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Question
Let A = {−2, −1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x2 − 2x − 3. Find:
(b) pre-images of 6, −3 and 5.
Solution
(b) Let x be the pre-image of 6.
Then,
f(6) = x2 − 2x − 3 = 6
⇒ x2 − 2x − 9 = 0
⇒ \[x = 1 \pm \sqrt{10}\]
Since
Let x be the pre-image of -3. Then,
f(− 3) ⇒ x2 − 2x − 3 = − 3
⇒ x2 − 2x = 0
⇒ x = 0, 2
Clearly
So, 0 and 2 are pre-images of −3.
Let x be the pre-image of 5. Then,
f(5) ⇒ x2 − 2x − 3 = 5
⇒ x2 − 2x − 8 = 0
⇒ (x − 4) (x + 2) = 0 ⇒ x = 4, − 2
Since
Hence,
pre-images of 6, − 3 and 5 are \[\phi, \left\{ 0, 2, \right\}, - 2\] respectively.
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