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.

 

Answer 1

(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

\[x = 1 \pm \sqrt{10} \not\in A\]  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  \[\phi, \left\{ 0, 2, \right\}, - 2\] respectively.