The range of the real function \(f(x) =\frac{1}{1-x^2}\) is(a) R – {

1.	The range of the real function \(f(x) =\frac{1}{1-x^2}\) is(a) R – {– 1, 1} (b) (0, ∞) (c) [1, ∞) (d) (– ∞, 0)
Answer» Answer : (c) [1, ∞) f(x) = \(\frac{1}{1-x^2}\) is not defined when 1 – x² = 0, i.e. x = ± 1 ∴ Domain of f (x) = R – {– 1, 1} Let y = \(\frac{1}{1-x^2}\) ⇒ (1-x²) = \(\frac{1}{y}\) ⇒ x2 = 1 - \(\frac{1}{y}\) ⇒ \(x\,\pm\sqrt{1-\frac{1}{y}}\) This shows that x will not be defined when 1 - \(\frac{1}{y}\) < 0, i.e., y < 1 ∴ Range of f : y ≥ 1, i.e. y ∈ [1, ∞)

The range of the real function \(f(x) =\frac{1}{1-x^2}\) is(a) R – {– 1, 1} (b) (0, ∞) (c) [1, ∞) (d) (– ∞, 0)

Answer»

Answer : (c) [1, ∞)

f(x) = \(\frac{1}{1-x^2}\) is not defined when 1 – x² = 0, i.e. x = ± 1

∴ Domain of f (x) = R – {– 1, 1}

Let y = \(\frac{1}{1-x^2}\) ⇒ (1-x²) = \(\frac{1}{y}\)

⇒ x2 = 1 - \(\frac{1}{y}\) ⇒ \(x\,\pm\sqrt{1-\frac{1}{y}}\)

This shows that x will not be defined when 1 - \(\frac{1}{y}\) < 0, i.e., y < 1

∴ Range of f : y ≥ 1, i.e. y ∈ [1, ∞)