Question -
Answer -
Given thatf: R − {3} → R − {2} given by f (x) = (x-2)/(x-3)
Now we have to showthat the given function is one-one and on-to
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)
⇒ x y – 3 x – 2 y+ 6 = x y – 3y – 2x + 6
⇒ x = y
So, f isone-one.
Surjectivity:
Let y be anyelement in the co-domain (R − {2}), such that f(x) = y for someelement x in R − {3} (domain).
f(x) = y
⇒ (x – 2) /(x – 3) = y
⇒ x – 2 = x y – 3y
⇒ x y – x = 3y – 2
⇒ x ( y – 1 ) = 3y – 2
⇒ x = (3y – 2)/ (y– 1), which is in R – {3}
So, for every elementin the co-domain, there exists some pre-image in the domain.
⇒ f is onto.
Since, f isboth one-one and onto, it is a bijection.