(i) Given f (x) = –|x| We know |x| is defined as non-negative real number for every x ∈ R. ∴ Domain of f = R Also, |x| ≥ 0 ∀ x ∈ R ⇒ – |x| ≤ 0 ∀ x ∈ R ∴ Range of f = (– ∞, 0] (ii) Given f (x) = $\sqrt{9-x^2}$ We know $\sqrt{9-x^2}$ ≥ 0 ⇒ 9 − $x^2$ ≥ 0 ⇒ $x^2$ ≤ 9 ⇒ |x| ≤ 3 ⇒ – 3 ≤ x ≤ 3 ∴ Domain of f = [– 3, 3] Let y = f (x) = $\sqrt{9-x^2}$ ⇒ $y^2$ = 9 – $x^2$, y ≥ 0 ...(i) ⇒ $x^2$ = 9 – $y^2$ ⇒ 9 – $y^2$ ≥ 0 (Since $x^2$ ≥ 0) ⇒ $y^2$ ≤ 9 ⇒ |y| ≤ 3 ⇒ – 3 ≤ y ≤ 3 But from (i), y ≥ 0 ⇒ 0 ≤ y ≤ 3. ∴ Range of f = [0, 3].