Question -
Answer -
Given:
f : R → R and f(x) = x2.
(i) rangeof f
Domain of f = R (set of real numbers)
We know that the square of a real number is always positive or equal tozero.
∴range of f = R+∪ {0}
(ii) {x:f(x) = 4}
Given:
f(x) = 4
we know, x2 = 4
x2 – 4 = 0
(x – 2)(x + 2) = 0
∴ x= ± 2
∴{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.
∴{y:f(y) = –1} = ∅