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