We know the square of a real number is never negative.

Clearly, f(x) takes real values only when 16 – x² ≥ 0

⇒ 16 ≥ x²

⇒ x² ≤ 16

⇒ x² – 16 ≤ 0

⇒ x² – 4² ≤ 0

⇒ (x + 4)(x – 4) ≤ 0

⇒ x ≥ –4 and x ≤ 4

∴ x ∈ [–4, 4]

In addition, f(x) is also undefined when 16 – x² = 0 because denominator will be zero and the result will be indeterminate.

16 – x² = 0 ⇒ x = ±4

Hence, x ∈ [–4, 4] – {–4, 4}

∴ x ∈ (–4, 4)

Thus, domain of f = (–4, 4)

Let f(x) = y

⇒ 1 = (16 – x²)y²

⇒ 1 = 16y² – x²y²

⇒ x²y² + 1 – 16y² = 0

⇒ (y²)x² + (0)x + (1 – 16y²) = 0

As x ∈ R, the discriminant of this quadratic equation in x must be non-negative.

⇒ 0² – 4(y²)(1 – 16y²) ≥ 0

⇒ –4y²(1 – 16y²) ≥ 0

⇒ 4y²(1 – 16y²) ≤ 0

⇒ 1 – 16y² ≤ 0 [∵ y² ≥ 0]

⇒ 16y² – 1 ≥ 0

⇒ (4y)² – 1² ≥ 0

⇒ (4y + 1)(4y – 1) ≥ 0

⇒ 4y ≤ –1 and 4y ≥ 1

However, y is always positive because it is the reciprocal of a non-zero square root.

Thus, range of f =