Question

If the solution set of $|x-k|<2$ is a subset of the solution set of the inequality $\frac{2x-1}{x+2}<1,$ then the number of possible integral value(s) of $k$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

$\frac{2x-1}{x+2}<1\Rightarrow \frac{2x-1}{x+2}-1<0\Rightarrow \frac{2x-1-x-2}{x+2}<0\Rightarrow \frac{x-3}{x+2}<0\Rightarrow -2<x<3\cdots \left(A\right)$

Also, $|x-k|<2$
$\Rightarrow -2<x-k<2\Rightarrow k-2<x<k+2\cdots \left(B\right)$

Given $B$ is subset of $A$.
So, $k-2\ge -2$ and $k+2\le 3$
$\Rightarrow k\ge 0$ and $k\le 1$
Hence, $k\in [0,1]$
Possible integral values of $k$ is $\{0,1\}$