Question

If $f\left(x\right)=\frac{1}{|x-1|-x^2}$ and $D_f,R_f$ denote the domain and range of the function respectively, then

• A. $D_f=R-\left\{\frac{-1\pm \sqrt{5}}{2}\right\}$
• B. $D_f=R-\left\{\frac{-1\pm \sqrt{3}}{2}\right\}$
• C. $R_f=(-\infty ,0)\cup [\frac{4}{5},\infty )$
• D. $R_f=(-\infty ,-\frac{4}{3})\cup [\frac{4}{5},\infty )$

Answer 1

Answer: (A), (C)

The correct options are
A $D_f=R-\left\{\frac{-1\pm \sqrt{5}}{2}\right\}$
C $R_f=(-\infty ,0)\cup [\frac{4}{5},\infty )$

$f\left(x\right)=\frac{1}{|x-1|-x^2}$
The function is defined when
$|x-1|-x^2\ne 0$
Now,
$|x-1|=x^2$
Squaring on the both sides,
$\Rightarrow (x-1{)}^2=x^4$
$\Rightarrow (x^2{)}^2-(x-1{)}^2=0$
$\Rightarrow (x^2+x-1)(x^2-x+1)=0$
$x^2-x+1>0$ as $\Delta <0$
$\therefore x^2+x-1=0$
$\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$

$\therefore D_f=R-\left\{\frac{-1\pm \sqrt{5}}{2}\right\}$

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

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

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 $\left(2\right)$
$\Rightarrow y\in (-\infty ,0)\cup [\frac{4}{5},\infty )$
$\therefore R_f=(-\infty ,0)\cup [\frac{4}{5},\infty )$