(i) Given f (x) = 2 – 3x, x ∈ R, x > 0 Since x > 0 ⇒ –3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f(x) < 2 ∴ The range of f (x) is (– ∞, 2). (ii) Given f (x) = $x^2$ + 2, x is a real number We know $x^2$ ≥ 0 ⇒ x2 + 2 ≥ 0 + 2 ⇒ $x^2$ + 2 ≥ 2 ⇒ f (x) ≥ 2 ∴ The range of f (x) is [2, ∞). (iii) Given f (x) = x, x is a real number. Let y = f (x) = x ⇒ y = x ∴ Range of f (x) = Domain of f (x) ∴ Range of f (x) is R.