\left( x-2 \right)=0 \\
 & \Rightarrow \left( x-2 \right)\left( x-3 \right)=0 \\
 & \Rightarrow x=2,3 \\
\end{align}\]
The conditions $ x\in \left( -\infty ,1 \right) $ and \[x=2,3\] can’t happen together. We don’t have any solution in the interval $ x\in \left( -\infty ,1 \right) $ .
Therefore, for $ x\in \left[ 1,2 \right) $ , the values are
\[\begin{align}
  & \left| {{x}^{2}}-3x+2 \right|=-\left( {{x}^{2}}-3x+2 \right)=-{{x}^{2}}+3x-2 \\
 & \left| x-1 \right|=x-1 \\
\end{align}\].
The equation $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ becomes
\[-{{x}^{2}}+3x-2+x-1=x-3\].
Simplifying we get
\[\begin{align}
  & -{{x}^{2}}+3x-2+x-1=x-3 \\
 & \Rightarrow {{x}^{2}}-3x=0 \\
 & \Rightarrow x\left( x-3 \right)=0 \\
 & \Rightarrow x=0,3 \\
\end{align}\]
The conditions $ x\in \left[ 1,2 \right) $ and \[x=0,3\] can happen together. We have solutions in the interval $ x\in \left[ 1,2 \right) $ .
Therefore, for $ x\in \left[ 2,\infty \right) $ , the values are
\[\begin{align}
  & \left| {{x}^{2}}-3x+2 \right|={{x}^{2}}-3x+2 \\
 & \left| x-1 \right|=x-1 \\
\end{align}\].
The equation $ \left| {{x}^{2}}-3x+2 \right|+\left| x-1 \right|=x-3 $ becomes
\[{{x}^{2}}-3x+2+x-1=x-3\].
Simplifying we get
\[\begin{align}
  & {{x}^{2}}-3x+2+x-1=x-3 \\
 & \Rightarrow {{x}^{2}}-3x+4=0 \\
 & \Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 4\times 1}}{2\times 1}=\dfrac{3\pm i\sqrt{7}}{2} \\
\end{align}\]
There is no real solution in $ x\in \left[ 2,\infty \right) $ .
Therefore, the solution is $ x\in \varnothing $ . The correct option is A
So, the correct answer is “Option A”.

Note: In mathematical notation we express it with modulus value. Let a number be x whose sign is not mentioned. The absolute value of that number will be $ \left| x \right| $ . We can say $ \left| x \right|\ge 0 $ .
We can express the function $ f\left( x \right)=\left| x \right| $ as $ f\left( x \right)=\left\{ \begin{align}
  & x\left( x\ge 0 \right) \\
 & -x\left( x<0 \right) \\
\end{align} \right. $ .
We can write $ f\left( x \right)=\left| x \right|=\pm x $ depending on the value of the number x.