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Question
Classify the following function as injection, surjection or bijection :
f : Q − {3} → Q, defined by `f (x) = (2x +3)/(x-3)`
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Solution
Injective Test:
x1 and x2 be any tow elements in the domain (Q - {3}), such that f(x1) = f(x2)
`(2x_1 + 3)/(x_1 - 3) = (2x_2 + 3)/(x_2 - 3)`
⇒ `(2x_1 + 3)(x_2 - 3) = (2x_2 + 3)(x_1 - 3)`
⇒`2x_1x_2 - 6x_1 + 3x_2 - 9 = 2x_1x_2 - 6x_2 + 3x_1 - 9`
⇒ `9x_1 = 9x_2`
⇒ `x_1 = x_2`
f is an injection.
Surjective test:
Let y be any elements in the co-domain (Q - {3}), such that f(x) = y for some elements ξnQ (Domain)
f(x) = y
`(2x + 3)/(x - 3) = y`
⇒ `2x + 3 = xy - 3y`
⇒ `2x - xy = -3y - 3`
⇒ `x(2 - y) = -3(y + 1)`
`x = (3(y + 1))/(y - 1)` Which is not defined at y = 2
So, f is not a surjection.
Bijection test:
Here, f is an injective but not surjective, then it is not bijective.
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