The range of the function f(x)=1/(1-x^2), x ≠±1 is :

The correct answer is Option (4) → $(-∞, 0) ∪ [1, ∞)$

$f(x)=\frac{1}{1-x^2},y=\frac{1}{1-x^2}$

so $\frac{1}{y}=1-x^2$

$x^2=\frac{y-1}{y}$

$⇒x=\sqrt{\frac{y-1}{y}}$

$⇒y∈R-[0,1)$

$≡y∈(-∞, 0) ∪ [1, ∞)$