Question:
Mark (√) against the correct answer in the following:
Let $f(x)=\frac{1}{\left(1-x^{2}\right)} \cdot$ Then, range $(f)=?$
A. $(-\infty, 1]$
B. $[1, \infty)$
C. $[-1,1]$
D. none of these
Solution:
$f(x)=\frac{1}{1-x^{2}}$
$\Rightarrow \mathrm{y}=\frac{1}{1-\mathrm{x}^{2}}$
$\Rightarrow \mathrm{y}-\mathrm{yx}^{2}=1$
$\Rightarrow \mathrm{y}-1=\mathrm{yx}^{2}$
$\Rightarrow x=\sqrt{\frac{y-1}{y}}$
$\Rightarrow \frac{y-1}{y} \geq 0$
$\Rightarrow \mathrm{y} \geq 1$
$\therefore$ range $(f)=[1, \infty)$
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