Question

If $f\left(x\right)=\frac{1}{|x-1|-x^2},$ then the domain of the function is

• A. $R-\left\{\frac{-1\pm \sqrt{5}}{2}\right\}$
• B. $R-\left\{\frac{-1\pm \sqrt{3}}{2}\right\}$
• C. $R-\left\{\frac{-1\pm \sqrt{7}}{2}\right\}$
• D. $R-\left\{\frac{-1\pm \sqrt{6}}{2}\right\}$

Answer 1

The correct option is A $R-\left\{\frac{-1\pm \sqrt{5}}{2}\right\}$

$f\left(x\right)=\frac{1}{|x-1|-x^2}$
The function is defined when
$|x-1|-x^2\ne 0$
Now,
$|x-1|=x^2$
When $x\ge 1$, we get
$x-1=x^2\Rightarrow x^2-x+1=0D=1^2-4=-3\le 0$
So, no real roots

When $x<1$, we get
$-x+1=x^2\Rightarrow x^2+x-1=0\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$

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