Mark (√) against the correct answer in the following: Let $\mathrm{f}(\mathrm{x})=\frac{1}{\left(1-\mathrm{x}^{2}\right)}$. Then, range $(\mathrm{f})=?$ A. $(-\infty, 1]$ B. $[1, \infty)$ C. $[-1,1]$ D. none of these

Home » Mark (√) against the correct answer in the following: Let . Then, range A. B. C. D. none of these

Mark (√) against the correct answer in the following: Let $\mathrm{f}(\mathrm{x})=\frac{1}{\left(1-\mathrm{x}^{2}\right)}$ . Then, range $(\mathrm{f})=?$
A. $(-\infty, 1]$
B. $[1, \infty)$
C. $[-1,1]$
D. none of these

CBSE | Class 12 | Functions | Maths | RS Aggarwal

Mark (√) against the correct answer in the following: Let $\mathrm{f}(\mathrm{x})=\frac{1}{\left(1-\mathrm{x}^{2}\right)}$ . Then, range $(\mathrm{f})=?$
A. $(-\infty, 1]$
B. $[1, \infty)$
C. $[-1,1]$
D. none of these

Solution:

Option(B) is correct.
$f(x)=\frac{1}{\left(1-x^{2}\right)}$
The range of $f(x)$ can be found out by putting $f(x)=y$
$\begin{array}{l} \mathrm{y}=\frac{1}{\left(1-x^{2}\right)} \\ \Rightarrow \mathrm{y}-\mathrm{yx}^2=1 \\ \Rightarrow \mathrm{yx}^2=\mathrm{y}-1 \\ \Rightarrow \mathrm{x}^2=\frac{y-1}{y} \end{array}$
$\begin{array}{l} \Rightarrow \mathrm{x}=\sqrt{\frac{y-1}{y}} \\ \Rightarrow \sqrt{\frac{y-1}{y}} \geq 0 \\ \Rightarrow \mathrm{y} \geq 1 \end{array}$
Therefore, Range of the $\mathrm{f}(\mathrm{x})=[1, \infty)$