Given f : R → R and f(x) = x².

i. range of f

Domain of f = R (set of real numbers)

We know that the square of a real number is always positive or equal to zero.

Hence, the range of f is the set of all non-negative real numbers.

Thus, range of f = R⁺∪ {0}

ii. {x: f(x) = 4}

Given f(x) = 4

⇒ x² = 4

⇒ x² – 4 = 0

⇒ (x – 2)(x + 2) = 0

∴ x = ±2

Thus, {x: f(x) = 4} = {–2, 2}

iii. {y: f(y) = –1}

Given f(y) = –1

⇒ y² = –1

However, the domain of f is R, and for every real number y, the value of y² is non-negative.

Hence, there exists no real y for which y² = –1.

Thus, {y: f(y) = –1} = ∅