Question

The domain and range of the function
$f=\left\{\left(\frac{1}{1-x^{2}}\right) : x \in R, x \ne \pm 1\right\}$ are respectively

• A. $R-\left\{-1, 1\right\}, \left(-\infty, 0\right) \cup [1, \infty)$
• B. $R, \left(-\infty, 0\right) \cup [1, \infty)$
• C. $R, [1, \infty)$
• D. None of these

Answer 1

Answer: (A)

We have, $f\left(x\right)=\frac{1}{1-x^{2}}$
Clearly, $f\left(x\right)$ is defined for all $x \in R$ except for which $x^{2}-1=0$
i.e., $x = \pm 1$.
Hence, Domain of $f=R-\left\{-1,1\right\}$.
Let $f\left(x\right)=y$. Then, $\frac{1}{1-x^{2}}=y$
$\Rightarrow 1-x^{2}=\frac{1}{y}$
$\Rightarrow x^{2}=1-\frac{1}{y}=\frac{y-1}{y}$
$\Rightarrow x=\pm\sqrt{\frac{y-1}{y-0}}$
Clearly, $x$ will take real values, if
$\frac{y-1}{y-0} \le 0$



$\Rightarrow y < 0$ or $y \ge 1$
$\Rightarrow y \in\left(-\infty, 0\right) \cup [1, \infty)$
Hence, range $\left(f\right)=\left(-\infty, 0\right) \cup [1, \infty)$