Question

If $\text{sgn}\left(\frac{x-2}{x+1}\right)\le \frac{x-1}{2}$, then $x\in$

• A. $[-1,\infty )$
• B. $(-1,2)\cup (3,\infty )$
• C. $(-1,2]\cup [3,\infty )$
• D. $(-1,\infty )$

Answer 1

The correct option is C $(-1,2]\cup [3,\infty )$

Given : $\text{sgn}\left(\frac{x-2}{x+1}\right)\le \frac{x-1}{2}$

Case $-\left(1\right):$If $\text{sgn}\left(\frac{x-2}{x+1}\right)=1$
$\Rightarrow \frac{x-2}{x+1}>0\Rightarrow x\in (-\infty ,-1)\cup (2,\infty )\cdots \left(i\right)$
Now the inequality becomes,
$1\le \frac{x-1}{2}\Rightarrow x\in [3,\infty )\cdots \left(ii\right)$
Taking $\left(i\right)\cap \left(ii\right)$, we get
$\Rightarrow x\in [3,\infty )\cdots \left(A\right)$

Case $-\left(2\right):$ If $\text{sgn}\left(\frac{x-2}{x+1}\right)=-1$
$\Rightarrow \frac{x-2}{x+1}<0\Rightarrow x\in (-1,2)\cdots \left(iii\right)$
Now the inequality becomes,
$-1\le \frac{x-1}{2}\Rightarrow x\ge -1\Rightarrow x\in (-1,\infty )\cdots \left(iv\right)$
Taking $\left(iii\right)\cap \left(iv\right)$
$\Rightarrow x\in (-1,2)\cdots \left(B\right)$

Case $-\left(3\right):$ If $\text{sgn}\left(\frac{x-2}{x+1}\right)=0$
$\Rightarrow \frac{x-2}{x+1}=0\Rightarrow x=2\cdots \left(v\right)$
Now the inequality becomes,
$0\le \frac{x-1}{2}\Rightarrow x\ge 1\cdots \left(vi\right)$
Taking $\left(v\right)\cap \left(vi\right)$
$\Rightarrow x\in \left\{2\right\}\cdots \left(C\right)$

From $\left(A\right)\cup \left(B\right)\cup \left(C\right)$, we get
$x\in (-1,2]\cup [3,\infty )$