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)

R-\left\\{-1, 1\right\\}, \left(-\infty, 0\right) \cup [1, \infty)

Answer 2

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 $\left(f\right)=\left(-\infty, 0\right) \cup [1, \infty)$