Question

The range of the function $y=\frac{1}{|x-1|-x^2}$ is

• A. $[-\frac{4}{3},\frac{4}{3}]-\left\{0\right\}$
• B. $(-\infty ,0)\cup [\frac{4}{5},\infty )$
• C. $(-\infty ,-\frac{4}{3})\cup [\frac{4}{5},\infty )$
• D. $[-\frac{4}{3},0]$

Answer 1

The correct option is B $(-\infty ,0)\cup [\frac{4}{5},\infty )$

$y=\frac{1}{|x-1|-x^2}$
$\Rightarrow x^2-|x-1|+\frac{1}{y}=0,y\ne 0$

Case $:1$ When $x\ge 1$
$\Rightarrow x^2-x+(\frac{1}{y}+1)=0$
$\Rightarrow \Delta \ge 0$
$\Rightarrow 1-4(\frac{1}{y}+1)\ge 0$
$\Rightarrow -\frac{4}{y}-3\ge 0$
$\Rightarrow \frac{4+3y}{y}\le 0,y\ne 0$
$\Rightarrow y\in [-\frac{4}{3},0)...\left(1\right)$

Case $:2$ When $x<1$
$\Rightarrow x^2+x+(\frac{1}{y}-1)=0$
$\Rightarrow \Delta \ge 0$
$\Rightarrow 1-4(\frac{1}{y}-1)\ge 0$
$\Rightarrow -\frac{4}{y}+5\ge 0$
$\Rightarrow \frac{5y-4}{y}\ge 0,y\ne 0$
$\Rightarrow y\in (-\infty ,0)\cup [\frac{4}{5},\infty )...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow y\in (-\infty ,0)\cup [\frac{4}{5},\infty )$