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Question
Show that the function f : R − {3} → R − {2} given by f(x) = `(x-2)/(x-3)` is a bijection.
Solution
f : R − {3} → R − {2} given by
`f (x) = (x-2)/(x-3)`
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
⇒ `(x-2)/(x-3) = (y-2)/(y-3)`
⇒ ( x-2 ) (y - 3) = ( y-2 ) ( x-3 )
⇒ xy - 3x - 2y + = xy = 3y - 2x + 6
⇒ x = y
So, f is one-one.
Surjectivity :
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element xin R − {3} (domain).
f(x) = y
⇒ `(x-2)/(x-3) =y`
⇒ x - 2 = xy - 3y
⇒ xy - x = 3y - 2
⇒ x ( y-1 ) = 3y - 2
⇒ x = `(3y - 2)/(y-1),`which is in R -{3}
So, for every element in the co-domain, there exists some pre-image in the domain. ⇒ f is onto.
Since, f is both one-one and onto, it is a bijection.
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