Question

|x+1| + |x-1| > 2 , find the solution for x € R​

Answer 1

Step-by-step explanation:

here's the answer.

hope it will be helpful

Answer 2

$\large\underline{\text{Theorems or Important Facts}}$

$\red{\bigstar}$Modulus functions.

The result of modulus functions is the distance between two points in the number line.

$\implies|x-1|=\begin{cases}x-1 &\text{ for }x>1 \\0 &\text{ for }x=1 \\-x+1 &\text{ for } x<1 \end{cases}$

$\implies|x+1|=\begin{cases}x+1 &\text{ for }x>-1 \\0 &\text{ for }x=-1 \\-x-1 &\text{ for } x<-1 \end{cases}$

However, the modulus function should give a positive number or 0, so it is important to dissect proper intervals.

$\large\underline{\text{Solution}}$

Let's dissect the intervals. To properly dissect the intervals it is easier to use a number line. For the shorter method, you can combine two intervals by considering 0 with positive. But here we will use the proper method.

$\large\underline{\text{Note}}$

If the inequation is true for all $x$ in the interval, the interval will be the solution set.

If the inequation is false for all $x$ in the interval, the interval is not the solution set.

For $x<-1$

$\implies(-x+1)+(-x-1)=-2x$

$\implies -2x>2$

$\implies\boxed{x<-1}$

For $x=-1$

$\implies(-x+1)+0=-x+1=2$

$\implies2>2$

$\implies\text{No solution.}$

For $-1<x<1$

$\implies(x-1)+(-x+1)=0$

$\implies0>2$

$\implies\text{No solution.}$

For $x=1$

$\implies0+(x+1)=x+1=2$

$\implies2>2$

$\implies\text{No solution.}$

For $x>1$

$\implies(x-1)+(x+1)=2x$

$\implies2x>2$

$\implies\boxed{x>1}$

$\large\underline{\text{Answer}}$

So, the solution is $x<-1$ or $x>1$.