(a) {(x, y) : y2 = x, x, y ∈ R}
(b) {(x, y) : y = |x|, x, y ∈ R}
(c) {(x, y) : x2 + y2 = 1, x, y ∈ R}
(d) {(x, y) : x2 − y2 = 1, x, y ∈ R}
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Solution
(b) {(x, y) : y = |x|, x, y ∈ R}
For every value of x ∈ R, there is a unique value y∈ R. i.e. there is a unique image for all values of x ∈ R. Also, values of x occur only once in the ordered pairs. Thus, it is a function.