Question

Solve the following inequality

$\frac{ |x - 1| }{x + 2} < 1$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm \: \dfrac{ |x - 1| }{x + 2} < 1$

We know, By definition of Modulus function, we have

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: if \: x \: < 0} \\ \\ &\sf{ \: \: x \: \: if \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition,

$\begin{gathered}\begin{gathered}\bf\: |x - 1| = \begin{cases} &\sf{ - x + 1 \: \: if \: x \: < 1} \\ \\ &\sf{ \: \: x - 1 \: \: if \: x \geqslant 1} \end{cases}\end{gathered}\end{gathered}$

Consider, Case - 1

$\rm \: When \: x < - 1$

So, given inequality can be rewritten as

$\rm \: \dfrac{ - x + 1 }{x + 2} < 1$

$\rm \: \dfrac{ - x + 1 }{x + 2} - 1 < 0$

$\rm \: \dfrac{ - x + 1 - x - 2}{x + 2} < 0$

$\rm \: \dfrac{ - 2x - 1}{x + 2} < 0$

$\rm \: \dfrac{ - (2x + 1)}{x + 2} < 0$

$\rm \: \dfrac{2x + 1}{x + 2} > 0$

$\rm\implies \:x < - 2 \: \: or \: \: x > - \dfrac{1}{2}$

$\rm \: As \: x \: < \: - \: 0$

$\rm\implies \: \: x \: \in \: ( - \infty , - 2) \: \cup \: \bigg( - \dfrac{1}{2},1 \bigg) - - - (1)$

Now, Consider Case - 2

$\rm \: When \: x \: \geqslant \: 1$

So, given inequality can be rewritten as

$\rm \: \dfrac{x - 1}{x + 2} < 1$

$\rm \: \dfrac{x - 1}{x + 2} - 1 < 0$

$\rm \: \dfrac{x - 1 - x - 2}{x + 2} < 0$

$\rm \: \dfrac{ - 3}{x + 2} < 0$

$\rm \: \dfrac{ 3}{x + 2} > 0$

$\rm\implies \:x > - 2$

$\rm \: As \: x \: \geqslant \: 1$

$\rm\implies \:x \: \in \: [1, \: \infty ) - - - (2)$

So, from equation (1) and (2), we concluded that

$\rm\implies \:x \: \in \: ( - \infty , - 2) \: \cup \: \bigg( - \dfrac{1}{2},1 \bigg) \cup \: [1, \infty )$

Or

$\rm\implies \:x \: \in \: ( - \infty , - 2) \: \cup \: \bigg( - \dfrac{1}{2}, \infty \bigg)$

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BASIC CONCEPT USED

If a and b are two positive real numbers, such that a < b, then

$\rm \: (x - a)(x - b) < 0 \: \: \rm\implies \:a < x < b \:$

$\rm \: (x - a)(x - b) \leqslant 0 \: \: \rm\implies \:a \leqslant x \leqslant b \:$

$\rm \: (x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x\leqslant b \: or \: x \geqslant a \:$

$\rm \: (x - a)(x - b) > 0 \: \: \rm\implies \:x < b \: or \: x > a \:$

Answer 2

Question:-

Solve the following inequality

$\sf \large \frac{ |x - 1| }{x + 2} < 1.$

Given:-

$\sf \large Inequality \: \frac{ |x - 1| }{x + 2} < 1.$

Solution:-

$\sf \large \frac{ |x - 1| }{x + 2} < 1$

$\sf \large \frac{ |x - 1| }{x + 2} - 1 < 0$

$\sf \large \frac{ |x - 1 | - x - 2}{x + 2} < 0$

$\sf \large Let,f(x) = \frac{ |x - 1| - x - 2}{x + 2}$

So, We have:

$\sf \large\begin{gathered}\begin{gathered}\begin{gathered}\: \sf \large f(x)= \begin{cases} &\sf \frac{ - 3}{x + 2} \: \: \: if \: \: x \geqslant 1\\ \\ &\sf \frac{2x + 1}{x + 3} \: \: \: if \: \: x < 1 \end{cases}\end{gathered}\end{gathered} \\ \end{gathered}$

☆ We want to know when f(x) < 0.

$\sf \large Case - I (f(x)when \: \: x \geqslant 1)$

$\sf \large f(x) = \frac{ - 3}{x + 2} < 0$

$\sf \large\Rightarrow x \in( - 2, \infty) ∩(1, \infty)$

$\sf \large \Rightarrow x \in(1, \infty)$________________________________________

$\sf \large Case-II(f(x)where \: \: x < 1)$

$\sf \large f(x) = \frac{2x + 1}{x + 2} < 0$

$\sf \large \Rightarrow \: x \in \bigg[( - \infty, - 2)∪( - \frac{1}{2}, \infty) \bigg]∩( - \infty,1)$

$\sf \large \Rightarrow \in( - \infty, - 2)∪( - \frac{1}{2} ,1)$

☆ Taking union of both sets We get:-

$\sf \large \: x\in( - \infty, - 2)∪( - \frac{1}{2} , \infty)$

Answer:-

${ \boxed{ \sf \large \red{ x \in( - \infty, - 2)∪( - \frac{1}{2},1).}}}$

Hope you have satisfied. ⚘