Let A = {−2, −1, 0, 1, 2} and f :

(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

(b) Let x be the pre-image of 6.

Then,

f(6) = x2 − 2x − 3 = 6

⇒ x2 − 2x − 9 = 0

= 4 – 7 = − 3

Thus, range of f(A) = (− 4, − 3, 0, 5).

$\Rightarrow x=1 \pm \sqrt{10}$

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

Let $x$ be the pre-image of $-3$. Then,

$f(-3) \Rightarrow x^{2}-2 x-3=-3$

$\Rightarrow x^{2}-2 x=0$

$\Rightarrow x=0,2$

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 $\phi,\{0,2,\},-$,2 respectively.