Given,f, (x) =\(\frac{1}{4-x^2}\)
Let y = f (x) y =\(\frac{1}{4-x^2}\)
x2 = \(\frac{4y-1}{y}\)
x = \(\sqrt{\frac{4y-1}{y}}\)
Clearly, x will assume real values, if
⇒ 4y – 1 ≥ 0 and y ≠ 0
4y ≥ 1 and y ≠ 0
⇒ y ≥ \(\frac{1}{4}\)and y ≠ 0
⇒ y ∈ (– ∞, 0) ∪ [ \(\frac{1}{1}\) , ∞]
∴ Range of y = (– ∞, 0) ∪ [ \(\frac{1}{1}\) , ∞]
