Question

The solution set of the inequality $|x+2|-|x-1|<x-\frac{3}{2}$ is

• A. $\left(\frac{9}{2},\infty \right)$
• B. $\left(-\infty ,\frac{3}{2}\right)$
• C. $\left(-2,-\frac{3}{2}\right)$
• D. $\left(-1,-\frac{3}{2}\right)$

Answer 1

Answer: (A)

The inequality is $|x+2|-|x-1|<x-\frac{3}{2}$
Dividing the problem into three intervals :
(i) If $x<-2$, then
But $-\frac{3}{2}>-2$, hence no common values
$\Rightarrow x\in \varphi$
(ii) If $-2\le x<1$, then
$(x+2)+(x-1)<x-\frac{3}{2}$
$\Rightarrow x<-\frac{5}{2}$
But $-\frac{5}{2}<-2$, hence no common values
$\Rightarrow x\in \varphi$
(iii) If $x\ge 1$ , then
$(x+2)-(x-1)<x-\frac{3}{2}\Rightarrow x>\frac{9}{2}$
$\because \frac{9}{2}>1.\Rightarrow$ common solution is
$x>\frac{9}{2}\Rightarrow x\in \left(\frac{9}{2},\infty \right)$
$\therefore$ Solution set is $x\in \left(\frac{9}{2},\infty \right)$