Question

If $|x-2|+|x+1|\ge 3$, then complete solution set of this inequation is

• A. $[1,\infty )$
• B. $(-\infty ,-2]$
• C. $R$
• D. $[-2,1]$

Answer 1

Answer: (C)

$R$

The inequation is

$|x-2|+|x+1|\ge 3$

Case (1)

For $-\infty <x<-1$ ............ (1)

the inequation is

$-(x-2)-(x+1)\ge 3$

$-x+2-x-1\ge 3$

$-2x+1\ge 3$

$-2x+1\ge 3$

$-2x\ge 2$

$x\le -1$ ................ (2)

From (1) and (2), the inequation is true for

$-\infty <x<-1$

Case (2)

For $-1\le x<2$ ,

the inequation is

$-(x-2)+x+1\ge 3$

$-x+2+x+1\ge 3$

$3\ge 3$ , which is true.

So the inequation is true for $-1\le x<2$

Case (3)

For $2\le x<\infty$ .............. (3)

the inequation is

$x-2+x+1\ge 3$

$2x-1\ge 3$

$2x\ge 4$

$x\ge 2$ ........... (4)

From (3) and (4),

$2\le x<\infty$

Therefore, the inequation is true for all real values of $x$