Classify the function as injection, surjection or bijection f: Q − {3} → Q, defined by f (x) = $$\dfrac{ (2x +3)}{(x-3)}$$

Asked by Aaryan | 1 year ago |  51

##### Solution :-

Given f: Q − {3} → Q, defined by f (x) = $$\dfrac{ (2x +3)}{(x-3)}$$

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

f(x) = f(y)

$$\dfrac{ (2x +3)}{(x-3)}=\dfrac{ (2y +3)}{(y-3)}$$

(2x + 3) (y − 3) = (2y + 3) (x − 3)

2xy − 6x + 3y − 9 = 2xy − 6y + 3x − 9

9x = 9y

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

$$\dfrac{ (2x +3)}{(x-3)}=y$$

2x + 3 = x y − 3y

2x – x y = −3y − 3

x (2−y) = −3 (y + 1)

x = $$\dfrac{-3(y + 1)}{(2 – y)}$$ which is not defined at y = 2.

So, f is not a surjection and f is not a bijection

Answered by Aaryan | 1 year ago

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