Question

Let A = {−2, −1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x² − 2x − 3. Find:

(a) range of f, i.e. f(A).
(b) pre-images of 6, −3 and 5.

Answer 1

(a) Given:
f (x) = x² − 2x − 3
f (−2) = (− 2)² − 2(− 2) − 3
= 4 + 4 – 3
= 8 − 3 = 5
f (−1) = (−1)² − 2(−1) − 3
= 1+ 2 − 3
= 3 − 3 = 0
f (0) = (0)² − 2(0) − 3
= 0 − 0 − 3
= − 3
f (1) = (1)² − 2(1) − 3
= 1 − 2 − 3
=1 − 5 = − 4
f (2) = (2)² – 2(2) − 3
= 4 − 4 – 3
= 4 – 7 = − 3
Thus, range of f(A) = (− 4, − 3, 0, 5).

(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 $x=1\pm \sqrt{10}\notin A$, there is no pre-image of 6.

Let x be the pre-image of $-$3. Then,
f(− 3) ⇒ x² − 2x − 3 = − 3
⇒ x² − 2x = 0
⇒ x = 0, 2
Clearly $0,2\in A$. 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 $-2\in A$, − 2 is the pre-image of 5.
Hence,
pre-images of 6, − 3 and 5 are $\varphi ,\left\{0,2,\right\},-2$ respectively.