If f : Q → Q is defined as f(x) = x2, then f−1 (9) is equal to
(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ
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Solution
(c) {−3, 3}
If f : A → B, such that y ∈ B, then { y }={x ∈ A: f (x) = y}.
In other words, { y} is the set of pre-images of y.
Let{9} = x
Then, f (x) = 9
⇒ x2 = 9
⇒ x = ± 3
∴{9} = {- 3, 3}.