Question

If the function $f:R-\{1,-1\}\to A$ defined by $f\left(x\right)=\frac{x^2}{1-x^2}$, is surjective, then $A$ is

• A. $R-[-1,0)$
• B. $R-\{-1\}$
• C. $R-(-1,0)$
• D. $[0,\infty )$

Answer 1

The correct option is A $R-[-1,0)$

$f:R-\{1,-1\}\to A$
$f\left(x\right)=\frac{x^2}{1-x^2}$ is surjective.
As $f$ is surjective, so every element in Co-domain must have a pre-image in domain.
Let $y=\frac{x^2}{1-x^2}$
$\Rightarrow y-y{x}^2=x^2\Rightarrow x^2(1+y)=y\Rightarrow x=\pm \sqrt{\frac{y}{1+y}}$
For $x$ to be defined,
$1+y\ne 0\Rightarrow y\ne -1\cdots \left(i\right)\text{and}\frac{y}{1+y}\ge 0\Rightarrow y\in (-\infty ,-1)\cup [0,\infty )\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$y\in R-[-1,0)$